I want to know how to prove that $$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=\frac{4G}{\pi }$$ Here $G$ denotes Catalan's constant, I obtained such result with the help of mathematica.

I also found that the integral equals a certain infinite series $$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=\sum _{n=0}^{+\infty }\frac{\binom{2n}{n}^2}{16^n\left(2n+1\right)}=\frac{4G}{\pi }$$ which can also be found in this link.

So I've $2$ questions

$1)$$¿$How can we transform the integral into the mentioned series?

$2)$$¿$Is there a simple way to evaluate the main integral without resorting to series expansion?

What I did for question $\#2$ is to employ the substitution $x=\ln\left(t\right)$ $$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=-2\int _1^{\infty }\frac{1-t^2}{\ln \left(t\right)\left(1+t^2\right)^2}\:\mathrm{d}t$$ But I'm not sure how to proceed.

  • 1
    $\begingroup$ For your second query, Here are several different approaches( see robjohn answer). $\endgroup$
    – Naren
    Jul 15, 2021 at 10:14
  • 4
    $\begingroup$ Just an idea: introduce $\displaystyle f(s)=\int _1^{\infty }\frac{(1-t^2)t^s}{\ln \left(t\right)\left(1+t^2\right)^2}\:\mathrm{d}t$ then derive under the integral sign w.r.t $s$ then perform the change of variable $y=\frac{1}{t}$. $\endgroup$
    – FDP
    Jul 15, 2021 at 12:40
  • $\begingroup$ How do you get the sum indices beneath the Sigma instead of to the right? I try to use the same syntaxes in my answer that I see in the question, but it only partially works. $\endgroup$ Jul 24, 2021 at 17:09
  • $\begingroup$ @tamayo Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Sep 25, 2021 at 16:49

6 Answers 6


Let $I(a)=\int _0^{\infty }\frac{\sinh (ax)}{x\cosh ^2x}{d}x$. Then $$I’(a)= \int _{-\infty}^{\infty }\frac{\cosh (ax)}{2\cosh ^2x}{d}x \overset{t=e^{2x}}=\int_0^\infty \frac{t^{-\frac a2}+t^{\frac a2}}{2(1+t)^2}dt =\frac{\pi a}{2\sin\frac{\pi a}2} $$ and $$\int _0^{\infty }\frac{\sinh x}{x\cosh ^2x}{d}x =\int_0^1 I’(a)da \overset{s=\tan\frac {\pi a}4} =\frac 4\pi \int_0^{1}\frac {\tan^{-1}s}{s}ds =\frac{4}{\pi}G \\ $$

  • 5
    $\begingroup$ I really need to learn how to do integration using the Leibnitz rule because this result is awesome! +1 $\endgroup$ Jul 16, 2021 at 14:48
  • 4
    $\begingroup$ (+1) Your answers best illustrates Feynman technique. $\endgroup$
    – User 1207
    Jul 17, 2021 at 18:19
  • 2
    $\begingroup$ You have not justified differentiating under the integral here (It is not difficult to do so, but it would help other readers to understand why this is legitimate for an integral from $0$ to $\infty$). And consider providing a reference on the evaluation of the integral that results in $\frac{\pi a}{2\sin(\pi a/2)}$. $\endgroup$
    – Mark Viola
    Jul 24, 2021 at 21:03
  • $\begingroup$ The integrand $\frac{t^{-1/a}+t^{1/a}}{2(1+t)^2}$ does not follow from the substitution $t=e^{2x}$ It should be $\frac{t^{a/2}+t^{-a/2}}{2(1+t)^2}$. You would have recognized this had you taken my advice to show how you arrived at $\frac{\pi a}{2\sin(\pi a/2)}$ $\endgroup$
    – Mark Viola
    Jul 25, 2021 at 14:58

I present another approach, utilising a well-known property of the Laplace Transform i.e. $$\int_{0}^{+\infty} f\left(x\right) g\left(x\right) \, dx = \int_{0}^{+\infty} \left(\mathcal{L} f\right)\left(y\right)\left(\mathcal{L}^{-1} g\right)\left( y\right) \, dy$$ Letting $f\left(x\right) = \tanh \left(x\right) \text{sech} \left(x\right)$ and $g(x) = \frac{1}{x}$: $$\left( \mathcal{L} f \right) (y) = 1 + \frac{1}{2} y \left( \psi^{(0)} \left(\frac{1+y}{4}\right)- \psi^{(0)} \left(\frac{3+y}{4}\right)\right)$$ $$\left( \mathcal{L}^{-1} g \right) \left(y \right) = 1$$ Where $\psi$ represents the polygamma function.

