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Feynman's Trick

a.k.a. Differentiation under the Integral Sign & Leibniz Integral
Rule

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Among a few other integral tricks and techniques, Feynman's trick was
a strong reason that made me love evaluating integrals, and although
the technique itself goes back to Leibniz being commonly known as the
Leibniz integral rule, it was Richard Feynman who popularized it,
which is why it is also referred to as Feynman's trick. Here's an
excerpt from his book, Surely You're Joking, Mr. Feynman:

"One thing I never did learn was contour integration. I had learned
to do integrals by various methods shown in a book that my high
school physics teacher Mr. Bader had given me.

One day he told me to stay after class. "Feynman," he said, "you talk
too much and you make too much noise. I know why. You're bored. So
I'm going to give you a book. You go up there in the back, in the
corner, and study this book, and when you know everything that's in
this book, you can talk again."

So every physics class, I paid no attention to what was going on with
Pascal's Law, or whatever they were doing. I was up in the back with
this book: Advanced Calculus, by Woods. Bader knew I had studied
Calculus for the Practical Man a little bit, so he gave me the real
works -- it was for a junior or senior course in college. It had
Fourier series, Bessel functions, determinants, elliptic functions --
all kinds of wonderful stuff that I didn't know anything about.

That book also showed how to differentiate parameters under the
integral sign -- it's a certain operation. It turns out that's not
taught very much in the universities; they don't emphasize it. But I
caught on how to use that method, and I used that one damn tool again
and again. So because I was self-taught using that book, I had
peculiar methods of doing integrals.

The result was, when guys at MIT or Princeton had trouble doing a
certain integral, it was because they couldn't do it with the
standard methods they had learned in school. If it was contour
integration, they would have found it; if it was a simple series
expansion, they would have found it. Then I come along and try
differentiating under the integral sign, and often it worked. So I
got a great reputation for doing integrals, only because my box of
tools was different from everybody else's, and they had tried all
their tools on it before giving the problem to me."

For me, employing this trick felt like I was using cheat codes to
deal with integrals. At the same time, it enabled a lot of creativity
and wishful thinking, which transformed integrals into puzzles.
Unfortunately, this also means that there is no clear path on how and
when to use this technique. In addition, what Feynman wrote still
applies today since the method isn't taught much, if at all, in
universities. Therefore, the trick can seem obscure and difficult to
grasp for newcomers.

In the following section, we will embark on a journey to develop some
rules of thumb to have at our disposal when using Feynman's trick.
These are merely some heuristics that I tend to use, so deviating
from them can be perfectly acceptable. However, I hope that they can
provide a path to follow when nothing obvious or intuitive occurs
when someone tries to use this trick, or even better, so that they
can serve as motivation for someone to start using the method.

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Hello, World!

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Feynman already provided a significant hint about the trick when he
mentioned differentiating under the integral sign, which is also an
alternative name for the technique. More explicitly, if \(f(x,t)\)
and \(\frac{\partial f(x,t)}{\partial t}\) is continuous with respect
to both variables over the \([a,b]\) interval, then the following
holds:

\[\frac{d}{dt} \int_a^b f(x, t)dx = \int_a^b \frac{\partial f(x, t)}
{\partial t}dx\]

This is nice, but not so useful by itself since it doesn't say
anything about how and when to apply it. Moreover, learning is not a
spectator sport and one has to get their hands dirty as there are no
shortcuts to it. Take for example chess, most people could read and
understand the rules in a few minutes, however, if they would go on
to play a game then most likely they would get stomped by a more
experienced player. This is because the other player, through
practice, learned some strategies to use when playing.

Thus, with the goal to develop some strategies here as well, we will
dive straight into action and approach Feynman's trick using
practical examples. As a "Hello, World!" introduction, let's take a
look at the following integral:

\[I=\int_0^1 \frac{x-1}{\ln x}dx\]

You are encouraged to try and evaluate the integral using basic
methods, but the logarithm being in the denominator makes this
integral quite stubborn to deal with. Feynman's trick aims to get rid
of this issue by differentiating under the intgeral sign, with
respect to a parameter, in order to obtain an integral that is easier
to evaluate.

Unfortunately in the integral from above we lack a parameter,
therefore the first step is to parameterise the integral, which can
even mean introducing a whole function, but for this example we will
simply consider:

\[I(t)=\int_0^1 \frac{x^t-1}{\ln x}dx\]

Keep in mind that our original integral is just \(I(1)\). Also,
surely we could've placed a parameter in many different places, such
as:

\[I(a)=\int_0^1 \frac{x-a}{\ln x}dx, \quad I(b)=\int_0^1 \frac{x-1}
{b-\ln x}dx, \quad I(c)=\int_0^1 \frac{x-1}{\ln (cx)}dx\ ...\]

However, the main idea behind the trick is to obtain an integral that
we can evaluate easier, after differentiating with respect to the new
parameter. Let's put this in action and see what happens to \(I(t)\).

\[I'(t)=\int_0^1 \frac{\partial}{\partial t}\left(\frac{x^t-1}{\ln x}
\right)dx = \int_0^1x^tdx = \frac{x^{t+1}}{t+1}\bigg|_0^1 = \frac{1}
{t+1}\]

Notice how easy it was to evaluate the integral \(I'(t)=\int_0^1x^tdx
\) from above, had we kept \(I(a)\), \(I(b)\) or \(I(c)\) the things
wouldn't had simplified at all after differentiating, and most
significantly is that we would still have the \(\ln x\) in the
denominator, a thing which made the integral hard to deal with in the
first place.

We can already sense that the following might be an important
question in the future: How to parameterise the integral when using
Feynman's Trick?

We will worry about that a bit later, for now let's finish the
integral as we only found \(I'(t)\). Since we are looking to find \(I
(1)\) we need to integrate \(I'(t)\) back and set \(t=1\) in order to
arrive there. Here it's useful to recall that:

\[\int_a^b f'(x) dx = f(b)-f(a)\]

For us, \(f(x)\) is just \(I(t)\) in the above expression. Luckily \
(I(0)=0\), and as we are looking for \(I=I(1)\) we have:

\[I= I(1)-I(0)=\int_0^1 I'(t)dt = \int_0^1 \frac{1}{1+t}dt=\ln(1+t)\
bigg|_0^1 = \ln 2\]

So that is the big picture of Feynman's trick - we have an integral
that is hard to evaluate in it's original form, therefore by
differentiating under the integral sign we attempt to transform the
integral so that it can be easier integrated, and in the end we go
back to undo the differentiation step.

---------------------------------------------------------------------

The parameter

---------------------------------------------------------------------

As emphasized above, the main goal of the technique is to obtain an
integral that is easier to evaluate after differentiating with
respect to a parameter, and one issue is that it is not always
obvious how to parameterise the integral. In order to make things
more intuitively we will play around with the integral from below.

\[I=\int_0^1 \frac{\ln(1+x)}{1+x^2}dx\]

The most annoying thing is the logarithm, so if we get rid of it
everything should be straightforward. There are a few parameter
possibilities which makes sense to consider, namely:

\[I(a)=\int_0^1\frac{\ln(a+x)}{1+x^2}dx, \quad I(b) = \int_0^1 \frac
{\ln(1+bx)}{1+x^2}dx\]

With the first one we are out of luck, as differentiating with
respect to \(a\) gives:

\[I'(a)=\int_0^1 \frac{1}{(a+x)(1+x^2)}dx = \frac{1}{1+a^2}\int_0^1\
left(\frac{1}{a+x}+\frac{a}{1+x^2}-\frac{x}{1+x^2}\right)dx\] \[=\
frac{\ln(1+a)}{1+a^2}-\frac{\ln a}{1+a^2} +\frac{\pi}{4}\frac{a}{1+a^
2}-\frac{\ln 2}{2}\frac{1}{1+a^2}\]

Therefore, if we would try to go back to what we're looking, which is
\(I=I(1)-I(0)\), we would end up with \(I=I+\text{other stuff}\).
This cancels out \(I\) and we wouldn't be able to recover it.
Unfortunately, there's no magic formula that tells a priori whether
placing a parameter in a specific place would succeed or fail in
evaluating an integral - and sometimes we are simply unlucky.

