Mathematica obtains

$$\int_0^{\infty}\frac{x}{\sinh(\pi x)}\ln(1+x^2) \ \mathrm{d}x=-\frac{3}{4}+\frac{\ln 2}{6}+9\ln(A)-\frac{\ln \pi}{2}$$ where $A$ is the Glaisher-Kinkelin constant.

A numerical approximation of the integral strongly suggests that this is incorrect. What is the correct value?

  • $\begingroup$ I'm curious, how do you know it's incorrect? $\endgroup$ – MasterOfBinary Nov 10 '13 at 23:33
  • $\begingroup$ Try to evaluate the integral numerically using Maple or Mathematics. $\endgroup$ – Mhenni Benghorbal Nov 10 '13 at 23:38
  • $\begingroup$ This expression seems to be around 1.03 and the value of the integral around 0.08, not even close! $\endgroup$ – Bennett Gardiner Nov 11 '13 at 0:02

Cody's answer gave me the idea to look at $\displaystyle \int_{0}^{\infty} \frac{\sin ( s \arctan t)}{(1+t^{2})^{s/2} \sinh \pi t} \, dt $.

First add the restriction $ \text{Re}(s) >1$.


$$ \begin{align} &\int_{0}^{\infty} \frac{\sin ( s \arctan t)}{(1+t^{2})^{s/2} \sinh \pi t} \, dt \\ &= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin(s \arctan t)}{(1+t^{2})^{s/2} \sinh \pi t} \, dt \\&= \frac{1}{2} \int_{-\infty}^{\infty} \text{Im} \ \frac{1}{(1-it)^{s} \sinh \pi t} \, dt \\ &= \frac{1}{2} \, \text{Im} \, \text{PV} \int_{-\infty}^{\infty} \frac{1}{(1-it)^{s} \sinh \pi t} \, dt \\ &= \frac{1}{2} \, \text{Im} \left(\pi i \ \text{Res}\left[\frac{1}{(1-iz)^{s} \sinh \pi z},0\right] + 2 \pi i \sum_{n=1}^{\infty} \text{Res} \left[\frac{1}{(1-iz)^{s} \sinh \pi z},in \right] \right) \\ &= \frac{1}{2} \text{Im} \left(\pi i \left(\frac{1}{\pi} \right) + 2 \pi i \sum_{n=1}^{\infty} \frac{(-1)^{n}}{\pi (1+n)^{s}} \right) \\ &= \frac{1}{2} \, \text{Im} \Big(i + 2i \big( (1-2^{1-s}) \zeta(s)-1\big) \Big) \\& = (1-2^{1-s})\zeta(s) - \frac{1}{2}. \end{align}$$

By analytic continuation, the result is valid for all complex values of $s$.

Differentiating under the integral sign and letting $s=-1$ we get

$$ \begin{align} \frac{1}{2} \int_{0}^{\infty} \frac{t \log(1+t^{2})}{\sinh \pi t} \, dt + \int_{0}^{\infty} \frac{\arctan t}{\sinh \pi t} \, dt &= 2^{1-s} \log 2 \ \zeta(s) + (1-2^{1-s}) \zeta'(s) \Bigg|_{s=-1} \\ &= 4 \log (2) \left(-\frac{1}{12} \right) - 3 \zeta'(-1) \\ &= - \frac{\log 2}{3} - 3 \zeta'(-1) . \end{align}$$

So we need to evaluate $ \displaystyle \int_{0}^{\infty} \frac{\arctan t}{\sinh \pi t} \, dt $.

