# Request for crazy integrals

I'm a sucker for exotic integrals like the one evaluated in this post. I don't really know why, but I just can't get enough of the amazing closed forms that some are able to come up with.

So, what are your favorite exotic integral identities, and how do you prove them?

• You may want to invest in this book. I am giving it to myself as a Christmas present amazon.com/dp/0521796369/…
– user150203
Dec 6, 2018 at 3:39
• You should refer to this question: math.stackexchange.com/questions/1096701/nice-book-on-integrals
– user150203
Dec 6, 2018 at 3:46
• I know what you mean - I can't get enough of 'em either! There's that one that Peter Borwein likes to use in his demostrations: a product of progressively scaled sinc functions, that's π/2 upto a certain point then starts to be a bit less. A very very tiny bit! Dec 8, 2018 at 2:32
• You can invent such integrals through "milking" techniques described here. Perhaps this one is the most remarkable they found.
– J.G.
Dec 3, 2020 at 8:13
• $$\int_{0}^{\infty} \frac{x\ln(\tanh(x))^2}{3\cosh(2x)-2e^{-2x}}\text{d}x = -\frac{\pi^4}{96\sqrt{3} } -\frac{\ln3\ln^3(2+\sqrt{3} )}{24\sqrt{3} } -\frac{\pi^2}{12\sqrt{3} }\operatorname{Li}_2\left (-\frac{1}{\sqrt{3}} \right ) +\frac{\pi^2}{12\sqrt{3} }\operatorname{Li}_2\left (\frac{1}{\sqrt{3}} \right) -\frac{\pi^2}{4\sqrt{3} }\operatorname{Li}_2\left (\sqrt{3}-2 \right) -\frac{1}{2\sqrt{3} }\operatorname{Li}_4\left (2-\sqrt{3} \right) +\frac{1}{2\sqrt{3} }\operatorname{Li}_4\left (-2+\sqrt{3} \right)$$ Sep 28, 2021 at 13:02

Here are some of my favorites: $$\int_0^\pi \sin^2\Big(x-\sqrt{\pi^2-x^2}\Big)dx=\frac{\pi}{2}$$ $$\int_0^\infty \frac{\ln(x)}{(1+x^{\sqrt 2})^\sqrt{2}}dx=0$$ $$\int_0^\infty \frac{dx}{(1+x^{1+\sqrt{2}})^{1+\sqrt{2}}}=\frac{1}{\sqrt{2}}$$ $$\int_{-\infty}^\infty \ln(2-2\cos(x^2))dx=-\sqrt{2\pi}\zeta(3/2)$$ $$\int_0^\infty \frac{\text{erf}^2(x)}{x^2}dx=\frac{4\ln(1+\sqrt{2})}{\sqrt{\pi}}$$ $$\int_0^\infty \frac{x^{3}\ln(e^x+\frac{x^3}{6}+\frac{x^2}{2}+x+1)-x^4}{\frac{x^3}{6}+\frac{x^2}{2}+x+1}=\frac{\pi^2}{2}$$ $$\int_0^{\pi/2} \ln(x^2+\ln^2(\cos(x)))dx=\pi\ln(\ln(2))$$ $$\int_0^\infty \frac{\arctan(2x)+\arctan(x/2)}{x^2+1}dx=\frac{\pi^2}{4}$$ $$\int_0^{\pi/2}\frac{\sin(x+100\tan(x))}{\sin(x)}dx=\frac{\pi}{2}$$ $$\int_0^1 \frac{x\ln(1+x+x^4+x^5)}{1+x^2}dx=\frac{\ln^2(2)}{2}$$ $$\int_0^{1/2}\sin(8x^4+x)\cos(8x^4-x)\cos(4x^2)xdx=\frac{\sin^2(1)}{16}$$

$$\int_0^{2\pi} \sqrt{2+\cos(x)+\sqrt{5+4\cos(x)}}dx=4\pi$$

And here are four extremely exotic scrumptious integrals:

$$\int_0^1 \frac{\sin(\pi x)}{x^x (1-x)^{1-x}}dx=\frac{\pi}{e}$$ $$\int_{-\infty}^\infty \frac{dx}{(e^x-x)^2+\pi^2}=\frac{1}{1+\Omega}$$