In contrast, things work out nicely with the second choice from
above.

\[I'(b)=\int_0^1\frac{x}{(1+bx)(1+x^2)}dx=\frac{1}{1+b^2}\int_0^1\
left(\frac{b}{1+x^2}+\frac{x}{1+x^2}-\frac{b}{1+bx}\right)dx\] \[=\
frac{\pi}{4}\frac{b}{1+b^2}+\frac{\ln 2}{2}\frac{1}{1+b^2}-\frac{\ln
(1+b)}{1+b^2}\]

Again, we are looking to find \(I(1)\), and as \(I(0) = 0\), we have:

\[I=\int_0^1 I'(b)db \Rightarrow 2I=\int_0^1 \left(\frac{\pi}{4}\frac
{b}{1+b^2}+\frac{\ln 2}{2}\frac{1}{1+b^2}\right)db \Rightarrow I=\
frac{\pi}{8}\ln 2\]

This works, but we can do even better. Looking at the Hello, World!
integral we can see that there we simplified the logarithm in the
denominator while performing \(\frac{\partial}{\partial t}x^t\). This
is also the first thing that I always attempt to look for when using
this technique - namely, to simplify something from the integrand
which is independent to the parameter when differentiating. Surely
for the current integral we got rid of the logarithm, but the
denominator remained intact.

In short this will be our first rule of thumb: if possible, place the
parameter so that something from the integral, which is not related
to the parameter, gets simplified.

In order to achieve this with our integral we would need to get rid
of \(1+x^2\), and by using \(\ln x=\frac12\ln(x^2)\) we can rewrite
the integral as:

\[I=\int_0^1\frac{\ln(1+x)}{1+x^2}dx=\frac12\int_0^1\frac{\ln(1+2x+x^
2)}{1+x^2}dx\]

Finally, in this form it's more natural to place the parameter so
that it simplifies \(1+x^2\) when differentiating with respect to \(t
\), namely we can consider:

\[I(t)=\frac12\int_0^1 \frac{\ln(2x+t(1+x^2))}{1+x^2}dx\Rightarrow I'
(t)=\frac12\int_0^1 \frac{1}{2x+t(1+x^2)}dx=\frac14\frac{\ln\left(\
frac{1+\sqrt{1-t^2}}{t}\right)}{\sqrt{1-t^2}}\]

Like for \(I(b)\) we are looking to find \(I(1)\), however here \(I
(0)\) is equal to \(\frac12\int_0^1\frac{\ln(2x)}{1+x^2}dx\) not \(0
\).

\[\Rightarrow I=I(1)-I(0)+I(0)=\frac14\int_0^1\frac{\ln\left(\frac{1+
\sqrt{1-t^2}}{t}\right)}{\sqrt{1-t^2}}dt+\frac12\int_0^1\frac{\ln
(2x)}{1+x^2}dx\] \[\int_0^1\frac{\ln\left(\frac{1+\sqrt{1-t^2}}{t}\
right)}{\sqrt{1-t^2}}dt\overset{\frac{1+\sqrt{1-t^2}}{t}=x}=-2\int_0^
1 \frac{\ln x}{1+x^2}dx\Rightarrow I = \frac12\int_0^1 \frac{\ln 2}
{1+x^2}dx=\frac{\pi}{8}\ln 2\]

For this specific integral we only avoided performing partial
fractions so there wasn't really a big improvement by simplifying the
denominator. However I want to emphasize the importance of this
because it will make things come way more natural when deciding where
place the parameter. Of course, in case there's not an appropiate or
immediate way to achieve this, it's perfectly fine to place the
parameter elsewhere too.

As mentioned previously, practicing is the best approach to get along
with new techniques, therefore below are more integrals to evaluate
alongside some hidden steps in case those will be needed. However, I
strongly recommend to try and deal with the integrals before looking
at any hints, and only check them afterwards for correctness.

\[\int_0^\frac{\pi}{2} \frac{\ln(1-\sin x)}{\sin x}dx = ?\]

Consider introducing the following parameter: \[I(t)=\int_0^\frac{\
pi}{2} \frac{\ln(1-t\sin x)}{\sin x}dx \Rightarrow I'(t)= -\frac{2\
arctan\left(\sqrt{\frac{1+t}{1-t}}\right)}{\sqrt{1-t^2}}\] This
should lead to: \[\int_0^\frac{\pi}{2} \frac{\ln(1-\sin x)}{\sin x}dx
= I(1) - I(0)=\int_0^1 I'(t) dt \overset{\sqrt{\frac{1-t}{1+t}}=x} =
-\frac{3\pi^2}{8}\] But it would be even better if the integral would
be parameterised as: \[I(t)=\int_0^\frac{\pi}{2} \frac{\ln(1-\sin t\
sin x)}{\sin x}dx\] That is because usually when having trigonometric
functions, parameterising the integral with another trigonometric
function, leads to a more smoother result.

\[\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx = ?\]

Consider introducing the following parameter: \[I(t)=\int_0^1 \frac{\
ln(1-t(x-x^2))}{x-x^2}dx\Rightarrow I'(t) = \frac{4\arctan\left(\sqrt
{\frac{t}{4-t}}\right)}{\sqrt{t(4-t)}}\] This should lead to: \[I(1)=
\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx = I(1) - I(0) = \int_0^1 I'(t)
dt \overset{\sqrt{\frac{4-t}{t}}= x}= -\frac{\pi^2}{9}\]

\[\int_0^\frac{\pi}{2} \frac{\arctan(\sin x)}{\sin x}dx = ?\]

Consider introducing the following parameter: \[I(t)=\int_0^\frac{\
pi}{2} \frac{\arctan(t\sin x)}{\sin x}dx\Rightarrow I'(t)=\frac{\pi}
{2\sqrt{1+t^2}}\] This should lead to: \[I(1)=\int_0^\frac{\pi}{2} \
frac{\arctan(t\sin x)}{\sin x}dx = I(1)-I(0) = \int_0^1 I'(t)dt = \
frac{\pi}{2}\ln(1+\sqrt 2)\] It will also work if the integral is
parameterised as: \[I(t)=\int_0^\frac{\pi}{2} \frac{\arctan(\tan t\
sin x)}{\sin x}dx\] However, in this case the first variant is simple
enough to integrate back.

\[\int_0^\infty x^2e^{-\left(4x^2+\frac{9}{x^2}\right)}dx = ?\]

Consider introducing the following parameter: \[I(t)=\int_0^\infty x^
2e^{-\left(4x^2+\frac{t}{x^2}\right)}dx\Rightarrow I'(t)=-\frac{\sqrt
\pi}{4} e^{-4\sqrt t}\] Where the above result follows by using
Glasser's master theorem alongside the Gaussian integral. This should
lead to: \[\int_0^\infty x^2e^{-\left(4x^2+\frac{9}{x^2}\right)}dx =
I(9)- I(0) + I(0) = \int_0^9 I'(t) dt +\frac{\sqrt \pi}{32}=\frac{13}
{32}\frac{\sqrt \pi}{e^{12}}\]

\[\int_0^1 \frac{\ln x}{1-x^2}dx = ?\]

Consider parameterising the integral as: \[I(t)=\frac12\int_0^1\frac
{\ln(1-t(1-x^2))}{1-x^2}dx\Rightarrow I'(t)=\frac{\arctan\left(\sqrt
{\frac{t}{1-t}}\right)}{2\sqrt{t(1-t)}}\] This should lead to: \[\
int_0^1 \frac{\ln x}{1-x^2}dx = I(1)- I(0) = \int_0^1 I'(t)dt \
overset{\sqrt{\frac{1-t}{t}} = x}= -\frac{\pi^2}{8}\]

\[\int_0^\infty \frac{e^{-x^2}}{1+x^2}dx = ?\]

Consider parameterising the integral as: \[I(t)=\int_0^\infty \frac{e
^{-t(1+x^2)}}{1+x^2}dx\Rightarrow I'(t) = -\frac{\sqrt \pi}{2\sqrt t}
e^{-t}\] This should lead to: \[\int_0^\infty \frac{e^{-x^2}}{1+x^2}
dx = e\left(I(1)-I(\infty)\right) = -e\int_1^\infty I'(t)dt= \frac{\
pi e}{2}\operatorname{erfc}(1)\] Where \(\operatorname{erfc}(x)\) is
the complementary error function.

\[\int_0^\infty \frac{\ln\left(\frac{1-x^2+x^4}{(1-x^2)^2}\right)}
{(1+x^2)^2}dx = ?\]

Since \(1-x^2+x^4=(1+x^2)^2-3x^2\), consider parameterising the
integral as: \[I(t)=\int_0^\infty \frac{\ln\left(\frac{t(1+x^2)^2-3x^
2}{(1-x^2)^2}\right)}{(1+x^2)^2}dx\Rightarrow I'(t)=\frac{\pi}{2\sqrt
{t(4t-3)}}\] And in order to go back it should be observed that \(\
frac34(1+x^2)^2-3x^2=\frac34(1-x^2)^2\). \[\int_0^\infty \frac{\ln\
left(\frac{1-x^2+x^4}{(1-x^2)^2}\right)}{(1+x^2)^2}dx=I(1)- I\left(\
frac34\right)+ I\left(\frac34\right)\] \[=\int_\frac34^1 I'(t)dt + \
frac{\pi}{4}\ln\left(\frac{3}{4}\right) = \frac{\pi}{2}\ln\left(\
frac32\right)\]

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Accelerated Feynman's trick

---------------------------------------------------------------------

The previous chapter emphasized to parameterise integrals so that
something from the integral, which is not related to the parameter,
gets simplified when differentiating (if possible). However there are
times when even though we can introduce a parameter to accomplish
that, it wouldn't be enough to finish the integral.