$$\implies I = \int_{0}^{\infty} \left(1 + \frac{1}{2} y \left( \psi^{(0)} \left(\frac{1+y}{4}\right)- \psi^{(0)} \left(\frac{3+y}{4}\right)\right)\right) \, dy$$ I will now proceed to integrate indefinitely, then take limits at the end.

$$\int 1 \, dy + \frac{1}{2} \int y \, \psi^{(0)} \left(\frac{1+y}{4}\right) \, dy - \frac{1}{2} \int y \, \psi^{(0)} \left(\frac{3+y}{4}\right) \, dy$$ We proceed with integrating by parts. $$=y + \frac{1}{2} \left( 4y \ln \Gamma \left( \frac{1+y}{4} \right) - 4 \int \ln \Gamma \left( \frac{1+y}{4} \right) \, dy \right)-\frac{1}{2} \left( 4y \ln \Gamma \left( \frac{3+y}{4} \right) - 4 \int \ln \Gamma \left( \frac{3+y}{4} \right) \, dy \right)$$ $$=y + 2 y \ln \Gamma \left( \frac{1+y}{4}\right) - 8 \psi^{(-2)} \left(\frac{1+y}{4}\right) - 2y \ln \Gamma \left( \frac{3+y}{4} \right) + 8 \psi^{(-2)} \left( \frac{3+y}{4} \right)$$ Taking limits to $\infty$ and $0$, and then subtracting gives us: $$2\ln (2\pi) + 8 \psi^{(-2)} \left( \frac{1}{4} \right)-8 \psi^{(-2)} \left( \frac{3}{4} \right)$$ $$=\boxed{\frac{4G}{\pi}}$$ (the equality was checked with WolframAlpha) as required.


We can use contour integration, but we have to be a little careful with it.

Begin by noting that the integrand is an even function, so we shall evaluate

$I=\int_{-\infty}^{+\infty}\dfrac{\sinh x}{x\cosh^2x}~dx$

and take half of this result. in this way the contour accesses infinity at both ends of the desired integration range, which facilitates using contour integration.

A semicircular contour that is often used with this doubly infinite integration range is problematic because the integrand will not be $o(1/z)$ over the entire semicircle. We violate the small-$o$ relation at the imaginary axis where there are infinitely many singularities.

We therefore design a more refined contour: a rectangle with corners at $-N, +N, +N+(2\pi N)i, -N+(2\pi N)i$. Traversing the rectangle counterclockwise and applying the Residue Theorem then gives

$I(1) + I(2) + I(3) + I(4)=2\pi i\sum _{k=0}^{2N} R[(2k+1)\pi i/2]$

where the integration path for $I(1)$ runs from $-N$ to $+N$, the path for $I(2)$ runs from $+N$ to $+N+(2\pi N)i$, and so on around the remaining sides of the rectangle for $I(3)$ and $I(4)$. Then the required integral is the limit of $I(1)$ as $N\to\infty$, which matches with the sum of residues provided that $I(2),I(3),I(4)$ go to zero in this limit.

To verify the zero limit for $I(2)$ render (limits represent the asymptotic behavior as $N\to\infty$):

$|\sinh(N+iy)|=\sqrt{\sinh^2N+\sin^2y}\to e^{N/2}$

$|\cosh(N+iy)|=\sqrt{\cosh^2N-\sin^2y}\to e^{N/2}$

Thus since the length of the side is $2\pi N$, the Triangle Inequality guarantees

$|I(2)|\le\dfrac{2\pi\sqrt{\sinh^2N+\sin^2y}}{(\cosh^2N-\sin^2y)}\to 2\pi e^{-N/2}\to0$

A similar analysis eliminates $I(4)$ as $N\to\infty$.

For $I(3)$ we use the fact that the hyperbolic functions are periodic, thus for the selected values of $y$ they match the real-argument values at $x$. Thereby

$|\dfrac{\sinh z}{z\cosh^2 z}|=|\dfrac{\sinh x}{z\cosh^2 x}|<\dfrac{1}{y\cosh x}<[1/(2\pi N)]e^{-|x|/2}$

$|I(3)|<[1/(2\pi N)]\int_{-N}^Ne^{-|x|/2}dx\to0$ (the integral is bounded as $N\to\infty$.)

So we have

$I=\lim_{N\to\infty}(I(1))=2\pi i\sum _{k=0}^\infty R[(2k+1)\pi i/2]$

and it remains to find the residues.