$$\int_0^\infty \frac{3\pi^2+4(z-\sinh(z))^2}{[3\pi^2+4(z-\sinh(z))^2]^2+16\pi^2(z-\sinh(z))^2}dz=\frac{1}{8+8\sqrt{1-w^2}}$$

$$\int_0^{\pi/2}\ln|\sin(mx)|\ln|\sin(nx)|dx=\frac{\pi^3}{24}\frac{\gcd^2(m,n)}{mn}+\frac{\pi \ln^2(2)}{2}$$

...where $$\Omega$$ is the Omega Constant, $$w$$ is the Dottie Number, and $$m,n\in\mathbb N$$.

• Just the user I was hoping would answer! Thank you for your contribution, those integrals look delicious! :) Dec 18, 2018 at 1:02
• Are you sure that the $6$-th integral you listed (the one that evaluates as $\pi^2$) converges? The graph of it tells me otherwise... Jan 24, 2019 at 17:30
• @clathratus My apologies; the $-x^2$ should have been $-x^4$. I have fixed my answer so that the integral converges. Jan 24, 2019 at 21:28

Here are some links to a few integrals: 1 (Big list, but not all of them got the right answer). From AoPS: 2, 3 , 4. Some that are solvable with Feynman's trick: here.

As for my favourites (most of them appeared on Romanian Mathematical Magazine), some are: $$I_1=\int_0^\frac{\pi}{2} \frac{\arctan(\tan x\sec x)}{\tan x +\sec x}dx=\frac{\pi}{2}\ln 2 -\frac{\pi}{6}\ln(2+\sqrt 3)$$ $$I_2=\int_0^\infty \exp\left(-\frac{3x^2+15}{2x^2+18}\right)\cos\left(\frac{2x}{x^2+9}\right)\frac{dx}{x^2+1}=\frac{\pi}{e}$$ $$I_3=\int_0^1 \frac{\ln^2 (1+x) (\ln^2 (1+x) +6\ln^2(1-x))}{x}dx=\frac{21}{4}\zeta(5)$$ $$I_4=\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2}\frac{dx}{\sqrt x}=-\frac{\pi}{24}$$ $$I_5=\int_0^\infty \frac{1-\cos x}{8-4x\sin x +x^2(1-\cos x)}dx=\frac{\pi}{4}$$ $$I_6=\int_0^\infty \frac{\arctan x}{x^4+x^2+1}dx=\frac{\pi^2}{8\sqrt{3}}-\frac{2}{3}G+\frac{\pi}{12}\ln(2+\sqrt{3})$$ $$I_7=\int_0^\infty \frac{\ln(1+x)}{x^4-x^2+1}dx=\frac{\pi}{6}\ln(2+\sqrt 3)+\frac23 G -\frac{\pi^2}{12 \sqrt 3}$$ $$I_8=\int_0^1 \frac{\ln(1-x^2)\ln(1+x^2)}{1+x^2}dx=\frac{\pi^3}{32}-3G\ln 2+\frac{\pi}{2}\ln^22.$$ $$I_{9}=\int_0^{\frac{\pi}{4}} \ln\left(2+\sqrt{1-\tan^2 x}\right)dx = \frac{\pi}{2}\ln\left(1+\sqrt{2}\right)+\frac{7\pi}{24}\ln2-\frac{\pi}{3}\ln\left(1+\sqrt{3}\right)-\frac{G}{6}$$ $$I_{10}=\int_{-\infty}^\infty \frac{\sin \left(x-\frac{1}{x}\right) }{x+\frac{1}{x}}dx=\frac{\pi}{e^2}$$ $$I_{11}=\int_{-\infty}^\infty \frac{\cos \left(x-\frac{1}{x}\right) }{\left(x+\frac{1}{x}\right)^2}dx=\frac{\pi}{2e^2}$$ $$I_{12}=\int_0^1 \frac{\ln(1-x)\ln(1-x^4)}{x}dx=\frac{67}{32}\zeta(3)-\frac{\pi}{2} G$$ $$I_{13}=\int_0^\frac{\pi}{2} x^2 \sqrt{\tan x}dx=\frac{\sqrt{2}\pi(5\pi^2+12\pi\ln 2 - 12\ln^22)}{96}$$ $$I_{14}=\int_0^\frac{\pi}{4} \operatorname{arcsinh} (\sin x) dx=G-\frac58\operatorname{Cl}_2\left(\frac{\pi}{3}\right)$$ $$I_{15}=\int_0^\frac{\pi}{2} x \arcsin \left(\sin x-\cos x\right)dx=\frac{\pi^3}{96}+\frac{\pi}{8}\ln^2 2$$ $$I_{16}=\int_0^\infty \int_0^\infty \frac{\ln(1+x+y)}{xy\left((1+x+y)(1+1/x+1/y)-1\right)}dxdy=\frac72 \zeta(3)$$ Where $$G$$ is Catalan's constant and $$\operatorname{Cl}_2 (x)$$ is the Clausen function.