In this chapter we will look at a different way to obtain this
simplification. Let's start by looking at a modified version of an
integral that was previously given as an exercise.

\[I=\int_{-\infty}^\infty \frac{e^{-x^2}}{1+x^4}dx\]

With \(\int_{-\infty}^\infty \frac{e^{-x^2}}{1+x^2}dx\) it was quite
direct to parameterise the integral as \(\int_{-\infty}^\infty \frac
{e^{-t(1+x^2)}}{1+x^2}dx\) since it simplifies the denominator,
however the similar way to do that for our integral, \(\int_{-\infty}
^\infty \frac{e^{-x^2-t(1+x^4)}}{1+x^4}dx\), doesn't seem to work as
it complicates things a bit too much.

There is however a way to simplify the denominator and in the same
time to obtain a decent integral afterwards. Without getting into too
much details I will parameterise the integral as:

\[I(t)=\int_{-\infty}^\infty \frac{e^{-x^2}}{1+x^4} e^{-tx^2}(x^2\sin
t+\cos t) dx\]

This will seem obscure, but fear not as we will never use this
approach again. The whole point is to simplify \(1+x^4\), and the
above function was created explicitly to achieve that, as \(\frac{\
partial}{\partial t}e^{-tx^2}(x^2\sin t+\cos t)\) is \(-(1+x^4)e^{-tx
^2}\sin t\). Note that even though we introduced a couple other
terms, those aren't disturbing.

\[I'(t)=-\sin t\int_{-\infty}^\infty e^{-x^2}e^{-tx^2}dx \overset
{(1+t)x^2\to x^2}= -\frac{\sin t}{\sqrt{1+t}} \int_{-\infty}^\infty e
^{-x^2}dx=-\sqrt \pi \frac{\sin t}{\sqrt{1+t}}\]

Here we are looking to find \(I=I(0)\), and we also have \(I(\infty)=
0\), therefore:

\[I=-\int_0^\infty I'(t)dt=\sqrt \pi \int_0^\infty \frac{\sin t}{\
sqrt{1+t}}dt\overset{\sqrt{1+t}=x}=2\sqrt \pi \int_1^\infty \sin(x^
2-1)dx\] \[=2\sqrt{\pi} \cos 1 \int_1^\infty\sin(x^2)dx-2\sqrt{\pi} \
sin 1 \int_1^\infty\cos(x^2)dx\] \[=\pi\cos 1\frac{1-2S\left(\sqrt{\
frac{2}{\pi}}\right)}{\sqrt 2}-\pi\sin 1\frac{1-2C\left(\sqrt{\frac
{2}{\pi}}\right)}{\sqrt 2}\]

Where \(S(x)\) and \(C(x)\) are the Fresnel integrals. However, the
approach is important here, not the result itself.

We can avoid the parametrisation from above by directly using \(\frac
{1}{1+x^4}=\int_0^\infty e^{-tx^2}\sin t \, dt\), and then switch to
double integrals, or put in other words: employ the accelerated
Feynman's trick (in which we skip the usual parameterisation step).

\[I=\int_{-\infty}^\infty \frac{e^{-x^2}}{1+x^4}dx = \int_{-\infty}^\
infty e^{-x^2}\int_0^\infty e^{-tx^2}\sin t \, dtdx\] \[=\int_{0}^\
infty \sin t\int_{-\infty}^\infty e^{-x^2}e^{-tx^2}dxdt\overset{(1+t)
x^2\to x^2}=\sqrt{\pi}\int_0^\infty \frac{\sin t}{\sqrt{1+t}}dt\]

The rest goes exactly as with the previous method, as all we did here
was to skip differentiation step and instead we switched to double
integrals.

A natural question that arises here is how did \(\frac{1}{1+x^4}=\
int_0^\infty e^{-tx^2}\sin t\, dt\) appear? Or even better, how can
someone come up with similar results for other integrals? In the case
from above, simply the Laplace transform of the sine function was
used, however in general it's useful to have a list of such
identities. There are tables of integral results that can be used -
for example: Table of Integrals, Series, and Products by Gradshteyn
and Ryzhik - but alternatively one can build up their own list of
results which tend to appear often while evaluating other integrals.

---------------------------------------------------------------------

Let's conclude this chapter by evaluating one of the most popular
integrals that appears when Feynman's trick gets into the
conversation.

\[I=\int_0^\infty \frac{\sin x}{x}dx\]

Since \(\int_0^\infty e^{-xt} dt = \frac{1}{x}\), we can make use of
this to rewrite the integral as:

\[I=\int_0^\infty \int_0^\infty \sin x e^{-xt}dtdx = \int_0^\infty \
int_0^\infty \sin x e^{-xt}dxdt=\int_0^\infty \frac{1}{1+t^2}dt =\
frac{\pi}{2}\]

Alternatively, we can also consider the parameter version of this
integral, \(\int_0^\infty \frac{\sin x}{x}e^{-xt}dx\), however I feel
like switching to double integrals is way more intuitively.

It might be worth to highlight again that this method should be used
preferable when parameterising the integral leads to nowhere. For the
above integral, the natural introduction of \(\int_0^\infty \frac{\
sin(tx)}{x}dx\) unfortunatelly does fail, as we obtain a divergent
integral after differentiating under the integral sign.

Like in the previous chapter below are more integrals alongside some
hints in order to practice with the accelerated variation of
Feynman's trick. However in this case I do recommend to peek at hints
faster in case nothing obvious comes to mind, and afterwards to
attempt and understand why the mentioned identity can be used.

\[\int_0^\infty \sin(x^2)dx=?\]

Start by substituting \(x^2\to x\) and then switch to double
integrals using: \[\int_0^\infty e^{-xt^2}dt = \frac{\sqrt \pi}{2\
sqrt x}\] Where the latter result is due to the Gaussian integral.
Also, this integral is one particular case of the Fresnel integral.

\[\int_0^\infty \frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}dx=?\]

Switch directly to double integrals by using: \[\int_0^1 \frac{\ln t}
{t-\frac{1}{x}}dt = \operatorname{Li}_2(x)\]

\[\int_0^1 \frac{\arctan x\ln(1+x^2)}{x(1+x)}dx = ?\]

Switch to double integrals by using the following result: \[\int_0^x
\frac{\arctan t}{1+xt}dt = \frac{\arctan x \ln(1+x^2)}{2x}\]

\[\int_{-\infty}^\infty \frac{x}{\pi^2+x^2}\frac{e^{2x}}{(1+e^{x})^3}
dx\]

Consider switching to double integrals with: \[\frac{x}{\pi^2+x^2}=\
Im\left(-\frac{1}{\pi+ix}\right)=-\Im\int_0^\infty e^{-(\pi+ix)t}dt\]
It's also really useful to try and see what happens when the Laplace
transform of the cosine function is used instead, or the equivalent:
\[\frac{x}{\pi^2+x^2}=\Re\left(\frac{1}{i\pi+x}\right)=\Re\int_0^\
infty e^{-(i\pi+x)t}dt\]

\[\int_0^\infty x\left(\operatorname{Ci}^2(x)+\operatorname{si}^2(x)\
right)\operatorname{Ci}(x)dx\]

Consider switching to double integrals using: \[\operatorname{Ci}^2
(x)+\operatorname{si}^2(x)=\int_0^\infty \frac{e^{-xy}\ln(1+y^2)}{y}
dy\]

Above \(\operatorname{Li}_2(x)\) denotes the dilogarithm function and
\(\operatorname{Ci}(x)\), \(\operatorname{si}(x)\) are the cosine and
the sine integral functions, defined as:

\[\operatorname{Li}_2(x)=\sum_{n=1}^\infty \frac{x^n}{n^2},\ \
operatorname{Ci}(x) = - \int_x^\infty \frac{\cos t}{t}dt,\ \
operatorname{si}(x) = - \int_x^\infty \frac{\sin t}{t}dt\]
---------------------------------------------------------------------

More Feynman's trick variants

---------------------------------------------------------------------

We already got familiar with a popular version of Feynman's trick in
the previous chapter. Similarly, now we will take a look at other
interesting variants of Feynman's trick, which although might appear
less often, they can still help to expand the applicability of the
technique.

---------------------------------------------------------------------

Differentiating under the integral sign

We will start by taking a look at a much simpler case of Feynman's
trick, namely, in the situation when it would be enough to simply
differentiate under the integral sign without performing that "undo"
step to integrate back.

