nice integral

Art of Problem Solving

AoPS Online

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

nice integral

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1535 posts

#1 h Mar 28, 2025, 8:09 PM • 1 Y

Y by Figaro

$\int_0^\infty \frac{\tanh x}{4x (1+\cosh(2x))} dx$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1169 posts

#2 h Mar 29, 2025, 6:42 PM

Y by

$\int_0^\infty\frac{\tanh x}{x(1+\cosh x)}=\ln 2.$

This post has been edited 1 time. Last edited by Entrepreneur, Mar 31, 2025, 7:07 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

773 posts

Figaro

#3 h Mar 30, 2025, 1:48 PM

Y by

$\int_0^\infty \frac{\tanh x}{4x (1+\cosh(2x))} dx=\frac{7 \zeta(3)}{8\pi^2}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1535 posts

#4 h Mar 30, 2025, 7:50 PM

Y by

Figaro wrote:

$\int_0^\infty \frac{\tanh x}{4x (1+\cosh(2x))} dx=\frac{7 \zeta(3)}{8\pi^2}$

Numerically correct. Can you post your solution? I don’t have it.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

773 posts

Figaro

#5 h Mar 31, 2025, 12:18 PM • 1 Y

Y by Svyatoslav

Certainly. The first two substitutions are $u=\cosh{2x}$ and then $t=u+\sqrt{u^2-1}$ . We arrive at $2 \int_1^\infty \frac{t-1}{(t+1)^3 \ln t}dt$ . Letting $s=1/t$ , we see that this is equal to $2 \int_0^1 \frac{s-1}{(s+1)^3 \ln s}ds$ . So the whole integral is equal to $\int_0^\infty \frac{t-1}{(t+1)^3 \ln t}dt$ .

This can be solved by differentiating under the integral sign. $f(a)=\int_0^\infty \frac{t^a-1}{(t+1)^3 \ln t}dt$ and $f'(a)=\int_0^\infty \frac{t^a}{(t+1)^3 }dt=$ (beta function) $=\frac{\pi a(1-a)}{2 \sin(a \pi)}$ . So $\int_0^\infty \frac{t-1}{(t+1)^3 \ln t}dt=f(1)=\int_0^1 \frac{\pi t(1-t)}{2 \sin(t \pi)}dt= \frac{7 \zeta(3)}{2 \pi^2}$ .

I know that's a rather/very short explanation; please don't hesitate to ask for specific details.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

540 posts

#6 h Mar 31, 2025, 5:38 PM • 1 Y

Y by MS_asdfgzxcvb

An elegant solution!
Using complex integration along the rectangular contour, one can show that for any real $a\geqslant0$
$-\,\frac i8\int_{-\infty}^\infty\frac{\tanh x}{(a-ix)\big(1+\cosh(2x)\big)}dx=\frac14\int_0^\infty\frac{x\tanh x}{(a^2+x^2)\big(1+\cosh(2x)\big)}dx=-\,\frac1{16\pi^2}\psi^{(2)}\Big(a+\frac12\Big)$ where $\psi^{(2)}(1+z)=-2\sum_{n=1}^\infty\frac1{(n+z)^3}$ - polygamma function.
Another complex integration approach is to consider the integral $\oint_C\frac{\tanh z}{1+\cosh(2z)}z^{-s}dz$ , where $s\in(1,2)$ , and the contour $C$ consists of the big circle of the radius $R\to \infty$ and the cut along the positive part of the axis $X$ . The residues at $z=\frac{\pi i}2+\pi i k, k=0,\,\pm1,\,\pm2...$ evaluated, and then the limit $s\to1$ taken.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a