6
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ You may want to invest in this book. I am giving it to myself as a Christmas present amazon.com/dp/0521796369/… $\endgroup$ – user150203 Dec 6 '18 at 3:39
  • 1
    $\begingroup$ You should refer to this question: math.stackexchange.com/questions/1096701/nice-book-on-integrals $\endgroup$ – user150203 Dec 6 '18 at 3:46
  • 1
    $\begingroup$ 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! $\endgroup$ – AmbretteOrrisey Dec 8 '18 at 2:32
6
$\begingroup$

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: $$I_1=\int_0^\frac{\pi}{2} \frac{\arctan(\tan x\sec x)}{\tan x +\sec x}dx=\frac{\pi}{2}\ln 2 -\frac{\pi}{6}\ln(2+\sqrt 3)$$ $$I_2=\int_0^\infty \exp\left(-\frac{3x^2+15}{2x^2+18}\right)\cos\left(\frac{2x}{x^2+9}\right)\frac{dx}{x^2+1}=\frac{\pi}{e}$$ $$I_3=\int_0^1 \frac{\ln^2 (1+x) (\ln^2 (1+x) +6\ln^2(1-x))}{x}dx=\frac{21}{4}\zeta(5)$$ $$I_4=\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2}\frac{dx}{\sqrt x}=-\frac{\pi}{24}$$ $$I_5=\int_0^\infty \frac{1-\cos x}{8-4x\sin x +x^2(1-\cos x)}dx=\frac{\pi}{4}$$ $$I_6=\int_0^\infty \frac{\arctan x}{x^4+x^2+1}dx=\frac{\pi^2}{8\sqrt{3}}-\frac{2}{3}G+\frac{\pi}{12}\ln(2+\sqrt{3})$$ $$I_7=\int_0^\infty \frac{\ln(1+x)}{x^4-x^2+1}dx=\frac{\pi}{6}\ln(2+\sqrt 3)+\frac23 G -\frac{\pi^2}{12 \sqrt 3}$$ $$I_8=\int_0^\infty \left(\sqrt{x}-\sqrt{\sqrt{1+x^2}-1}\right)\sin{x}dx=\frac12 \sqrt{\frac{\pi}{2}}\frac{e-1}{e}$$ $$I_{9}=\int_0^{\frac{\pi}{4}} \ln\left(2+\sqrt{1-\tan^2 x}\right)dx = \frac{\pi}{2}\ln\left(1+\sqrt{2}\right)+\frac{7\pi}{24}\ln2-\frac{\pi}{3}\ln\left(1+\sqrt{3}\right)-\frac{G}{6}$$ $$I_{10}=\int_{-\infty}^\infty \frac{\sin \left(x-\frac{1}{x}\right) }{x+\frac{1}{x}}dx=\frac{\pi}{e^2}$$ $$I_{11}=\int_{-\infty}^\infty \frac{\cos \left(x-\frac{1}{x}\right) }{\left(x+\frac{1}{x}\right)^2}dx=\frac{\pi}{2e^2}$$ $$I_{12}=\int_0^1 \frac{\operatorname{arcsinh}\sqrt{\frac{1}{x+2}}}{\sqrt{3-2x-x^2}}dx=\frac23 G$$ $$I_{13}=\int_0^\frac{\pi}{2} x^2 \sqrt{\tan x}dx=\frac{\sqrt{2}\pi(5\pi^2+12\pi\ln 2 - 12\ln^22)}{96}$$ $G$ stands for Catalan's constant. $I_8$ and $I_{12}$ are one of Cornel Ioan Valean's integral, I heard that his book is coming out soon which surely contains some exotic ones. I can't prove them here, but if you desire to evaluate one of those integrals and get stuck feel free to ask for help/hints.

$\endgroup$
  • 5
    $\begingroup$ These are really nice integrals. Thanks. $\endgroup$ – clathratus Dec 6 '18 at 0:58
  • 2
    $\begingroup$ Cheers @Zacky! I too will be working on these! $\endgroup$ – user150203 Dec 6 '18 at 3:37
  • $\begingroup$ Could I have a starter tip on $I_6$ and $I_7$? I'm very lost $\endgroup$ – clathratus Dec 13 '18 at 3:09
  • 1
    $\begingroup$ @Zacky Oh yeah, looks like you have one. Nice! $\endgroup$ – Frpzzd Jan 5 at 19:53
  • 1
    $\begingroup$ @Frpzzd Here is another one that you might like: $$\int_{-\infty}^\infty \frac{\cos\left(x-\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2}dx$$ $\endgroup$ – Zacky Jan 5 at 20:04
4
$\begingroup$

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 $m,n\in\mathbb N$.

$\endgroup$
  • 1
    $\begingroup$ Just the user I was hoping would answer! Thank you for your contribution, those integrals look delicious! :) $\endgroup$ – clathratus Dec 18 '18 at 1:02
  • $\begingroup$ 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... $\endgroup$ – clathratus Jan 24 at 17:30
  • $\begingroup$ @clathratus My apologies; the $-x^2$ should have been $-x^4$. I have fixed my answer so that the integral converges. $\endgroup$ – Frpzzd Jan 24 at 21:28
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Hypergeometric function for not, thanks for the answer. $\endgroup$ – clathratus Dec 6 '18 at 0:59
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Really interesting! where can I learn more? $\endgroup$ – clathratus Dec 8 '18 at 2:43
  • 1
    $\begingroup$ 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 ... ! $\endgroup$ – AmbretteOrrisey Dec 8 '18 at 2:51
  • 1
    $\begingroup$ @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! $\endgroup$ – AmbretteOrrisey Jan 2 at 6:09
  • 1
    $\begingroup$ Use eis.aen.hypaegon@gmail.com ¶ (It's Koine Greek ... it means to which they had gone.) $\endgroup$ – AmbretteOrrisey Jan 2 at 6:23
  • 1
    $\begingroup$ @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. $\endgroup$ – AmbretteOrrisey Jan 3 at 17:40
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – AmbretteOrrisey Dec 15 '18 at 5:27
  • $\begingroup$ 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 $$\int_1^\infty\frac{x\log\frac{1+1/x}{1-1/x}}{1+x^2}dx=2\sum_{n=0}^\infty {I_n\over 2n+1}$$$$=$$$$2\sum_{n=0}^\infty{1\over 2n+1}\sum_{k=0}^\infty{(-1)^k\over 2(n+k)+1}$$ $\endgroup$ – AmbretteOrrisey Dec 15 '18 at 6:54
  • $\begingroup$ 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 $$\sum_{n=1}^\infty{n\over n^2-{1\over4}}\sum_{k=0}^\infty{(-1)^k\over 2(n+k)+1}={\pi\over8}(\pi-2) .$$ $\endgroup$ – AmbretteOrrisey Dec 15 '18 at 9:11
  • $\begingroup$ 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. $\endgroup$ – AmbretteOrrisey Dec 15 '18 at 9:20
  • $\begingroup$ @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. $\endgroup$ – James Arathoon Dec 15 '18 at 16:24
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ cool stuff! I really don't know anything about PDE's but this is sure exotic! Thanks for the answer. $\endgroup$ – clathratus Dec 24 '18 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.