8
$\begingroup$

I'd like to compute: $$ I_n = \int_{-\infty}^{+\infty}\mathrm{d}x\frac{x^{2n}}{\cosh^2 x}. $$ We have, quite easily: $$ I_0 = \int_{-\infty}^{+\infty}\mathrm{d}x\frac{1}{\cosh^2 x}=\left[\tanh x\right]_{-\infty}^{+\infty}=2. $$ So to begin with, I gave a try at $$ I_1 = \int_{-\infty}^{+\infty}\mathrm{d}x\frac{x^{2}}{\cosh^2 x}. $$ My idea was to use the following contour $\Gamma_{\varepsilon, M}$ to integrate on in the complex plane: the rectangle with vertices $M$, $M+i\frac{\pi}{2}$, $-M+i\frac{\pi}{2}$, $-M$, and an indent of radius $\varepsilon$ at $i\frac{\pi}{2}$, where $\cosh z=0$. Then by Residues Theorem: $$ \oint_{\Gamma_{\varepsilon, M}} \mathrm{d}z\frac{z^{2}}{\cosh^2 z} = 0. $$

Problem is, when I make the calculations the integrand is still singular $O(\frac{1}{\varepsilon})$ on my semicircle. More explicitly: $$ \int_{0}^{-\pi} \mathrm{d}\theta i\varepsilon e^{i\theta}\frac{(i\frac{\pi}{2}+\varepsilon e^{i\theta})^{2}}{\cosh^2 (i\frac{\pi}{2}+\varepsilon e^{i\theta})} = \int_{0}^{-\pi} \mathrm{d}\theta i\varepsilon e^{i\theta} \frac{-\frac{\pi^2}{4} + O(\varepsilon)}{\varepsilon^2e^{2i\theta}+O(\varepsilon^3)}. $$ To avoid this problem I tried: $$ \oint_{\Gamma_{\varepsilon, M}} \mathrm{d}z\frac{(z-i\frac{\pi}{2})^{2}}{\cosh^2 z}=0. $$

But then I only get: $$ \int \mathrm{d}x\frac{(x-i\frac{\pi}{2})^{2}}{\cosh^2x}-\int \mathrm{d}x\frac{x^{2}}{\cosh^2(x-i\frac{\pi}{2})}=0 $$ $$ \int \mathrm{d}x\left(\frac{x^2}{\cosh^2x}+\frac{x^2}{\sinh^2x}\right) = \frac{\pi^2}{2} $$

since $\cosh(x+i\frac{\pi}{2}) = i\sinh x$.

$\endgroup$
7
$\begingroup$

Why not simply go up to $\Im{z} = \pi$ so that the rectangle $C$ has vertices $\pm R$ and $\pm R+i \pi$? Then we consider

$$\oint_C dz \frac{z^{2 n+1}}{\cosh^2{z}}$$

which is equal to

$$\int_{-R}^R dx \frac{x^{2 n+1}}{\cosh^2{x}} + i \int_0^{\pi} dy \frac{(R+i y)^{2 n+1}}{\cosh^2{(R + i y)}} \\ + \int_R^{-R} dx \frac{(x+i \pi)^{2 n+1}}{\cosh^2{x}} + i \int_{\pi}^0 dy \frac{(-R+i y)^{2 n+1}}{\cosh^2{(-R + i y)}}$$

As $R \to \infty$, the second and fourth integrals vanish. Thus the contour integral is equal to

$$\int_{-\infty}^{\infty} dx \frac{x^{2 n+1}-(x+i \pi)^{2 n+1}}{\cosh^2{x}}$$

Note that the highest power in the numerator is $x^{2 n}$, and that all odd powers vanish upon integration. Thus the contour integral is expressible in terms of integrals of lower powers:

$$-i \pi (2 n+1)\int_{-\infty}^{\infty} dx \frac{x^{2 n}}{\cosh^2{x}} +i \sum_{k=0}^{n-1} (-1)^k \binom{2 n+1}{2 k} \pi^{2 (n-k)+1}\int_{-\infty}^{\infty} dx \frac{x^{2 k}}{\cosh^2{x}}$$