• These are really nice integrals. Thanks. Dec 6, 2018 at 0:58
• Cheers @Zacky! I too will be working on these!
– user150203
Dec 6, 2018 at 3:37
• Could I have a starter tip on $I_6$ and $I_7$? I'm very lost Dec 13, 2018 at 3:09
• @Zacky Oh yeah, looks like you have one. Nice! Jan 5, 2019 at 19:53
• @Frpzzd Here is another one that you might like: $$\int_{-\infty}^\infty \frac{\cos\left(x-\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2}dx$$ Jan 5, 2019 at 20:04

You might find a lot of crazy integrals and series in the book, (Almost) Impossible Integrals, Sums, and Series. A few examples of integrals,

$$\int_0^{\pi/2} \cot (x) \log (\cos (x)) \log ^2(\sin (x)) \operatorname{Li}_3\left(-\tan ^2(x)\right) \textrm{d}x$$ $$=\frac{109}{128}\zeta(7)-\frac{23}{32}\zeta(3)\zeta(4)+\frac{1}{16}\zeta(2) \zeta(5);$$ $$\int_0^{\log(1+\sqrt{2})} \coth (x) \log (\sinh (x)) \log \left(2-\cosh ^2(x)\right)\text{Li}_2\left(\tanh ^2(x)\right) \textrm{d}x$$ $$=\frac{73}{128}\zeta(5)-\frac{17}{64}\zeta(2)\zeta(3);$$ $$\int_0^1 \frac{\displaystyle\log^2(1-x)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{1+x} \textrm{d}x$$ $$=\frac{1}{36} \log ^6(2)-\frac{1}{6}\log ^4(2)\zeta (2)+\frac{7}{24} \log ^3(2) \zeta (3)+\frac{5}{8}\log ^2(2) \zeta (4)-\frac{581}{48} \zeta (6)$$ $$-\frac{7}{8} \log (2) \zeta (2)\zeta (3)-\frac{79}{64} \zeta^2 (3);$$

$$\sin (\theta)\sin\left(\frac{\theta}{2}\right)\int_0^1 \frac{\displaystyle x}{(1-x) \left(1-2 x \cos (\theta)+x^2\right)} (\zeta (m+1)-\text{Li}_{m+1}(x)) \textrm{d}x$$ $$=(-1)^{m-1} \sum_{k=1}^{\infty}\frac{H_{k+1}}{(k+1)^{m+1}}\sin\left(\frac{k \theta}{2}\right)\sin\left(\frac{(k+1)\theta}{2}\right)$$ $$+(-1)^{m-1}\sum_{i=2}^{m} (-1)^{i-1}\zeta(i)\sum_{k=1}^{\infty}\frac{\displaystyle \sin\left(\frac{k\theta}{2}\right)\sin\left(\frac{(k+1) \theta}{2}\right)}{(k+1)^{m-i+2}};$$ $$\sin\left(\frac{\theta}{2}\right)\int_0^1\frac{x(\cos(\theta)-x)}{(1-x)(1-2x\cos(\theta)+x^2)}(\zeta (m+1)-\text{Li}_{m+1}(x))\textrm{d}x$$ $$=(-1)^{m-1}\sum_{k=1}^{\infty}\frac{H_{k+1}}{(k+1)^{m+1}}\sin\left(\frac{k\theta}{2}\right)\cos\left(\frac{(k+1)\theta}{2}\right)$$ $$+(-1)^{m-1}\sum_{i=2}^{m}(-1)^{i-1} \zeta(i)\sum_{k=1}^{\infty} \frac{\displaystyle \sin\left(\frac{k\theta}{2}\right)\cos\left(\frac{(k+1)\theta}{2}\right)}{ (k+1)^{m-i+2}}.$$