As a small note, it's true that "differentiating under the integral
sign" tends to be used as an alternative name for Feynman's trick,
however I prefer to keep this for the variant where only the
differentiating process takes part, or as mentioned above, when
there's no need to integrate back the result, and the name describes
quite literally what we are doing.

Let's make this more clear by looking at the following integral:

\[I=\int_0^1 x^3 \ln^2 x \, dx \]

We are already aware from the Hello, World! integral how \(\ln x\)
can be simplified, since \(\frac{\partial}{\partial a}x^a = x^a \ln x
\). However, by introducing the parameter in that original form as \
(x^a \ln^2 x\), we would just produce a third logarithm, so that's
going in the opposite direction.

Fortunatelly, if we take a step back, we can observe that after we
find the result of \(\int_0^1 x^a dx\), then differentiating it
w.r.t.\(a\) would give us as many logarithms as we want. So, let's
put that integral to use.

\[\mathcal J(a)=\int_0^1 x^a dx = \frac{1}{a+1}\] \[\Rightarrow \
mathcal J'(a) = \int_0^1 x^a \ln x \, dx = \left(\frac{1}{a+1}\right)
' = - \frac{1}{(a+1)^2}\] \[\Rightarrow \mathcal J''(a) = \int_0^1 x^
a \ln^2 x \, dx = \left(- \frac{1}{(a+1)^2}\right)' = \frac{2}{(a+1)^
3}\] \[\Rightarrow I= \mathcal J''(3) = \int_0^1 x^3 \ln^2 x \, dx =
\frac{1}{32}\]

Of course the integral itself was quite simple this time, however the
important part that should be highlighted is that not always we need
to perform that "undo" step after differentiating under the integral
sign - and sometimes knowing a general integral result can provide us
more useful integrals by differentiating it.

---------------------------------------------------------------------

Feynman's trick & indefinite integrals

Further, we will take a look at how Feynman's trick can be applied to
indefinite integrals. Let's consider:

\[\int \frac{1}{\sqrt{x^3}} \exp\left({-\frac{(a-bx)^2}{2x}}\right)
dx\]

In this form it makes no sense to differentiate the integral with
respect to any parameter, but we can extend the integral with
temporary bounds by writing:

\[\int f(x)dx = F(x) + C = F(x) - F(0) = \int_0^x f(t)dt\] \[\
Rightarrow I(a,b,t)=\int_0^t\frac{1}{\sqrt{x^3}} \exp\left({-\frac
{(a-bx)^2}{2x}}\right) dx\]

After this we can go on apply Feynman's trick, however, first we are
going get rid of the square root via the substitution \(\frac{1}{\
sqrt x}\to x\).

\[I(a,b,t)=2\int_\frac{1}{\sqrt t}^\infty \exp\left(-\frac12 \left
(ax-b/x\right)^2\right)dx\]

Here, we can notice that the derivative of \(ax-\frac{b}{x}\) is \(a+
\frac{b}{x^2}\) so it would be quite helpful if we had that
additional term. In the same time if we differentiate the integrand
with respect to \(b\) we'll produce \(a-\frac{b}{x^2}\), which is
really useful as \((ax-b/x)^2\) is equal to \((ax+b/x)^2+4ab\) and
the derivative of \(ax+\frac{b}{x}\) is \(a-\frac{b}{x^2}\). So let's
differentiate as mentioned above:

\[\frac{\partial}{\partial b}I(a,b,t)=2\int_\frac{1}{\sqrt t}^\infty
\exp\left(-\frac12 \left(ax-b/x\right)^2\right)\left(a-\frac{b}{x^2}\
right)dx\] \[=2e^{2ab}\int_\frac{1}{\sqrt t}^\infty \exp\left(-\
frac12 (ax+b/x)^2\right)\left(a-\frac{b}{x^2}\right)dx\] \[\overset
{ax+b/x\to x}=2e^{2ab}\int_{\frac{a}{\sqrt t}+b\sqrt t}^\infty \exp{\
left(-\frac{x^2}{2}\right)}dx=\sqrt{2\pi}e^{2ab}\operatorname{erfc}\
left(\frac{\frac{a}{\sqrt t}+b\sqrt t}{\sqrt 2}\right)\]

Where \(\operatorname{erfc}(x)\) is the complementary error function.
Now we'll go back to \(I(a,b,t)\), but we should be careful to
replace the dummy variable \(b\), with something else as the \(b\)
parameter does also appear in the bounds.

\[\lim_{b\to -\infty}I(a,b,t)=0\Rightarrow I(a,b,t)=\sqrt{2\pi}\int_
{-\infty}^be^{2ax}\operatorname{erfc}\left(\frac{\frac{a}{\sqrt t}+x\
sqrt t}{\sqrt 2}\right)dx\] \[\overset{IBP}=\frac{\sqrt{\pi}}{\sqrt
2a}e^{2ax}\operatorname{erfc}\left(\frac{\frac{a}{\sqrt t}+x\sqrt t}
{\sqrt 2}\right)\bigg|_{-\infty}^b+\frac{\sqrt t}{a}\int_{-\infty}^b
e^{2ax}\exp\left(-\frac{(a+xt)^2}{2t}\right)dx\] \[=\sqrt{\frac{\pi}
{2}}\frac{1}{a}e^{2ab}\operatorname{erfc}\left(\frac{\frac{a}{\sqrt
t}+b\sqrt t}{\sqrt 2}\right)+\frac{\sqrt t}{a}\int_{-\infty}^b \exp\
left(-\frac12\left(x\sqrt{t}-\frac{a}{\sqrt{t}}\right)^2\right)dx\] \
[\overset{x\sqrt t-\frac{a}{\sqrt t}\to -x}=\sqrt{\frac{\pi}{2}}\frac
{1}{a}e^{2ab}\operatorname{erfc}\left(\frac{\frac{a}{\sqrt t}+b\sqrt
t}{\sqrt 2}\right)+\frac{1}{a}\int^{\infty}_{\frac{a}{\sqrt t}-b\sqrt
t}\exp\left(-\frac{x^2}{2}\right)dx\] \[=\sqrt{\frac{\pi}{2}}\frac{1}
{a}\left(e^{2ab}\operatorname{erfc}\left(\frac{\frac{a}{\sqrt t}+b\
sqrt t}{\sqrt 2}\right)+\operatorname{erfc}\left(\frac{\frac{a}{\sqrt
t}-b\sqrt t}{\sqrt 2}\right)\right)\]

Or for the indefinite integral, this would lead to:

\[\small \int \frac{1}{\sqrt{x^3}} \exp\left({-\frac{(a-bx)^2}{2x}}\
right) dx=\sqrt{\frac{\pi}{2}}\frac{1}{a}\left(e^{2ab}\operatorname
{erfc}\left(\frac{\frac{a}{\sqrt x}+b\sqrt x}{\sqrt 2}\right)+\
operatorname{erfc}\left(\frac{\frac{a}{\sqrt x}-b\sqrt x}{\sqrt 2}\
right)\right)+C\]
---------------------------------------------------------------------

Feynman's trick & power series

Next, we will take a look at how to combine Feynman's trick with
power series. For this we are going to look at:

\[I=\int_{0}^{1}\int_{0}^{1}\frac{x\ln x\ln y}{1-xy}\frac{dxdy}{\ln
(xy)}\]

We are already got familiar with what to do when there is a logarithm
in the denominator as we saw that we can get rid of them by using \(\
frac{d}{dt} x^t = x^t\ln x\), however here also the \(1-xy\) term
appears. In order to solve this issue we'll make use of the geoemtric
series, namely \(\frac{1}{1-x}=\sum_{n=0}^\infty x^n\), but we will
expand into series a bit later and for now continue with the
following integral:

\[I(n)=\int_{0}^{1}\int_{0}^{1}\frac{(xy)^{n-1}x\ln x\ln y}{\ln(xy)}
dxdy\Rightarrow I'(n)=\int_0^1\int_0^1 (xy)^{n-1} x \ln x \ln y dxdy
\] \[=\int_0^1 x^{n} \ln x \, dx \int_0^1 y^{n-1} \ln y\, dy=\frac{1}
{(n+1)^2}\frac{1}{n^2}\]

Now we have to to get back to \(I(n)\):

\[I(n)=-(I(\infty)-I(n))=-\int_n^\infty \frac{1}{(x+1)^2 x^2 } dx=-\
frac{1}{n}-\frac{1}{n+1}+2\ln\left(1+\frac{1}{n}\right)\]

And finally, we'll put the geometric series to use.

\[\int_0^1\int_0^1 \frac{ x\ln x\ln y}{(1-xy)\ln(xy)}dx\ dy=\sum_{n=
1}^\infty \int_0^1\int_0^1 \frac{(xy)^{n-1} x\ln x\ln y}{\ln(xy)}dxdy
\] \[\small =\sum_{n=1}^\infty \left(-\frac{1}{n}-\frac{1}{n+1}+2\ln\
left(1+\frac{1}{n}\right)\right)=\sum_{n=1}^\infty \left(\underbrace
{\frac{1}{n}-\frac{1}{n+1}}_{1}-2\left(\underbrace{\frac{1}{n}-\ln\
left(1+\frac{1}{n}\right)}_{\gamma}\right)\right)\]

So the result is simply \(1-2\gamma\), where \(\gamma\) is the
Euler-Mascheroni constant.