By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue of the integrand at the pole $z=i \pi/2$, which is a double pole.

$$\operatorname*{Res}_{z=i \pi/2} \frac{z^{2 n+1}}{\cosh^2{z}} = \lim_{z\to i \pi/2} \frac{d}{dz}\frac{(z-i \pi/2)^2 z^{2 n+1}}{\cosh^2{z}} = (-1)^{n+1} (2 n+1) \left ( \frac{\pi}{2}\right )^{2 n}$$

In order to verify this you can proceed as follows: $$ \lim_{z\to i \pi/2} \frac{d}{dz}\frac{(z-i \pi/2)^2 z^{2 n+1}}{\cosh^2{z}} = \lim_{u\to 0} \frac{d}{dz}\frac{u^2 (u+i\pi/2)^{2 n+1}}{-\sinh^2{u}} = \lim_{u\to 0} (u+i\pi/2)^n\ \frac{-\left((2n+1)u^2+2(u+i\pi/2)u\right)2\sinh u+2(u+i\pi/2)u^2\cosh u}{\sinh^3u} = \lim_{u\to 0} (u+i\pi/2)^n\ \frac{-(2n+3)u^3-i\pi u +2u^3 + i\pi u + O(u^4)}{u^3+O(u^4)}= (-1)^{n+1}(2n+1)\left(\frac{\pi}{2}\right) ^{2n}. $$

You then need to find the integrals of lower powers. Actually, you can get around this by setting up a system of equations for each $k$ from $0$ to $n$. Let

$$I_k = \int_{-\infty}^{\infty} dx \frac{x^{2 k}}{\cosh^2{x}}$$

and

$$R_k = (-1)^{k} (2 k+1) \left ( \frac{\pi}{2}\right )^{2 k}$$

Then we have

$$(2 n+1) I_n - \sum_{k=0}^{n-1} (-1)^k \binom{2 n+1}{2 k} \pi^{2 (n-k)} I_k = 2 R_n$$ $$(2 n-1) I_{n-1} - \sum_{k=0}^{n-2} (-1)^k \binom{2 n-1}{2 k} \pi^{2 (n-1-k)} I_k = 2 R_{n-1}$$ $$\cdots$$ $$3 I_1 - \pi^2 I_0 = 2 R_1$$ $$I_0 = 2 $$

$\endgroup$
  • $\begingroup$ Thanks Ron. I'm having problems with the computation of the residue at the double pole: shouldn't it be second derivative? And if so: I've done the calculations a couple of times but it doesn't turn out correct. Thanks in advance! $\endgroup$ – Brightsun Feb 17 '14 at 10:44
  • $\begingroup$ @Brightsun: Double pole is a derivative; in general for an $n$-tuple pole, it is the $n-1$th derivative. The calculation is messy. $\endgroup$ – Ron Gordon Feb 17 '14 at 13:33
  • $\begingroup$ Totally right dude, thanks. $\endgroup$ – Brightsun Feb 17 '14 at 13:37
7
$\begingroup$