A few examples of series (which you may also transform into some fancy integrals if you wish to),

$$\sum_{n=1}^{\infty}\frac{H_n}{n^2}\left(\frac{ H_1}{1^3}+\frac{H_2}{2^3}+\cdots +\frac{H_n}{n^3} \right)=10\zeta(7)+\frac{9}{2}\zeta(2)\zeta(5)-\frac{23}{2}\zeta(3)\zeta(4);$$ $$\sum_{n=1}^{\infty}\frac{H_n}{n^3}\left(\frac{H_1}{1^2}+\frac{H_2}{2^2}+\cdots +\frac{H_n}{n^2} \right)=\frac{23}{2}\zeta(3)\zeta(4)-\frac{11}{2}\zeta(2)\zeta(5)-4\zeta(7);$$ $$\sum_{n=1}^{\infty}\frac{H_n^2}{n^2}\left(\frac{H_1}{1^2}+\frac{H_2}{2^2}+\cdots +\frac{H_n}{n^2} \right)=\frac{45}{16}\zeta(7)-\frac{7}{2}\zeta(2)\zeta(5)+\frac{17}{2}\zeta(3)\zeta(4);$$ $$\sum_{n=1}^{\infty}\frac{H_n}{n^2}\left(\frac{H_1^2}{1^2}+\frac{H_2^2}{2^2}+\cdots +\frac{H_n^2}{n^2} \right)=\frac{93}{8} \zeta(7)+\frac{11}{2}\zeta(2)\zeta(5)-\frac{51}{4}\zeta(3)\zeta(4);$$ $$\zeta(4)$$ $$=\frac{8}{5}\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^2}+\frac{64}{5}\sum _{n=1}^{\infty } \frac{ \left(H_{2 n}\right)^2}{ (2 n+1)^2}+\frac{64}{5}\sum _{n=1}^{\infty } \frac{H_{2 n}}{(2 n+1)^3}$$ $$-\frac{8}{5}\sum _{n=1}^{\infty } \frac{\left(H_{2 n}\right){}^2}{ n^2}-\frac{32}{5}\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n+1)^2}-\frac{64}{5}\log(2)\sum _{n=1}^{\infty } \frac{H_{2 n}}{(2 n+1)^2}-\frac{8}{5}\sum _{n=1}^{\infty } \frac{H_{2 n}^{(2)}}{ n^2}.$$

Extremely crazy integrals you may also find in the paper The derivation of eighteen special challenging logarithmic integrals by Cornel Ioan Valean.

I'm sure a lot of crazy integrals you'll also meet in the sequel of the book (Almost) Impossible Integrals, Sums, and Series since the author prepares a continuation of this book.

• WOAH!! This is very impressive! I will have to buy that book! Thanks for the great answer (+1) Aug 2, 2019 at 17:45

$$\int_{-\infty}^\infty\prod_{k=1}^n\operatorname{sinc}{\theta\over(2k-1)}d\theta=\pi ,$$provided $$n\in{1 ... 7}$$ ... for $$n\geq8$$, it starts being $$<π$$ by the most miniscule amounts!