---------------------------------------------------------------------

Feynman's trick & differential equations

In what's to come we are going to take a look at a combination
between Feynman's trick and differential equations. Let's consider
the following integral:

\[I=\int_0^\infty \frac{\cos x}{1+x^2}dx\]

We can start by parameterising the cosine function and then employ
the accelerated Feynman's trick:

\[I(t)=\int_0^\infty \frac{\cos(tx)}{1+x^2}dx=\int_0^\infty \int_0^\
infty \cos(tx)\sin y e^{-xy}dydx\] \[\small =\int_0^\infty \sin y \
int_0^\infty \cos(tx)e^{-xy}dxdy=\int_0^\infty \frac{y\sin y}{t^2+y^
2}dy\overset{y\to tx}=\int_0^\infty \frac{x\sin(tx)}{1+x^2}dx\]

We haven't made much progress above, since we simply arrived at
another integral with \(x\sin(tx)\) instead of \(\cos(tx)\), thus
complexity is the same. However, as \(\frac{\partial}{\partial t}\cos
(tx)\) is \(x\sin(tx)\), differentiating \(I(t)\) gives us a
differential equation to work with, namely:

\[I'(t)=- \int_0^\infty \frac{x\sin(tx)}{1+x^2}dx = - I(t) \
Rightarrow \frac{I'(t)}{I(t)}=-1\Rightarrow I(t) = C e^{-t}\] \[I(0)=
\int_0^\infty \frac{1}{1+x^2}dx=\frac{\pi}{2} \Rightarrow I(t)=\frac
{\pi}{2}e^{-t}\] \[ I = I(1) \Rightarrow I = \frac{\pi}{2e}\]

As a small note for the starting step, although employing the
accelerated Feynman's trick was rather obvious as to get rid of the
denominator, the additional introduction of the \(t\) parameter might
be weird first. However performing the same steps without this
parameter gives us:

\[\int_0^\infty \frac{\cos x}{1+x^2}dx=\int_0^\infty \frac{x\sin x}
{1+x^2}dx\]

Which indicates that one might put to use the fact that \(I(1)=-I'(1)
\), by adding the additional \(t\) parameter.

---------------------------------------------------------------------

Generalizing Feynman's trick

So far we've seen the Feynman's trick applied only when the parameter
was inside the integrand, however it can also be used when the bounds
are parameterised as well. More generally, the following holds:

\[\frac{d}{dt} \int_{a(t)}^{b(t)} f(x, t)dx = \frac{d}{dt}b(t)f(x,b
(t))-\frac{d}{dt}a(t)f(x,a(t))+\int_{a(t)}^{b(t)} \frac{\partial f(x,
t)}{\partial t}dx\]

We'll put this to use with the integral from below.

\[I=\int_\frac{1}{\sqrt 2}^1 \frac{\operatorname{arccosh}\left(\sqrt
2 x\right)}{\sqrt{1-x^2}}dx\]

Above we can see that the same \(\sqrt 2\) appears in both the lower
bound and the \(\operatorname{arccosh}\) function, so we'll
parameterise the integral as:

\[I(t)=\int_\frac{1}{t}^1 \frac{\operatorname{arccosh}\left(t x\
right)}{\sqrt{1-x^2}}dx \Rightarrow I'(t) = \frac{1}{t^2}\frac{\
cancelto{0}{\operatorname{arccosh}\left(t\frac{1}{t}\right)}}{\sqrt
{1-\frac{1}{t^2}}}+\int_\frac{1}{t}^1\frac{x}{\sqrt{1-t^2x^2}\sqrt
{1-x^2}}dx\] \[\overset{1-x^2\to x^2}=\frac1t\int_0^{\sqrt{1-\frac{1}
{t^2}}}\frac{1}{\sqrt{1-\frac{1}{t^2}-x^2}}dx=\frac1t\arcsin\left(\
frac{x}{\sqrt{1-\frac{1}{t^2}}}\right)\bigg|_0^\sqrt{1-\frac1{t^2}}=\
frac{\pi}{2t}\]

We're looking to find \(I=I\left(\sqrt 2\right)\), and since \(I\left
(1\right)=0\), we have:

\[I=\int_1^\sqrt 2I'(t)dt = \frac{\pi}{2} \int_1^\sqrt 2 \frac{1}{t}
dt=\frac{\pi}{4}\ln 2\]
---------------------------------------------------------------------

Generating integrals using Feynman's trick

Now we'll take a look at a fancier way to use Feynman's trick,
especially in order to generate new integrals, for this we're
considering:

\[I(t)=\int_0^\frac{\pi}{2} \arctan\left(\frac{\sin x-\tan\frac{t}
{2}}{\cos x}\right)dx\]

Note that we are not trying to evaluate the above integral, instead
we are simply using it in order to build up new integrals with the
result that follows after differentiating w.r.t. \(t\).

\[I'(t)=-\frac12\int_0^\frac{\pi}{2}\frac{\cos x}{1-\sin t\sin x}dx=\
frac12 \frac{\ln(1-\sin t)}{\sin t}\]

We also have that \(I(\pi)=-\frac{\pi^2}{4}\) and \(I(0)=\frac{\pi^2}
{8}\), therefore:

\[I(\pi)-I(0)=-\frac{\pi^2}{4}-\frac{\pi^2}{8}=\frac12\int_0^\pi \
frac{\ln(1-\sin x)}{\sin x}dx\] \[\Rightarrow \int_0^\pi \frac{\ln(1-
\sin x)}{\sin x}dx=-\frac{3\pi^2}{4}\]

In retrospect, this integral also appeared as an exercise in the
second chapter, and with the same suggestion from there, we can
evaluate the integral by applying Feynman's trick to:

\[I(t)=\int_0^\pi \frac{\ln(1-\sin t\sin x)}{\sin x}dx\]

Admittedly, following this parameterisation is much more intuitevely
than what we've shown with the new variation, however it's also
useful to have this trick in the bag.

---------------------------------------------------------------------

To keep the practice going, underneath are listed some integrals that
can be evaluated with one version of Feynman's trick described in
this chapter.

\[\int_1^\infty \int_1^\infty (x+y)^2 e^{-(x+y)}dxdy = ?\]

Start by showing that: \[I(t)=\int_1^\infty \int_1^\infty e^{-t(x+y)}
dxdy = \left(\frac{e^{-t}}{t}\right)^2\] Then differentiate both
sides two times with respect to \(t\) and set \(t=1\).

\[\int x^4\cos(nx)dx = ?\]

Differentiate four times with respect to \(n\) the following extended
indefinite integral: \[ I(n,t) = \int_0^t \cos(nx) dx \]

\[\int_0^1 \int_0^1 \frac{dxdy}{(1+xy)\ln(xy)} = ?\]

Combine the geometric series \(\sum\limits_{n=0}^\infty (-1)^n x^n =
\frac{1}{1+x}\) with: \[I(t)=\int_0^1 \int_0^1 \frac{(xy)^t}{\ln(xy)}
dxdy\]

\[\int_0^\infty \frac{\sin^2 x}{x^2(1+x^2)}dx = ?\]

Solve the resulting differential equation after differentiating twice
the following integral: \[ I(t) = \int_0^\infty \frac{\sin^2 (tx)}{x^
2(1+x^2)}dx \]

\[\int_0^1 \ln x \left(\frac{1}{t+x}+\frac{1}{\frac{1}{t}+x}\right)dx
= ?\]

Split the integral in two parts, then substitute \(x\to tx\) and
respectively \(x\to \frac{x}{t}\) in the resulting integrals to
obtain: \[I(t)= \int_0^1 \ln x \left(\frac{1}{t+x}+\frac{1}{\frac{1}
{t}+x}\right)dx = \int_0^\frac{1}{t} \frac{\ln(tx)}{1+x}dx + \int_0^t
\frac{\ln\left(\frac{x}{t}\right)}{1+x}dx\] Now employ the Feynman's
trick (in the generalized variant).