Let $ \displaystyle I(z) = \int_{0}^{\infty} \frac{\arctan \frac{x}{z}}{\sinh \pi x} \, dx $.


$$ \begin{align} I(z) &= \int_{0}^{\infty} \frac{1}{\sinh \pi x} \int_{0}^{\infty} \frac{\sin tx}{t} e^{-zt} \, dt \, dx \\ &= \int_{0}^{\infty} \frac{e^{-zt}}{t} \int_{0}^{\infty} \frac{\sin t x}{\sinh \pi x} \, dx \, dt \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{e^{-zt}}{t} \tanh \left( \frac{t}{2} \right) \, dt . \end{align} $$

Now differentiate under the integral sign.

$$ \begin{align} I'(z) &= - \frac{1}{2} \int_{0}^{\infty} \tanh \left(\frac{t}{2} \right) e^{-zt} \, dt \\ &= - \frac{1}{2} \int_{0}^{\infty} \frac{1-e^{-t}}{1+e^{-t}} e^{-zt} \, dt \\ &= - \frac{1}{2} \int_{0}^{\infty} \left( -1 + 2 \sum_{n=0}^{\infty} (-1)^{n} e^{-tn}\right) e^{-zt} \, dt \\ &= \frac{1}{2z} - \sum_{n=0}^{\infty} (-1)^{n} \frac{1}{z+n} \\ &= \frac{1}{2z} - \frac{1}{2} \psi \left(\frac{z+1}{2} \right) + \frac{1}{2} \psi \left( \frac{z}{2} \right) \tag{1} \end{align}$$

Integrating back and using Stirling's formula to determine the value of the constant of integration, we find

$$ I(z) = \frac{\ln z}{2} - \log \Gamma \left(\frac{z+1}{2} \right) + \log \Gamma \left(\frac{z}{2} \right) - \frac{\log 2}{2} .$$


$$ \int_{0}^{\infty} \frac{\arctan x}{\sinh \pi x} \, dx = I(1) = \frac{\log \pi}{2} - \frac{\log 2}{2} .$$

Combining this result with the first result we have

$$ \begin{align} \int_{0}^{\infty} \frac{x \log(1+x^{2})}{\sinh \pi x} \, dx &= 2 \left(-\frac{\log 2}{3} - 3 \zeta'(-1) - \frac{\log \pi}{2} + \frac{\log 2}{2} \right) \\&=\frac{\log 2}{3} - \log \pi - 6 \zeta'(-1) \\ &= \frac{\log 2}{3} - \log \pi - \frac{1}{2} + 6 \log A \\ &\approx 0.0788460364 . \end{align}$$

$ $

$(1)$ http://mathworld.wolfram.com/DigammaFunction.html (5)

  • $\begingroup$ There ya' go, buddy. :):) Very nice indeed. $\endgroup$ – Cody Nov 13 '13 at 20:45

I'll conjecture that the correct value is : $$I:=\frac{\ln\;2}3-\ln\, \pi-6\,\zeta'(-1)$$ ($\zeta'(-1)=\frac 1{12}-\ln\,A\;$ with $A$ is the Glaisher-Kinkelin constant)

Let's obtain this solution using the Abel-Plana formula (a little as proposed by Cody) :

$$\sum_{n=0}^\infty (-1)^n\,f(n)=\frac{f(0)}2+i\int_0^\infty\frac{f(it)-f(-it)}{2\,\sinh(\pi\,t)}dt$$

Setting $\,f(z):=z\,\ln(1+z)\,$ returns quickly the equivalent sum $\,\displaystyle 2\sum_{k=1}^\infty (-1)^k\,(k-1)\,\ln(k)$. Unfortunately this sum is divergent ('regularization' returns the correct value) so let's try :

$\,\displaystyle f(z):=z\;\ln\left(1+\frac 1z\right)-1\,$ and get (observing that $\,\lim_{z\to 0^+} f(z)=-1$) :

\begin{align} S&:=-1+\sum_{n=1}^\infty (-1)^n\,\left(n\,\ln\left(1+\frac 1n\right)-1\right)\\ &=-\frac 12+i\int_0^\infty\frac{it\ln\left(1+\frac 1{it}\right)+it\ln\left(1+\frac 1{-it}\right)}{2\,\sinh(\pi\,t)}dt\\ &=-\frac 12-\frac 12\int_0^\infty\frac{t\ln\left(1+\frac 1{t^2}\right)}{\sinh(\pi\,t)}dt\\ &=-\frac 12-\frac 12\int_0^\infty\frac{t\ln(1+t^2)}{\sinh(\pi\,t)}dt+\int_0^\infty\frac{t\ln(t)}{\sinh(\pi\,t)}dt\\ &=-\frac 12(1+I)+J \end{align}