• Really interesting! where can I learn more? Dec 8, 2018 at 2:43
• It's actually called the Borwein integral. I strongly recommend the works of Peter Borwein in this connection - you'd love it! There's heaps of these crazy integrals in his works, & he's a specialist in unlimited precision arithmetic. He has an algorithm for finding closed-form expressions given a decimal expansion, that finds with a certain probability - rapidly increasing with number of digits - the peak-probability closed-form ... like if you were to say "hmmm - that looks like the square-root of ζ(3)!", or something - but systematised. And Ising integrals ... and ... and ... ! Dec 8, 2018 at 2:51
• @Clathratus -- I presume you mean me? I like the sound of the idea! Quite likely ... if it means I get an endless supply of the "meat that you know not of", to paraphrase a certain prophet perched on the edge of a well in Samaria! Jan 2, 2019 at 6:09
• Use eis.aen.hypaegon@gmail.com ¶ (It's Koine Greek ... it means to which they had gone.) Jan 2, 2019 at 6:23
• @Clathratus -- what I'll do for now, I think, is get it sorted just why that Borwein integral has it's property - afterall, I've quoted the result without any proof! Got sufficient material on it - and it appears within my measure. It's immediately less mysterious reflecting that the Fourier transform of sinc is the top-hat function ... so analysing it - if you have the integral of sinc×cos(a) (a being the radius of the top-hat) it suddenly becomes 0 the instant you have more than a full cycle of cos inside the central 'lobe' of sinc. And that is indeed what it's essentially about. Jan 3, 2019 at 17:40

I'm partial to the one in this question What is the Centroid of $$z=\frac{1}{(1-i\tau)^{i+1}},\ \ \tau\in (-\infty,\infty)$$ .

I found a solution, but it was hardly elegant. A solution that doesn't use hypergeometric functions in the middle of the solution would be nice.

• Hypergeometric function for not, thanks for the answer. Dec 6, 2018 at 0:59

This might not be a difficult integral but it made me come up with a new method to solve it so I think it's quite exotic.

Let’s do the general integral $$\displaystyle I(a,b)=\int_{0}^{\infty}e^{-(ax^{-2}+bx^{2})}dx$$

Differentiate with respect to a

$$\displaystyle \frac{\partial I}{\partial a}=\int_{0}^{\infty}x^{-2}e^{-(ax^{-2}+bx^{2})}dx$$

Now differentiate with respect to b $$\displaystyle \frac{\partial^2 I}{\partial a \partial b}=\int_{0}^{\infty}x^{-2}x^{2}e^{-(ax^{-2}+bx^{2})}dx$$

$$\displaystyle \frac{\partial^2 I}{\partial a \partial b}=\int_{0}^{\infty}e^{-(ax^{-2}+bx^{2})}dx$$

$$\displaystyle \frac{\partial^2 I}{\partial a \partial b}=I$$

Thus our integral satisfies this PDE.This is a hyperbolic homogenous PDE. It is a second order PDE but it is first order with respect to each of the variables so we’ll need two boundary conditions to determine a unique solution.(In this case two asympotic BCs and one Drichlet boundary condition will be used).Keep this in mind we’ll need it later.

Let’s complete the square of expression in the exponential.

$$\displaystyle I(a,b)=\int_{0}^{\infty}e^{-(ax^{-2}+bx^{2}-2\sqrt{ab}+2\sqrt{ab})}dx$$

$$\displaystyle I(a,b)=\int_{0}^{\infty}e^{-(\sqrt{a}x^{-1}-\sqrt{b}x)^{2}-2\sqrt{ab}}dx$$

$$\displaystyle I(a,b)=e^{-2\sqrt{ab}}\int_{0}^{\infty}e^{-(\sqrt{a}x^{-1}-\sqrt{b}x)^{2}}dx$$

Now let’s explore more of it’s properties.One thing to note is that this integral diverges(blows up) at b=0 but at a=0 it has a well known value. It is the Gaussian integral so

$$\displaystyle I(0,b)=\int_{0}^{\infty}e^{-(bx^{2})}dx=\frac{1}{2}\sqrt{\frac{\pi}{b}}$$

The negative exponential was extracted from the integral rather than the positive one beacause

$$\displaystyle \lim_{a\to\infty}\int_{0}^{\infty}e^{-(ax^{-2}+bx^{2})}dx=0$$

and

$$\displaystyle \lim_{a\to\infty}e^{-2\sqrt{ab}}=0$$

So let’s assume that we assume that the solution to our PDE is of the form

$$\displaystyle I(a,b)=e^{-2\sqrt{ab}}K(b)$$

where K is a function of b(and diverges at b=0)