---------------------------------------------------------------------

In disguise this exercise is a reformulation of the following
identity: \[ \operatorname{Li}_2(-t)+\operatorname{Li}_2\left(-\frac
{1}{t}\right) = -\frac{\pi^2}{6} - \frac{1}{2}\ln^2 t \] Where \(\
operatorname{Li}_2(x)\) is the dilogarithm function.

\[\int_0^\infty e^{-x^2}dx = ?\]

This integral can be generated starting from: \[I(t)=\int_0^\infty \
frac{e^{-t(1+x^2)}}{1+x^2}dx\Rightarrow I'(t)=-e^{-t}\int_0^\infty e^
{-tx^2} dx\overset{\sqrt t x \to x}= -\frac{e^{-t}}{\sqrt t}\int_0^\
infty e^{-x^2} dx\] Afterwards it should be observed that \(I(0)=\
frac{\pi}{2}\) and \(I(\infty)=0\), therefore: \[\frac{\pi}{2} = \
int_0^\infty \frac{e^{-t}}{\sqrt t} dt \int_0^\infty e^{-x^2} dx \
overset{t\to t^2} = 2\left(\int_0^\infty e^{-x^2}dx\right)^2 \]

---------------------------------------------------------------------

Feynman's trick in practice

---------------------------------------------------------------------

In this last chapter we'll dive into some "real-world" integrals and
observe how Feynman's trick can be adapted for them. Additionally, we
will also attempt to build up some heuristics that'll help us
manipulate the integrals up to a point where we can introduce an
useful parameter. This is due to the fact that with most of the
previous examples we had the luxury to parameterise the integrals as
they appeared, however often this might not be the case.

---------------------------------------------------------------------

Breaking the rules

In the second chapter we've observed how parameterising the integral
so that it also simplifies some parts of the integral can make things
more intuitively. However, we can always look for better, especially
when our approach doesn't seem elegant enough.

Let's see how can we break this rule with the following integral:

\[I=\int_0^1 \frac{\ln(1-x^2+x^4)}{1-x^2} dx\]

As with our first rule of thumb it's straightforward to introduce the
new parameter as:

\[I(a)=\int_0^1 \frac{\ln(a(1-x^2)+x^4)}{1-x^2} dx \Rightarrow I'(a)=
\int_0^1 \frac{1}{a(1-x^2)+x^4}dx\]

You are encouraged to proceed forward with the above integral,
however computing \(I'(a)\) might not give the most pleasant result.
So to overcome this, we'll manipulate the integral a bit before
parametersing it.

One extremely useful substitution (which deserves its own special
chapter) is \(x\to \frac{1-x}{1+x}\). This one is great here
especially since it has the following property (among many others):

\[\int_0^1 \frac{f(x)}{1-x^2}dx \overset{x\to\frac{1-x}{1+x}}= \
frac12\int_0^1 \frac{f\left(\frac{1-x}{1+x}\right)}{x}dx\] \[\
Rightarrow I \overset{x\to\frac{1-x}{1+x}}= \frac12\underbrace{\int_0
^1 \frac{\ln(1+14x^2+x^4)}{x}dx}_{x^2\to x} - \frac12 \int_0^1 \frac
{\ln((1+x)^4)}{x}dx\] \[= \frac14\int_0^1 \frac{\ln(1+14x+x^2)}{x}dx
- 2 \int_0^1 \frac{\ln(1+x)}{x}dx = \frac14 I(\operatorname{arccosh}
7) - I(0)\]

Above, although we could have introduced a simple parameter, for a
smoother result we parameterised the integral as:

\[I(t)=\int_0^1 \frac{\ln(1+2\cosh t+x^2)}{x} dx \Rightarrow I'(t)= \
int_0^1 \frac{2\sinh t}{1+2\cosh t+x^2}dx\] \[= \int_0^1 \frac{2\sinh
t}{(x+\cosh t)^2-\sinh^2 t}dx = -2\operatorname{arctanh}\left(\frac
{x+\cosh t}{\sinh t}\right)\bigg|_0^1 = t\]

Finally, to find to find \(I\) we can combine \(I(\operatorname
{arccosh} 7)\) with \(I(0)\).

\[I = \frac14 \int_0^{\operatorname{arccosh 7}} t \, dt - \frac34 I
(0) = \frac{\ln^2(2+\sqrt 3)}{2} - \frac{\pi^2}{8}\]
---------------------------------------------------------------------

The result from above follows since includes \(I(0) = \frac{\pi^2}{6}
\), which is also the Basel problem in disguise, but you're further
encouraged to approach it by employing Feynman's trick.

\[\int_0^1 \frac{\ln(1+x)}{x}dx=\frac{\pi^2}{12}\]

One idea is to rewrite the integral as: \[ I=\int_0^1 \frac{\ln(1+x)}
{x}dx\overset{x\to x^3}=3\int_0^1 \frac{\ln(1+x^3)}{x}dx\] \[\
Rightarrow \frac13I-I=\int_0^1 \frac{\ln\left(\frac{1+x^3}{1+x}\
right)}{x}dx\Rightarrow I = -\frac32 \int_0^1 \frac{\ln(1-x+x^2)}{x}
dx \] Now put Feynman's trick to use for: \[I(t)=\int_0^1 \frac{\ln
(1-tx+x^2)}{x}dx\] But also take into account that: \[I(0)=\int_0^1 \
frac{\ln(1+x^2)}{x}dx\overset{x^2\to x}=\frac12 \int_0^1 \frac{\ln
(1+x)}{x}dx=\frac12I\]

---------------------------------------------------------------------

Switching to rational functions

Maybe it's just a personal preference, but for me working with
rational functions tends to give a higher visibility on how to
parameterise the integrals. We'll see what is meant by this when
dealing with the following integral:

\[I=\int_0^\frac{\pi}{2}\ln(2+\tan^2 x)dx\]

In this form, two immediate ways to parameterise the integral are as
follows:

\[I(a)=\int_0^\frac{\pi}{2}\ln(a+\tan^2 x)dx,\quad I(b)=\int_0^\frac
{\pi}{2}\ln(2+b\tan^2 x)dx\]

It turns out that both variants work, but we can do better. Another
rule of thumb that I follow is to use almost exclusive rational
functions as they often provide the most visibility to work with.
Let's do that for our integral too.

\[I=\int_0^\frac{\pi}{2}\ln(2+\tan^2 x)dx\overset{\tan x\to x}=\int_0
^\infty \frac{\ln(2+x^2)}{1+x^2}dx\]

And at this point it should be obvious where to place the parameter
so that we can simplify the denominator.

\[I(t)=\int_0^\infty \frac{\ln(1+t(1+x^2))}{1+x^2}dx\Rightarrow I'(t)
=\int_0^\infty \frac{1}{1+t+tx^2}dx=\frac{\pi}{2}\frac{1}{\sqrt t\
sqrt{1+t}}\] \[\Rightarrow I=\int_0^1 I'(t)dt=\frac{\pi}{2}\int_0^1 \
frac{1}{\sqrt t\sqrt{1+t}}dt\overset{t=x^2}=\pi\int_0^1\frac{1}{\sqrt
{1+x^2}}dx=\pi\ln(1+\sqrt 2)\]

Obviously, if you're already more used to trigonometric functions,
hyperbolic functions or anything else, then this can be ignored. I
however recommend switching to rational functions, a few exceptions
being when there's a clear way to parameterise the integral directly
or when at least one of the bounds is \(\infty\).

---------------------------------------------------------------------

For more practice you can also attempt to tackle the following
integral:

\[\int_0^\frac{\pi}{2} \ln(\sec^2 x +\tan^4 x)dx=?\]

Start by substituting \(\tan x \to x\) then employ Feynman's trick.

---------------------------------------------------------------------

Cleaning up the functions

Before parameterising the integrals it's quite useful to clean the
disturbing functions as much as possible before any parameterisation.
Let's demonstrate this by considering:

\[I=\int_0^\frac{\pi}{2}\arctan\left(\frac{2\sin x}{2\cos x -1}\
right)\frac{\sin\left(\frac{x}{2}\right)}{\sqrt{\cos x}}dx\]

As we have trigonometric functions all over the place, we will
perform the Weierstrass substitution \(\tan\left(\frac{x}{2}\right)\
to x\) in order to obtain only rational functions, as per the
previous heuristic.

\[\Rightarrow I = 2\int_0^1 \frac{x\arctan\left(\frac{4x}{1-3x^2}\
right)}{\sqrt{1-x^2}(1+x^2)}dx\]

It would be great to parameterise the integral so that we get rid of
the arctangent function, but unfortunately here it is a bit
overloaded, therefore we'll clean it by splitting it into two. In
case it's not obvious how to do that directly, we can differentiate
it, perform partial fractions and then integrate back.

\[\frac{d}{dx}\arctan\left(\frac{4x}{1-3x^2}\right)=\frac{12x^2+4}{9x
^4+10x^2+1}=\frac{1}{1+x^2}+\frac{3}{1+9x^2}\] \[\Rightarrow \int \
left(\frac{1}{1+x^2}+\frac{3}{1+9x^2}\right) dx = \arctan x + \arctan
(3x) + C\]