The integral $\,\displaystyle J:=\int_0^\infty\frac{t\ln(t)}{\sinh(\pi\,t)}dt\;$ may be evaluated using the integral representation of the $\eta$ function (see $(*)$) $\quad\displaystyle \frac{\eta(s)}{\sin(s\pi/2)}=-\int_0^\infty \frac{t^{-s}}{\sinh(\pi\,t)}dt\,$
by computing the derivative under the integral sign at $s=-1$ using $t^{-s}=e^{-s\ln(t)}\,$ to get $$J=\dfrac{\ln(2)}3+3\,\zeta'(-1)$$

while $I$ will be given by : $$I=-1-2(S-J)$$ with \begin{align} S&=-1+\sum_{n=1}^\infty (-1)^n\,\left(n\,\ln\left(1+\frac 1n\right)-1\right)\\ &=-1+\sum_{n=1}^\infty (-1)^n\,\left(-1+n\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k\;n^k}\right)\\ &=-1+\sum_{n=1}^\infty (-1)^n\,\sum_{k=2}^\infty\frac{(-1)^{k-1}}{k\;n^{k-1}}\\ &=-1+\sum_{m=1}^\infty\frac{(-1)^m}{m+1}\sum_{n=1}^\infty \frac{(-1)^n}{n^m}\quad\text{for}\;m:=k-1\\ &=\frac{\ln(2)}2-1-\sum_{m=2}^\infty\frac{(-1)^m}{m+1}\left(1-2^{1-m}\right)\zeta(m)\quad(*)\\ &=\frac{\ln(2)}2-1+\sum_{m=2}^\infty\frac{\zeta(m)}{m+1}\left((-1)^{m+1}-4\left(\frac {-1}2\right)^{m+1}\right)\\ \end{align} $(*)$ : from the Dirichlet eta definition and property $\,\displaystyle \eta(m)=-\sum_{n=1}^\infty \frac{(-1)^n}{n^m}=\left(1-2^{1-m}\right)\zeta(m)$

But for $\psi$ the digamma function we have $\,\displaystyle\displaystyle \psi(1+x)=-\gamma-\sum_{m=2}^\infty \zeta(m)\;(-x)^{m-1}$
and $\,\displaystyle\int (\psi(1+x)+\gamma)x\,dx=\int \sum_{m=2}^\infty \zeta(m)\;(-x)^{m} dx=-\sum_{m=2}^\infty \frac{\zeta(m)}{m+1}\;(-x)^{m+1}$

If we observe that $\,\displaystyle\int \psi(x)\,x\,dx=[\ln(\Gamma(x))\,x]-\int\ln(\Gamma(x))\,dx$ we may use the formulae obtained by Glaisher, Gosper and Adamchik for the 'Negapolygammas' $(11)$ with $\zeta(s,q)$ the Hurwitz zeta function : $$K(q):=\int_0^q\ln(\Gamma(x))\,dx=\frac{(1-q)q}2+\frac q2\ln(2\pi)-\zeta'(-1)+\zeta'(-1,q)$$ with the special values $\,\displaystyle K(1)=\frac 12\ln(2\pi)\,$ and $\,\displaystyle K\left(\frac 12\right)=\frac 18+\frac 5{24}\ln(2)+\frac 14\ln(\pi)-\frac 32\zeta'(-1)$