Let’s put this in the PDE

$$\displaystyle \frac{\partial I}{\partial a}=-\sqrt{\frac{b}{a}}e^{-2\sqrt{ab}}K(b)$$

$$\displaystyle \frac{\partial^2 I}{\partial a \partial b}=-\sqrt{\frac{b}{a}}e^{-2\sqrt{ab}}K^{'}(b)-\frac{1}{2\sqrt{ab}}e^{-2\sqrt{ab}}K(b)+\sqrt{\frac{b}{a}}\sqrt{\frac{a}{b}}e^{-2\sqrt{ab}}K(b)$$

$$\displaystyle \frac{\partial^2 I}{\partial a \partial b}=e^{-2\sqrt{ab}}(-\sqrt{\frac{b}{a}}K^{'}(b)-\frac{K(b)}{2\sqrt{ab}}+K(b))$$

As $$\displaystyle \frac{\partial^2 I}{\partial a \partial b}=I$$

So

$$\displaystyle e^{-2\sqrt{ab}}(-\sqrt{\frac{b}{a}}K^{'}(b)-\frac{K(b)}{2\sqrt{ab}}+K(b))=e^{-2\sqrt{ab}}K(b)$$

$$\displaystyle -\sqrt{\frac{b}{a}}K^{'}(b)-\frac{K(b)}{2\sqrt{ab}}+K(b)=K(b)$$

$$\displaystyle -\sqrt{\frac{b}{a}}K^{'}(a)=\frac{K(b)}{2\sqrt{ab}}$$

$$\displaystyle K^{'}(b)=-\frac{K(b)}{2b}$$

This is a separable ODE.Let’s solve it

$$\displaystyle \frac{1}{K}dK=-\frac{1}{2}\frac{1}{b}db$$

Let’s integrate

$$\displaystyle \int \frac{1}{K}dK=-\frac{1}{2}\int \frac{1}{b}db$$

$$\displaystyle \ln(K)=-\frac{1}{2}\ln(b)+C$$

$$\displaystyle \ln(K)=\ln(b^{-\frac{1}{2}})+C$$

$$\displaystyle K=e^{C}b^{-\frac{1}{2}}$$

Let $$\displaystyle v=e^{C}$$

So

$$\displaystyle K(b)=vb^{-\frac{1}{2}}$$

Thus the solution is $$\displaystyle I(a,b)=ve^{-2\sqrt{ab}}b^{-\frac{1}{2}}$$

This expression diverges at b=0 which is exactly what we wanted. Now let’s determine the constant v. As

$$\displaystyle I(0,b)=\frac{1}{2}\sqrt{\frac{\pi}{b}}$$

So $$\displaystyle \frac{1}{2}\sqrt{\frac{\pi}{b}}=vb^{-\frac{1}{2}}e^{0}$$ $$v=\frac{\sqrt{\pi}}{2}$$

Thus the integral is

$$\displaystyle \boxed{I(a,b)=\frac{1}{2}\sqrt{\frac{\pi}{b}}e^{-2\sqrt{ab}}} (0\leqslant a,b)$$

• cool stuff! I really don't know anything about PDE's but this is sure exotic! Thanks for the answer. Dec 24, 2018 at 18:49

I like

$$\int_{-\infty}^{\infty } \frac{r \log \left(\frac{\frac{\frac{D^2}{4}+r^2}{D r}+1}{\frac{\frac{D^2}{4}+r^2}{D r}-1}\right)}{\frac{D^2}{4}+r^2} \, dr=\pi^2$$

where $$D>0$$ (no proof supplied).

If you make the mistake of trying to convert the $$\log$$ term to its series form, to attempt to integrate term by term, this integral becomes really crazy, an infinite almost fractal cascade of further self similar integrals with the series for $$\pi/2$$ gradually appearing out of the fog

$$1+\frac{1}{3}\left(\frac{1}{2}\right)+\frac{1}{5}\left(\frac{1}{2}\frac{3}{4}\right)+\frac{1}{7}\left(\frac{1}{2}\frac{3}{4}\frac{5}{6}\right)+...=\frac{\pi}{2}$$

You miss all this underlying structure sensibly driving via the mathematical motorway.