However since we're integrating over the \((0,1)\) interval we have
that \(x \cdot 3 x>1\) for \(x > \frac{1}{\sqrt 3}\), so we need to
rewrite the integral as:

\[I=2\int_0^1 \frac{x(\arctan x + \arctan(3x))}{\sqrt{1-x^2}(1+x^2)}
dx - 2 \int_\frac{1}{\sqrt 3}^1 \frac{x}{\sqrt{1-x^2}(1+x^2)}dx\] \[=
2 \mathcal J - \frac{\pi}{\sqrt 2}\ln(2+\sqrt 3)\]

At this point there's only left to evaluate the first integral, \(\
mathcal J\), for which we'll employ Feynman's trick. There isn't any
way to place the parameter so that we get rid of anything from the
denominator so we'll simply introduce the following parameterisation:

\[\mathcal J(t)=\int_0^1 \frac{x(\arctan x + \arctan(tx))}{\sqrt{1-x^
2}(1+x^2)}dx\Rightarrow \mathcal J'(t) = \int_0^1 \frac{x^2}{\sqrt
{1-x^2}(1+x^2)(1+t^2 x^2)}dx\]

In general when we have rational functions it is prefered to
integrate over \((0,\infty)\), if possible, as it drastically reduces
the result, and when there's the derivative of \(\arcsin x\) in the
denominator one way to map \((0,1)\) to \((0,\infty)\) is to directly
substitute \(x \to \frac{1}{\sqrt{1+x^2}}\).

\[\Rightarrow \mathcal J=\int_0^\infty \frac{1}{(2+x^2)(1+t^2+x^2)}dx
= \frac{\pi}{2(1-t^2)}\left(\frac{1}{\sqrt{1+t^2}}-\frac{1}{\sqrt 2}\
right)\]

Now in order to go back to \(\mathcal J(3)\), we'll make use of \(\
mathcal J(-1)=0\), so:

\[\mathcal J = \frac{\pi}{2}\int_{-1}^3 \frac{1}{(1-t^2)}\left(\frac
{1}{\sqrt{1+t^2}}-\frac{1}{\sqrt 2}\right)dt\]

Finally, to finish this integral we'll subsitute \(t = \frac{1-x}
{1+x}\).

\[\Rightarrow \mathcal J = \frac{\pi}{4\sqrt 2}\int_{-\frac12}^\infty
\frac{1}{x}\left(\frac{1+x}{\sqrt{1+x^2}}-1\right)dx=\frac{\pi}{\sqrt
2}\ln \left(\frac{1+\sqrt 5}{2}\right)\]

As such, we can conclude that:

\[\int_0^\frac{\pi}{2}\arctan\left(\frac{2\sin x}{2\cos x -1}\right)\
frac{\sin\left(\frac{x}{2}\right)}{\sqrt{\cos x}}dx=\sqrt 2 \pi \ln \
left(\frac{1+\sqrt 5}{2}\right)- \frac{\pi}{\sqrt 2}\ln(2+\sqrt 3)\]
---------------------------------------------------------------------

Similarly you can try and tackle the following integral:

\[\int_0^\pi \arctan\left(\frac{1+2\cos x}{\sqrt 3}\right)dx=?\]

Start by substituting \(\tan\left(\frac{x}{2}\right) \to x\), then
split the arctangent function into two parts and use Feynman's trick.

---------------------------------------------------------------------

Preparing better integral bounds

Another useful thing to consider before using Feynman's trick is to
manipulate the bounds prior to parameterising the integral so that
they get rid of any complicated functions for the differentiated
integral, \(I'(t)\). This will almost always give a smoother and
easier integral that we have to undo. We will show this in action
with the integral from below.

\[I=\int_0^1 \frac{x^2 \ln(1-x^2)}{1+x^4}dx\]

With the previous heuristic we've already seen that it is useful to
integrate over \((0,\infty)\) when there are some rational functions
- therefore we will attempt to do the same with this integral. You
are encouraged to try and see what kind of mess it would be produced
by parameterising the integral as it is, when the bounds are \((0,1)
\). However we will jump straightforward to get our bounds to \((0,\
infty)\).

In order to obtain that we can notice that the integrand is even, so
we can move the bounds to \((-1,1)\) and then substitute \(x\to \frac
{1-x}{1+x}\) again, which is another useful way to get our bounds at
\((0,\infty)\).

\[I=\frac12\int_{-1}^1 \frac{x^2\ln(1-x^2)}{1+x^4}dx\overset{x\to\
frac{1-x}{1+x}}=\frac12\int_0^\infty \frac{(1-x)^2\ln\left(\frac{4x}
{(1+x)^2}\right)}{1+6x^2+x^4}dx\]

Now we can split the logarithm into three parts and use that:

\[\int_0^\infty \frac{(1-x)^2}{1+6x^2+x^4}dx = \frac{\pi}{2\sqrt 2}-\
frac{\ln(1+\sqrt 2)}{\sqrt 2}\] \[\int_0^\infty \frac{(1-x)^2\ln(x)}
{1+6x^2+x^4}dx\overset{x\to\frac{1}{x}} = -\int_0^\infty \frac{(1-x)^
2\ln(x)}{1+6x^2+x^4}dx = 0\]

Therefore our integral is:

\[I=\frac{\pi\ln 2}{2\sqrt 2}-\frac{\ln 2\ln(1+\sqrt 2)}{\sqrt 2}-\
mathcal J, \quad \mathcal J = \int_0^\infty \frac{(1-x)^2\ln(1+x)}
{1+6x^2+x^4}dx\]

To evaluate the emerging integral we will perform Feynman's trick.
There's not an obvious way to place the parameter as to simplify the
denominator, so we'll parameterise the integral as:

\[\mathcal J(a) = \int_0^\infty \frac{(1-x)^2\ln(1+ax)}{1+6x^2+x^4}dx
\Rightarrow \mathcal J'(a)\int_0^\infty \frac{x(1-x)^2}{(1+ax)(1+6x^
2+x^4)}dx\] \[=-\frac{(1+a)^2\ln a}{1+6a^2+a^4}-\frac{\ln(1+\sqrt 2)}
{\sqrt 2}\frac{a(3-a+a^2)+1}{1+6a^2+a^4}+\frac{\pi}{2\sqrt 2}\frac{a
(3+a+a^2)-1}{1+6a^2+a^4}\]

The partial fraction was ommited above, as what's really important
here is that we're left only with a simple \(\ln a\) as a
"disturbing" function. In contrast, if the bounds were \((0,1)\)
things would have been way more complicated.

Let's finish this integral as we still have to undo the
differentiating step.

\[\mathcal J(1)=\mathcal J(1)-\mathcal J(0)=\int_0^1 \mathcal J'(a)da
\] \[=-\int_0^1\frac{(1+a)^2\ln a}{1+6a^2+a^4}da-\frac{\pi\ln(1+\sqrt
2)}{8\sqrt 2}-\frac{3\ln 2\ln(1+\sqrt 2)}{4\sqrt 2}-\frac{\pi^2}{16\
sqrt 2}+\frac{3\pi\ln 2}{8\sqrt 2}\] \[\int_0^1\frac{(1+x)^2\ln x}
{1+6x^2+x^4}dx=\int_0^1\left(\frac{1}{2\sqrt 2}\frac{(1+\sqrt 2)-x}
{(1+\sqrt 2)^2+x^2}-\frac{1}{2\sqrt 2}\frac{(1-\sqrt 2)-x}{(1-\sqrt
2)^2+x^2}\right)\ln x \, dx\] \[\small =\frac{1}{8\sqrt 2}\left(\
operatorname{Li}_2\left(-(1+\sqrt 2)^2\right)-\operatorname{Li}_2\
left(-(1-\sqrt 2)^2\right)\right)+\frac{1}{2\sqrt 2}\left(\
operatorname{Ti}_2\left(-(1+\sqrt 2)\right)-\operatorname{Ti}_2\left
(-(1-\sqrt 2)\right)\right)\]

The result from above follows since:

\[\int_0^1 \frac{x\ln x}{a^2+x^2}dx\overset{x^2\to x}=\frac14\int_0^1
\frac{\ln x}{a^2+x}dx=\frac14\operatorname{Li}_2\left(-\frac{1}{a^2}\
right)\] \[\int_0^1 \frac{a\ln x}{a^2+x^2}dx\overset{x\to ax}=\int_0^
\frac1a\frac{\ln a +\ln x}{1+x^2}dx\overset{IBP}=-\operatorname{Ti}_2
\left(\frac1a\right)\]

Where \(\operatorname{Li}_2(x)\) is the dilogarithm and \(\
operatorname{Ti}_2(x)\) is the inverse tangent integral.