Note that: $$\int _{-\infty }^{\infty }{\frac {{x}^{2\,n}}{ \cosh^2 \left( x \right) }}\,{\mathrm{d}x}=2\,\int _{0}^{\infty }\!{\frac {{x}^{2\,n }}{\cosh^2 \left( x \right) }}\,{\mathrm{d}x}\tag{1}$$ then, consider the following integral for $\Re(s)>1$: $$\begin{aligned} \int _{0}^{\infty }{\frac {{x}^{s}}{ \cosh^2 \left( x \right) }}{dx}&=-2\,\int _{0}^{\infty }\!{x}^{s}{\frac {\mathrm{d}}{\mathrm{d}x}} \left( \frac{1}{ 1+{{\rm e}^{2\,x}}}\right)\, {\mathrm{d}x},\\ \mbox{integration by parts...}\quad&=2\,s\int _{0}^{\infty }\!{\frac {{x}^{s-1}}{1+{{\rm e}^{2\,x}}}}\,{\mathrm{d}x},\\ \mbox{partial fractions...}\quad&=2\,s\int _{0}^{\infty }\!2\,{\frac {{x}^{s-1}}{1-{{\rm e}^{4\,x}}}}\,{\mathrm{d}x}-2\,s\int _{0}^{\infty }\!{\frac {{x}^{s-1}}{1-{{\rm e}^{2\,x}}}}\,{\mathrm{d}x},\\ \mbox{rescale the variables...}\quad&=-s \left( {2}^{-2\,s+2}-{2}^{1-s} \right) \int _{0}^{\infty }\!{\frac {{x}^{s-1}}{-1+{{\rm e}^{x}}}}\,{\mathrm{d}x},\\ \mbox{geometric series...}\quad&=-s \left( {2}^{-2\,s+2}-{2}^{1-s} \right) \sum _{n=1}^{\infty } \left( \int _{0}^{\infty }\!{x}^{s-1}{{\rm e}^{-xn}}\,{\mathrm{d}x} \right),\\ \mbox{rescale the variable...}\quad&=-s \left( {2}^{-2\,s+2}-{2}^{1-s} \right) \left(\sum _{n=1}^{\infty } \frac{1}{n^s}\right)\left(\int _{0}^{\infty }\!{x}^{s- 1}{{\rm e}^{-x}}\,{\mathrm{d}x}\right),\\ \mbox{function definitions...}\quad&=-s \left( {2}^{-2\,s+2}-{2}^{1-s} \right) \Gamma \left( s \right) \zeta \left( s \right),\\ \mbox{absorb the s into}\,\, \Gamma \mbox{...}\quad&=\left( -{2}^{-2\,s+2}+{2}^{1-s} \right) \zeta \left( s \right) \Gamma \left( s+1 \right) \end{aligned} \tag{2}$$ where $\zeta$ is the Riemann zeta function. Then, if $s=2n,\, n\in \mathbb{Z}$: $$\Gamma \left( 2\,n+1 \right) = \left( 2\,n \right) !,\quad\zeta \left( 2\,n \right) ={\frac { \left( -1 \right) ^{n+1}B_{2n} \left( 2\,\pi \right) ^{2\,n}}{2\left( 2\,n \right) !}} \tag{3}$$ where $B$ denotes the Bernoulli number (or polynomial later) and thus: $$\int _{-\infty}^{\infty }\!{\frac {{x}^{2\,n}}{ \cosh^2 \left( x \right) }}{dx}= \left( 2-2^{2-2n} \right)\left( -1 \right) ^{n+1}B_{2n}\, {\pi }^{2\,n} \tag{4}$$ or if you prefer: $$\int _{-\infty}^{\infty }\!{\frac {{x}^{2\,n}}{ \cosh^2 \left( x \right) }}{dx}=2B_{2n}\left(\frac{1}{2} \right) \left( i\pi \right) ^{2\,n} \tag{5}$$

$\endgroup$
  • $\begingroup$ +1 . Nice job. It's well known in Statistical Physics with Bose-Einstein distribution. $\endgroup$ – Felix Marin Feb 16 '14 at 22:36
  • $\begingroup$ Oh cool, cheers! :) $\endgroup$ – Graham Hesketh Feb 16 '14 at 22:54
  • $\begingroup$ Actually I had those integrals jumping out of Sommerfeld's Expansions for highly degenerate metals! $\endgroup$ – Brightsun Feb 17 '14 at 18:27
3
$\begingroup$

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I_{n}\equiv\int_{-\infty}^{\infty}{x^{2n} \over \cosh^{2}\pars{x}}\,\dd x:\ {\large ?}.\qquad n = 0,1,2,3,\ldots}$.

\begin{align} \color{#c00000}{I_{n}}&=8\int_{0}^{\infty}{x^{2n}\expo{2x} \over \pars{\expo{2x} + 1}^{2}}\,\dd x =-4\int_{x\ =\ 0}^{x\ \to\ \infty}x^{2n}\,\dd\pars{1 \over \expo{2x} + 1} \\[3mm]&=2\delta_{n0} + 8n \color{#00f}{\int_{0}^{\infty}{x^{2n - 1}\expo{-2x} \over 1 + \expo{-2x}}\,\dd x} \end{align}

