Request for crazy integrals

Question

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?

Answer 1

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 $m,n\in\mathbb N$.

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

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

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

Answer 7

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.

Answer 8

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.