Finally, collecting all the results yields:

\[I=\frac{\pi^2}{16\sqrt 2}+\frac{\pi \ln 2}{8\sqrt 2}+\frac{\pi\ln
(1+\sqrt 2)}{8\sqrt 2}-\frac{\ln 2\ln(1+\sqrt 2)}{4\sqrt 2}\] \[\
small +\frac{1}{8\sqrt 2}\left(\operatorname{Li}_2\left(-(1+\sqrt 2)^
2\right)-\operatorname{Li}_2\left(-(1-\sqrt 2)^2\right)\right)+\frac
{1}{2\sqrt 2}\left(\operatorname{Ti}_2\left(-(1+\sqrt 2)\right)-\
operatorname{Ti}_2\left(-(1-\sqrt 2)\right)\right)\]
---------------------------------------------------------------------

With the same idea one can attempt to calculate the following
integral:

\[\int_0^1 \frac{\ln^2 x\ln(1+x)}{1+x^2}dx=?\]

Map the bounds from \((0,1)\) to \((0,\infty)\) by substituting \(x\
to\frac{1}{x}\) - here it is necessary to add the resulting integral
with the original one - afterwards use Feynman's trick.

---------------------------------------------------------------------

Multiple parameters

We mostly got familiar to apply Feynman's trick by introducing a
parameter somewhere, however sometimes even multiple parameters can
be used when encountering new integrals. To exemplify such a
situation, let's take a look at the following unit square integral
arising in geometric probability:

\[I=\int_0^1\int_0^1 \left(\frac{-\ln(xy)}{1-xy}\right)^mdxdy\]

In order to generate the \(\ln(xy)\) part it's straightforward to
consider the following integral:

\[I(a)=\int_0^1\int_0^1 \frac{(xy)^a}{(1-xy)^m}dxdy\]

Differentiating with respect to \(a\), \(m\) times followed by
setting \(a=0\) will gives us the desired integral, ignoring the \
((-1)^m\) term. However the denominator is still troublesome, and to
deal with that we will introduce one more parameter:

\[I(a,z)=\int_0^1\int_0^1 \frac{(xy)^a}{z-xy}dxdy\]

This is perfect now, as we can recover our original integral by
differentiating with respect to \(a\), \(m\) times, and with respect
to \(z\), \(m-1\) times, followed by setting \(a=0\) and respectively
\(z=1\) - ignoring some coefficients.

One way to evaluate \(I(a,z)\) is to expand the denominator into
geometric series as:

\[I(a,z)=\int_0^1\int_0^1 \frac{(xy)^a}{z-xy}dxdy=\frac{1}{z}\sum_{n=
0}^\infty \int_0^1\int_0^1(xy)^a\left(\frac{xy}{z}\right)^ndxdy\] \[=
\sum_{n=0}^\infty \frac{1}{z^{n+1}}\int_0^1x^{n+a}dx\int_0^1y^{n+a}dy
=\sum_{n=1}^\infty \frac{1}{z^{n+2}}\frac{1}{(n+a)^2}\]

Now we will take \(m\) derivatives with respect to \(a\) and then set
it to \(0\).

\[\frac{\partial^m}{\partial a^m}I(a,z) = \int_0^1\int_0^1 \frac{\ln^
m(xy)}{z-xy} dxdy =(-1)^m (m+1)!\sum_{n=1}^\infty \frac{1}{z^{n+2}}\
frac{1}{n^{m+2}}\]

This can be also rewriten in terms of the polylogarithm function as:

\[\int_0^1\int_0^1 \frac{(-\ln(xy))^m}{z-xy} dxdy = (m+1)!\
operatorname{Li}_{m+2}\left(\frac{1}{z}\right)\]

Finally, we can arrive at our original integral by taking \(m-1\)
derivatives with respect to \(z\) and setting it to \(1\).

\[(-1)^m(m-1)!\int_0^1\int_0^1 \frac{(-\ln(xy))^m}{(1-xy)^m}dxdy=
(m+1)!\frac{d^{m-1}}{dz^{m-1}}\operatorname{Li}_{m+2}\left(\frac{1}
{z}\right)\bigg|_{z=1}\]

Although the derivation from above was the important part since it
shows the main idea on how to differentiate in order to produce the
desired integral, by using \(\frac{\partial}{\partial z} \
operatorname{Li}_n(z) = \frac{\operatorname{Li}_{n-1}(z)}{z}\) the
result can be also written, with the help of OEIS, as:

\[\int_0^1\int_0^1 \left(\frac{-\ln(xy)}{1-xy}\right)^m dxdy =m(m+1)\
sum_{k=1}^{m-1}|s(m-1,m-k)|\zeta(k+2)\]

Where \(s(n,m)\) is the Stirling number of the first kind and \(\zeta
(z)\) is the Riemann zeta function.

---------------------------------------------------------------------

A similar idea can be applied for the following integral:

\[\int_0^\infty \frac{\cos(3x)}{(1+x^2)^4}dx=?\]

Make use of a more general integral that was evaluated in the fourth
chapter, namely: \[\int_0^\infty \frac{\cos(tx)}{a^2+x^2}dx = \frac{\
pi}{2a}e^{-at}\] Then differentiate \(3\) times w.r.t. \(a\).

---------------------------------------------------------------------

Cascaded Feynman's trick

Sometimes to enable an application of Feynman's trick we needed to
actually apply another Feynman's trick. Let's look at the following
integral:

\[I = \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\cot x \csc^2 y\ln(\cos
y)\ln(1-2\sin x+\sin^2 x\csc^2y)dxdy\]

The first step should be pretty obvious by now, namely to get rid of
the trigonometric functions.

\[I \overset{\large \sin x\to x \atop \large \cot y \to y}=-\frac12\
int_0^\infty \int_0^1 \frac{\ln\left(1+\frac{1}{y^2}\right)\ln((1-x)^
2+x^2y^2)}{x}dxdy\]

Now we can notice that we have two logarithms and only one of them
contains the \(x\) term, however since the bounds are \((0,1)\)
dealing with that integral won't produce much success (as the result
would be quite complicated).

However we also have the bounds as \((0,\infty)\) for the \(y\)
integral, and it would be even better if we could have a single
logarithm. To further obtain such a favorable integral form, we can
use the following result:

\[\int_0^1 \frac{t}{t^2+y^2}dt=\frac12\ln\left(1+\frac{1}{y^2}\right)
\] \[\Rightarrow I=-\int_0^\infty \int_0^1 \int_0^1 \frac{t\ln((1-x)^
2+x^2y^2)}{x(t^2+y^2)}dtdxdy \] \[\overset{y\to ty}=-\int_0^1 \frac
{1}{x}\int_0^1 \int_0^\infty \frac{\ln((1-x)^2+x^2t^2y^2)}{1+y^2}
dydtdx\]
---------------------------------------------------------------------

In the third chapter we saw how it's useful to have a list of
integral results. One more such useful integral that tend to appear
quite often is:

\[\int_0^\infty \frac{\ln(a^2+b^2 x^2)}{1+x^2}dx= \pi \ln(|a|+|b|)\]

You can differentiate either \(I(a)\) or \(I(b)\), and even directly
employ the accelerated Feynman's trick by writing the logarithm as an
integral.

---------------------------------------------------------------------

Now back to our integral, by using the above result, we can easily
finish the integral.

\[I=-\pi\int_0^1 \frac{1}{x}\int_0^1 \ln((1-x)+xt)dtdx\] \[=\pi\
underbrace{\int_0^1 \left(\frac1x+\frac{\ln(1-x)}{x^2}\right)dx}_{\
large -1}-\pi\underbrace{\int_0^1\frac{\ln(1-x)}{x}dx}_{\large -\frac
{\pi^2}{6}}=\frac{\pi^3}{6}-\pi\]
---------------------------------------------------------------------

Although this marks the conclusion of the essay, this isn't a static
website, and I might update it when I encounter new interesting
integrals that are worth to be shown. So far, the integrals comes
from my posts on Mathematics Stack Exchange, combined with some of
the most popular integrals - thus you can also check them directly
there.

For further exercises, I can recommend you to explore math forums and
magazines such as Art of Problem Solving, Mathematics Stack Exchange,
The American Mathematical Monthly, Crux Mathematicorum, or the
Romanian Mathematical Magazine, where dozens of fascinating integrals
are often posted or published. Additionally, delving into other
fields like Statistics, Physics, or Quantum Physics will present you
with many remarkable integrals -- some of which might be computed
using Feynman's trick.

---------------------------------------------------------------------

I would appreciate a notice on my email address (rxzacky@gmail.com)
in case you find any mistakes or if something feels unclear in this
essay - and even better if you have some further ideas or
suggestions.

This work is licensed under a Creative Commons Attribution 4.0
International License, and it can be cited as:
Zaharia Burghelea, "Feynman's Trick," https://zackyzz.github.io/
feynman.