\begin{align} &\color{#00f}{\int_{0}^{\infty}{x^{2n - 1}\expo{-2x} \over 1 + \expo{-2x}}\,\dd x} =\sum_{k = 0}^{\infty}\pars{-1}^{k}\int_{0}^{\infty} x^{2n - 1}\expo{-\pars{2k + 2}x}\,\dd x \\[3mm]&=\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{2k + 2}^{2n}} \int_{0}^{\infty}x^{2n - 1}\expo{-x}\,\dd x =-\,{\Gamma\pars{2n} \over 2^{2n}} \sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k^{2n}} \\[3mm]&=-{\Gamma\pars{2n} \over 2^{2n}} \bracks{{1 \over 2^{2n}}\sum_{k = 1}^{\infty}{1 \over k^{2n}} -\sum_{k = 1}^{\infty}{1 \over \pars{2k - 1}^{2n}}} \\[3mm]&=-{\Gamma\pars{2n} \over 2^{2n}} \bracks{{1 \over 2^{2n}}\sum_{k = 1}^{\infty}{1 \over k^{2n}} -\sum_{k = 1}^{\infty}{1 \over k^{2n}} + \sum_{k = 1}^{\infty}{1 \over \pars{2k}^{2n}}} \\[3mm]&=-{\Gamma\pars{2n} \over 2^{2n}}\pars{{1 \over 2^{2n - 1}} - 1} \sum_{k = 1}^{\infty}{1 \over k^{2n}} ={\Gamma\pars{2n} \over 2^{4n - 1}}\pars{2^{2n - 1} - 1}\zeta\pars{2n} \end{align}

$$\color{#66f}{\large I_{n}} \equiv\int_{-\infty}^{\infty}{x^{2n} \over \cosh^{2}\pars{x}}\,\dd x =\color{#66f}{\large2\delta_{n0} + n{2^{2n - 1} - 1 \over 2^{4n - 4}}\,\Gamma\pars{2n} \zeta\pars{2n}} $$

$\endgroup$
0
$\begingroup$

I would like to post an answer based on a slight variation of @Graham Hesketh's method, both because I find it a bit simpler and because it reflects the steps leading to the integrals temselves in the context where they appear.

For $t>0$ and $s>0$, we consider $$ \int_0^\infty \frac{x^{s-1}}{e^{tx}-1}dx=\sum_{n=1}^\infty\int_0^\infty x^{s-1}e^{-ntx}dx=t^{-s}\sum_{n=1}^\infty n^{-s} \int_0^\infty x^{s-1}e^{-x}dx=\frac{\zeta(s)\Gamma(s)}{t^s}\,, $$ and note that $$ \int_0^\infty \frac{x^{s-1}}{e^{tx}-1}dx -\int_0^\infty\frac{x^{s-1}}{e^{tx}+1}dx=2\int_0^\infty \frac{x^{s-1}}{e^{2tx}-1}dx = 2^{1-s}\int_0^\infty\frac{x^{s-1}}{e^{tx}-1}dx $$ so, by comparison, $$ \int_0^\infty \frac{x^{s-1}}{e^{tx}+1}dx=\frac{1-2^{1-s}}{t^s}\zeta(s)\Gamma(s)\,. $$ The derivative of both sides of this equation, with respect to $t$, yields $$ \int_0^\infty \frac{x^s}{\cosh^2(\frac{tx}{2})}dx = \frac{4s(1-2^{1-s})}{t^{s+1}}\zeta(s)\Gamma(s)\,. $$ In particular, for $s=2n$ with $n\in\mathbb N$, and $t=2$ we retrieve the integrals we wanted $$ I_{n}=\int_{-\infty}^{+\infty} \frac{x^{2n}}{\cosh^2 x}dx = 2^{1-2n}(1-2^{1-2n})\,4n\,\zeta(2n)\Gamma(2n)\,. $$ E.g. $$ I_0=2\,,\qquad I_1=\frac{\pi^2}{6}\,,\qquad I_2=\frac{7\pi^4}{120}\,. $$ (The case $n=0$ requires formally an analytic continuation $\epsilon \Gamma(\epsilon)=1+\mathcal O(\epsilon)$ and $\zeta(0)=-1/2$).

$\endgroup$

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.