Examples of "difficult" integrals with are easier to solve with a series?

Question

Yesterday someone posted an extremely elegant solution to a seemingly bizarre series where the integral:

$$\int_{0}^{1} x^{m}\ dx = \frac{1}{m + 1}$$

was utilized.

Oftentimes one will also interchange the summation from inside the integral to the outside in the case of uniformly convergent series.

Are there situations where an otherwise complicated looking integral has a "trivial" solution as a series?

I suppose that you are referring to this answer. (Which was not posted yesterday, but the last activity in that thread was, indeed, yesterday; which is why the question was bumped.) — Martin Sleziak, Jul 18 '14 at 09:38
Yes, these are precisely the sort of answers I was looking for! — Kayle of the Creeks, Jul 18 '14 at 10:34

score 7 · Answer 1 · answered Jul 18 '14 at 13:13

7

The following double integral is an example.

$$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}\,dx\,dy=\int_{0}^{1}\int_{0}^{1}\sum_{k=0}^{\infty}(xy)^{k-1}\,dx\,dy\\=\sum_{k=0}^{\infty}\int_{0}^{1}\int_{0}^{1}(xy)^{k-1}\,dx\,dy=\sum_{k=0}^{\infty}\int_{0}^{1}x^{k-1}\,dx\int_{0}^{1}y^{k-1}\,dy=\sum_{k=0}^{\infty}\frac1{k^2}=\frac{\pi^2}{6}$$

answered Jul 18 '14 at 13:13

RRL

90,707

This is from Noam Elkies' paper, found here at http://arxiv.org/pdf/math/0101168v5.pdf, in case you need context. – Mathguy Jul 19 '14 at 14:19
I haven't taken vector analysis yet so my understanding of two variable integrals is very superficial, mostly to the extent that I have had to use them on my physics courses! If I understand correctly, there is some generalization of uniform convergence to n-dimensions which allows you to change the orders of integration and limits with this double integral. – Kayle of the Creeks Jul 21 '14 at 10:04
@Mathguy: Thanks for the reference. – RRL Jul 21 '14 at 13:21
@user30562: Each summand is non-negative and the series converges monotonically which justifies the switch. In general we have bounded convergence, monotone and dominated convergence theorems to work with. – RRL Jul 21 '14 at 13:33

score 2 · Answer 2 · answered Jul 18 '14 at 17:23

In statistical physics, integrals of the form $\int_0^\infty dx\, {x^n}(e^{x}\pm 1)^{-1}$ are common, and the standard approach is to expand the reciprocal in powers of $e^{-x}$: \begin{align} \int_0^\infty dx\, \frac{x^n}{e^{x}\pm 1} &= \int_0^\infty dx\, \frac{x^n e^{-x}}{1\pm e^x}\\ &= \int_0^\infty dx\, x^n \sum_{k=0}^\infty (\mp 1)^k e^{-(k+1)x}\\ &= \sum_{k=0}^\infty (\mp 1)^k \int_0^\infty dx\, x^n e^{-(k+1)x}\\ &= \sum_{k=0}^\infty (\mp 1)^k \frac{n!}{(k+1)^{n+1}} = \left\{\array{n!\,\eta(n+1)\\ n!\, \zeta(n+1)}\right\} \end{align} where $\eta(n)$ and $\zeta(n)$ are the Dirichlet eta and Riemann zeta functions respectively. (Note that such integrals are really particular cases of the polylogarithm.)

score 1 · Answer 3 · answered Jul 18 '14 at 08:08

Here's an integral I saw on this AoPS thread a while ago: $\displaystyle\int_{1}^{\infty}\left(\ln\left(1-\dfrac{1}{x}\right)\right)^2\,dx$

WolframAlpha tells us the answer is $\dfrac{\pi^2}{3}$, which should remind you of $\zeta(2) = \displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$.

Indeed, using an infinite series is the way to go. Specifically, this one: $\displaystyle\sum_{n = 1}^{\infty}ny^{n} = \dfrac{y}{(1-y)^2}$.

Of course, evaluating that integral required a few more tricks. See that thread for a few solutions.

And how can one find the even zeta values remaining in the series framework? — Start wearing purple, Jul 18 '14 at 10:34

Examples of "difficult" integrals with are easier to solve with a series?

3 Answers3

Linked