Zelaron Gaming Forum - Hilariously hideous integrals

Zelaron Gaming Forum (http://zelaron.com/forum/index.php)

- Science and Art (http://zelaron.com/forum/forumdisplay.php?f=344)

- - Hilariously hideous integrals (http://zelaron.com/forum/showthread.php?t=58496)

Hilariously hideous integrals

I got this idea from an integral on Wikipedia:

$\reverse \int_{-\infty}^{\infty} \frac{e^{itx}}{x^2+1}\,\mathrm{d}x = \pi e^{-|t|}$

This integral is quite difficult to solve with standard techniques from elementary calculus. Instead, the usual approach is to rewrite it as $\reverse \int_{-\infty}^{\infty} \frac{e^{itx}}{(x-i)(x+i)}\,\mathrm{d}x$ , then evaluate it as a contour integral by using the following contour that surrounds the pole $\reverse x=i$ :

https://i.imgur.com/u7HpV8X.jpg

The point of this approach is that it leaves the other pole $\reverse x=-i$ outside the nice semicircle contour no matter the value of a, which lets you convert the problem into a "simple" calculation of limits (e.g. by using the residue theorem). Compared to integration, the calculation of limits is a relatively algorithmic process that usually works without any "art".

So I figured, why not add more poles to the lower half-plane? After all, this shouldn't affect the contour, and you should still be able to use the residue theorem on a single pole. In other words, you should, at least in principle, still be able to find exact solutions to such integrals by calculating a finite number of limits, even if they might be somewhat... messy.

Well, as far as I can tell, they are. I played around a bit with this one:

$\reverse I_n = \int_{-\infty}^{\infty} \frac{e^{itx}}{(x^2+1)\prod_{k=1}^{n}(x-k+i)(x+k+i)}\,\mathrm{d}x$

Obviously $\reverse I_n$ has a single pole $\reverse x=i$ in the upper half-plane, and its remaining $\reverse 2n+1$ poles in the lower half-plane.

Apparently I failed to evaluate any such integrals (except $\reverse I_0$ ) manually, but Mathematica returned some... interesting results for n=1 and n=2 after half an hour or so:

http://zelaron.com/buljong/i1rr.png

http://zelaron.com/buljong/i2rr.png

Have you seen any other particularly hideous integrals?

Have you seen the following post on Stackexchange: https://math.stackexchange.com/quest...tegral-milking

I think you'll enjoy it.

Quote:

Originally Posted by Demosthenes (Post 711273)

Have you seen the following post on Stackexchange: https://math.stackexchange.com/quest...tegral-milking

I think you'll enjoy it.

Nice ones! The techniques used to turn

$\reverse \int_{0}^{\pi}\cos(mx)\cos(nx)\,\mathrm{d}x = \frac{\pi}{2}\delta_{mn}$

into

$\reverse \int_{0}^{\pi/2}\ln|\sin(mx)|\cdot\ln|\sin(nx)|\,\mathrm{d}x = \frac{\pi^3}{24}\cdot\frac{\gcd^2(m,n)}{mn} + \frac{\pi\ln^2(2)}{2}$

look promising for a number of identities for orthogonal polynomials. For example, the Legendre polynomials $\reverse P_n(x)$ that I mentioned in another thread satisfy

$\reverse \int_{-1}^{1} P_m(x)P_n(x)\,\mathrm{d}x = \frac{2}{2n+1}\delta_{mn}.$

Also, I like this one from the book "Irresistible Integrals" (page 190) that was mentioned in the Math Underflow thread:

$\reverse \int_{0}^{1}(-\ln x)^{-\mu}\,\mathrm{d}x = \frac{\pi}{\mathrm{\Gamma}(\mu)\,\sin(\mu\pi)}$

It looks quite a bit like the functional equation for the Riemann zeta function, I think (where I replaced replaced its usual parameter $\reverse s$ with $\reverse 1-\mu$ and did a little algebra):

$\reverse 2^{1-\mu}\frac{\mathrm{\zeta(\mu)}}{\mathrm{\zeta(1-\mu)}}\, = \frac{\pi^{\mu}}{\mathrm{\Gamma(\mu)}\,\sin\left(0 .5(1-\mu)\pi\right)}$