obtaining (since $\,\psi(1+x)=\psi(x)+\frac 1x\,$) : $$\sum_{m=2}^\infty \frac{\zeta(m)}{m+1}\;(-x)^{m+1}=-x-\frac{\gamma}2x^2-\int \psi(x)\,dx=x-\frac{\gamma}2x^2+K(x)-\ln(\Gamma(x))\,x$$ getting (very laboriously I'll admit...) : \begin{align} S&=\frac{\ln(2)}2-1+\sum_{m=2}^\infty\frac{\zeta(m)}{m+1}\left((-1)^{m+1}-4\left(\frac {-1}2\right)^{m+1}\right)\\ &=\frac{\ln(2)}2-1+1+\frac 12\ln(2\pi)-4\left(\frac 18+\frac 5{24}\ln(2)+\frac 14\ln(\pi)-\frac 32\zeta'(-1)-\frac 12\ln\sqrt{\pi}\right)\\ &=-\frac 12+\frac 16\ln(2)+\frac {\ln(\pi)}2+6\zeta'(-1) \end{align} And finally $$I=-1-2\left(-\frac 12+\frac 16\ln(2)+\frac {\ln(\pi)}2+6\zeta'(-1)-\dfrac{\ln(2)}3-3\,\zeta'(-1)\right)\\ \boxed{\displaystyle I=\frac {\ln(2)}3-\ln(\pi)-6\,\zeta'(-1)}$$

Of course much more direct derivations exist but an interesting journey anyway!

  • 1
    $\begingroup$ could maybe be a comment? $\endgroup$ – Bennett Gardiner Nov 11 '13 at 0:08
  • 1
    $\begingroup$ @Bennett: Isn't the correct value the question? $\endgroup$ – Raymond Manzoni Nov 11 '13 at 0:09
  • $\begingroup$ I suppose. My apologies, I'm looking for a proof as well for my curiosity. $\endgroup$ – Bennett Gardiner Nov 11 '13 at 0:10
  • $\begingroup$ @Bennett: Of course a proof would be a much better answer ! $\endgroup$ – Raymond Manzoni Nov 11 '13 at 0:12
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    $\begingroup$ (+1) nice answer which answers the question in consideration. I believe that you put some effort to come up with the right answer. So, you deserve a credit for doing that. $\endgroup$ – Mhenni Benghorbal Nov 11 '13 at 4:01

If I may add something that may help.

It would appear this integral can be tied to the integrals of log-Gamma or Barnes G in some form or another.

If we write the $\displaystyle csch(\pi x)$ in terms of its exponential, it is then very similar to the known result that can be found by using Hermite's formula for the Hurwitz zeta.

Hermite's formula for $\zeta(s,1)$ is:

$\displaystyle \zeta(s,1)=1/2+\frac{1}{s-1}+2\int_{0}^{\infty}\frac{\sin(s\cdot \tan^{-1}(x))}{(x^{2}+1)^{s/2}(e^{2\pi x}-1)}dx$.

Which can be differentiated w.r.t s, leading to:

$\displaystyle \int_{0}^{\infty}\frac{x\log(x^2+1)}{e^{2\pi x}-1}dx=1/2\log(2\pi)-2/3-\log(A)$.

The integral in question is also equal to the series:

$\displaystyle \sum_{n=2}^{\infty}(-1)^{n+1}\log\left(1-\frac{1}{n}\right)+\sum_{n=2}^{\infty}(-1)^{n+1}n\cdot \log\left(1+\frac{1}{n-1}\right)$

I will look into it later when I find more time. But, if anyone can use these to derive the solution, please feel free.

  • $\begingroup$ Thanks @Cody ! I had found this one too as a consequence of the Abel-Plana formula but a $e^{\pi\,x}\;$ factor is still missing in the integral. Perhaps that for another $f$ function... (the derivation doesn't look easy). $\endgroup$ – Raymond Manzoni Nov 13 '13 at 16:58

Noting the tabulated cosine transforms of $\ln(1+1/x^2)$ and $x/\sinh(\pi x)$, by the Parseval relation for this transform $$\int_0^{\infty}\frac{x \ln(1+1/x^2)}{\sinh(\pi x)}dx=\int_0^{\infty}\frac{dy}{y}\frac{1+\sinh y -\cosh y}{1+\cosh y}$$ which has the known value $12\ln A-\ln \pi -1-\frac{1}{3}\ln 2$. It is also known that $$\int_0^{\infty}\frac{x \ln x^2}{\sinh(\pi x)}dx=\frac{1}{2}+\frac{2}{3}\ln 2-6\ln A =.$$ By combining these results one gets the desired value. This points to a disturbing bug in Mathematica.


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