To find these residues $R[(2k+1)\pi i/2]$, it is convenient to define a difference parameter $\delta$ at each of these singularities. Since the singularities are second-order poles, we need to carry the Laurent series expansion to two terms.

Therefore, at each singularity $z=(2k+1)\pi i/2$, we render the following Laurent series:

$\sinh z = (-1)^ki+0\delta+O(\delta^2)$

$\dfrac{1}{\cosh^2 z} = 1/\delta^2+0/\delta+O(1)$


And upon multiplying theses:

$\dfrac{\sinh z}{z\cosh^2z}=\dfrac{2(-1)^k(2k+1)}{\pi\delta^2}\color{blue}{-\dfrac{4(-1)^ki}{(2k+1)^2\pi^2\delta}}+O(1)$

from which we read off the residues

$R[(2k+1)\pi i/2]=\dfrac{-4(-1)^ki}{(2k+1)^2\pi^2}$


$I=\int_{-\infty}^{+\infty}\dfrac{\sinh x}{x\cosh^2x}~dx=(8/\pi)\color{blue}{\sum _{k=0}^\infty \dfrac{(-1)^k}{(2k+1)^2}}$

where the blue series directly defines the Catalan constant $G$. Thus $I=8G/\pi$ and the halved value in the original problem becomes

$\int_0^\infty\dfrac{\sinh x}{x\cosh^2x}~dx=4G/\pi.$

  • 4
    $\begingroup$ I really need to learn contour integration because this is fascinating (+1) $\endgroup$
    – Henry Lee
    Jul 16, 2021 at 2:12
  • 3
    $\begingroup$ @henrylee You might like what I commented under Quanto's answer. $\endgroup$ Jul 16, 2021 at 14:49
  • 4
    $\begingroup$ Haha we need to switch places then, I've done a lot of applied calculus (on the real side) but not so much beyond complex numbers and planes $\endgroup$
    – Henry Lee
    Jul 16, 2021 at 16:27
  • $\begingroup$ So ... how can this be improved? $\endgroup$ Jul 22, 2021 at 18:19
  • 1
    $\begingroup$ @OscarLanzi I believe that $|\tan(z)|$ is bounded on $|z|=n\pi$ for all $n$ (See this for $\cot(\pi z)$). I haven't worked the details, , but if so, then one could integrate over a semicircular contour with radius $N\pi$ and let $N\to \infty$. $\endgroup$
    – Mark Viola
    Jul 24, 2021 at 22:54

For $\Re(s)>0$, the Dirichlet beta function is defined by the infinite series $$\beta(s) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{s}}. $$

The Dirichlet beta function can be extended to rest of the complex plane by its functional equation, from which we see that $\beta(s)$ has zeroes at the negative odd integers.

An integral representation of the Dirichlet beta function is $$\beta(s) = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{1+e^{-2x}} \, \mathrm dx = \frac{1}{2\Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1}}{\cosh (x)} \, \mathrm dx \, , \quad \Re(s) >0.$$

Integrating by parts, we get $$\beta (s) = \frac{1}{2 \Gamma(s+1)} \int_{0}^{\infty} \frac{x^{s} \sinh (x) }{\cosh^{2} (x)} \, \mathrm dx . \tag{1}$$

Since the Dirichlet beta function and the reciprocal gamma function are entire functions, and since the Mellin transform defines an analytic function where it converges absolutely, the identity theorem says that equation $(1)$ holds for $\Re(s) >-2$.

Now if we multiply both sides of $(1)$ by $2 \Gamma(s+1)$ and let $s \to -1$, we get $$\int_{0}^{\infty} \frac{\sinh (x)}{x \cosh^{2} (x)} \, \mathrm dx = \lim_{s \to -1} 2 \beta(s) \Gamma(s+1) = 2 \lim_{s \to -1} \beta(s) \left(\frac{1}{s+1} + \mathcal{O}(1) \right) = 2\beta'(-1).$$

But from differentiating both sides of the functional equation, we see that $$-\beta'(-1) = -\frac{2}{\pi} \beta(2) = -\frac{2}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} = -\frac{2 G}{\pi}.$$

Therefore, $$\int_{0}^{\infty} \frac{\sinh (x)}{x \cosh^{2} (x)} \, \mathrm dx = \frac{4 G}{\pi}. $$