• I don't get this. How is this integral any different from $$\int_{-\infty}^\infty\frac{x\log\frac{x+1}{x-1}}{1+x^2}dx=? {\pi^2\over2} ?$$ Right! I agree through 'cheating' (evaluating it numerically) that it's so ... but I still say all that "D" business & putting it in a perversely complex form is a total red-herring! I get how there is a logarithmic singularity at x=±1 & how the imaginary part is an odd function between them. Dec 15, 2018 at 5:27
• You can derive using partial fraction representation the general case $$I_n\equiv\int_1^\infty\frac{dx}{x^{2n}(1+x^2)}=(-1)^n\bigg({\pi\over4}+\sum_{k=1}^n{(-1)^k\over2k-1}\bigg)=\bigg|{\pi\over4}+\sum_{k=1}^n{(-1)^k\over2k-1}\bigg|=\sum_{k=0}^\infty{(-1)^k\over 2(n+k)+1}$$ whence $$\int_1^\infty\frac{x\log\frac{1+1/x}{1-1/x}}{1+x^2}dx=2\sum_{n=0}^\infty {I_n\over 2n+1}$$$$=$$$$2\sum_{n=0}^\infty{1\over 2n+1}\sum_{k=0}^\infty{(-1)^k\over 2(n+k)+1}$$ Dec 15, 2018 at 6:54
• Likewise for the integral from 0 to 1 ... substitute $x=1/y$ & limits from 1 to $\infty$ & we get a similar series that begins at $n=1 \therefore$ with $I_1$ in the numerator & $2n-1$ in the denominator. So we get $$\int_0^\infty\frac{x\log\big(\frac{x+1}{x-1}\big)}{1+x^2}$$$$=$$$$\frac{\pi}{2}+2\sum_{n=1}^\infty{n\over n^2-{1\over4}}\sum_{k=0}^\infty{(-1)^k\over 2(n+k)+1}$$$$=$$$$\frac{π^2}{4} ,$$whence the integral over $(-\infty,\infty)$ is $\pi^2/2$. So we could numerically verify that $$\sum_{n=1}^\infty{n\over n^2-{1\over4}}\sum_{k=0}^\infty{(-1)^k\over 2(n+k)+1}={\pi\over8}(\pi-2) .$$ Dec 15, 2018 at 9:11
• Just one last thing, and then I'm through - as this is only taking delight in seeing explicitly how the series all pan-out & fit together, really - I'm sure the proper way to do it is something like Feynmann's trick or a contour integral, or whatever... but the value of the integral is π²/2 ... did you not put that in the first place & then edit it!? I'm almost sure you did. Dec 15, 2018 at 9:20
• @AmbretteOrrisey: The origin of this is an attempt to derive Coulombs law in a different way that is off-topic here. The physical integral is between $0$ and $\infty$, so was equal to $\pi^2/2$. $D$ is the charge separation. At the heart of my attempt at a physical model I ended up with a final integral that is independent of $D$ (the charge separation) which you have investigated in your comments. Physically the integral can be simplified because you can move the origin of the original volume integral from halfway between the charges to co-locate with one of the charges. Dec 15, 2018 at 16:24

We have $$\int_0^1x\left[{_2}F_1(\tfrac12,1;2;x)\right]^2dx=12-16\ln2,$$ $$\int_0^1 x\left[{_2}F_1(\tfrac13,\tfrac23;\tfrac32;x)\right]^3dx=\frac{27}{32},$$ $$\int_0^1x\left[{_3}F_2(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43;x)\right]^4dx=\frac{1792}{2187},$$ and $$\int_0^1\int_0^1\int_0^1\frac{dtdxdz}{x^{1/2}(1-x)^{1/2}t^{1/4}(1-t)^{5/12}(1-txz)^{1/4}}=\frac{2^{9/4}\cdot11\sqrt{\pi}}{3^{19/8}}\frac{\Gamma(\tfrac13)}{\Gamma(\tfrac14)}\sqrt{\sqrt{3}-1},$$ all from here.