You can integrate the function: $$ f(z):=\frac{\psi^{(0)}\left [ -iz/(2\pi)\right ]\sinh z}{\cosh^2z} $$ Along a rectangular contour with vertices $(\pm\infty,\pm\infty+2\pi i)$ Then you can get: $$ -2\pi i\int_{-\infty}^{\infty} \frac{\sinh z}{z\cosh^2z}\text{d}z = 2\pi i\sum_{z_k=\pi i/2,3\pi i/2}\text{Res}(f(z),z_k) $$ It equivalent to say: $$ \int_{0}^{\infty} \frac{\sinh z}{z\cosh^2z} \text{d}z =-\frac{1}{2} \left [ -\frac{1}{2\pi}\psi^{(1)} \left ( \frac{1}{4} \right ) + \frac{1}{2\pi}\psi^{(1)} \left ( \frac{3}{4} \right ) \right ] $$ And we have some special values(trival) $$ \begin{aligned} &\psi^{(1)} \left ( \frac{1}{4} \right )=\pi^2+8G\\ & \psi^{(1)} \left ( \frac{3}{4} \right )=\pi^2-8G \end{aligned} $$ Finally, $$ \int_{0}^{\infty} \frac{\sinh z}{z\cosh^2z} \text{d}z =\frac{1}{2} \cdot\frac{1}{2\pi} \cdot16G=\frac{4G}{\pi} $$ Completed the proof.

Here I write some other results,they're all trivial. $$\begin{aligned} &\int_{0}^{\infty} \frac{\sinh z}{z\cosh^4z} \text{d}z =\frac{2}{3\pi}G +\frac{16}{\pi^3} \beta(4)\\ &\int_{0}^{\infty} \frac{\sinh^3 z}{z\cosh^4z} \text{d}z =\frac{10}{3\pi}G -\frac{16}{\pi^3} \beta(4)\\ &\int_{0}^{1} x\ln\left ( \frac{2-x^2}{1-x^2} \right ) \arctan\left ( \frac{x}{\sqrt{2-2x^2} } \right ) \text{d}x =\frac{\pi}{2} (\sqrt{2}-1 )\ln(1+\sqrt{2} ) \end{aligned}$$


Enforcing the substitution $x\mapsto \frac12\log(x)$, we find that

$$\begin{align} \int_{0}^\infty \frac{\sinh(x)}{x\cosh^2(x)}\,dx&=\int_0^\infty \color{blue}{\left(\frac{x-1}{\log(x)}\right)}\frac1{x^{1/2}(x+1)^2}\,dx\\\\ &=\int_0^\infty \color{blue}{\left(\int_0^1 x^t\,dt\right)} \frac{1}{x^{1/2}(x+1)^2}\,dx\\\\ &=\int_0^1\int_0^\infty \frac{x^{t-1/2}}{(x+1)^2}\,dx\,dt\tag1\\\\ &=\int_0^1\frac{\pi(t-1/2)}{\sin(\pi (t-1/2))}\,dt\tag2\\\\ &=\frac1\pi\int_{-\pi/2}^{\pi/2} \frac{t}{\sin( t)}\,dt\tag3\\\\ &=\frac{4G}{\pi}\tag4 \end{align}$$

as was to be shown!


In arriving at $(1)$, we applied the Fubini-Tonelli Theorem.

To go from $(1)$ to $(2)$, we evaluated the inner integral, $\int_0^\infty \frac{x^{t-1/2}}{(1+t)^2}\,dt$. To do so, one can use complex analysis and integrate $\frac{z^{t-1/2}}{(1+z)^2}$ over the classical keyhole contour and apply the reside theorem. Alternatively one can use real analysis only. To proceed accordingly, we enforce the substitution $x\mapsto \tan^2(x)$ to find that

$$\begin{align} \int_0^\infty \frac{x^{t-1/2}}{(1+t)^2}\,dt&=2\int_0^{\pi/2}\sin^{2t}(x)\cos^{2-2t}(x)\,dx\\\\ &=B(1/2+t,3/2-t)\tag{N1} \end{align}$$

Then, applying the identity $B(1+x,y)=\frac{x}{x+y}B(x,y)$ (with $x=t-1/2$) and the reflection formula for the Gamma function to $(N1)$ reveals

$$\begin{align} \int_0^\infty \frac{x^{t-1/2}}{(1+t)^2}\,dt&=(t-1/2)B(t-1/2,1-(t-1/2))\\\\ &=(t-1/2)\Gamma(t-1/2)\Gamma(1-(t-1/2))\\\\ &\frac{\pi(t-1/2)}{\sin(\pi(t-1/2))} \end{align}$$

In going from $(2)$ to $(3)$, we enforced the substitution $\pi(t-1/2)\mapsto t$.

Finally, in arriving at $(4)$, we relied on the well-known representation for Catalan's Constant $G$, as given HERE.

  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Sep 25, 2021 at 16:49

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.