How can I prove that: $$\int_{0}^{\infty}\ln(x)\,\operatorname{sech}(x)\,dx=\int_{0}^{\infty}\frac{2\ln(x)}{e^x+e^{-x}}\,dx\\=\pi\ln2+\frac{3}{2}\pi\ln(\pi)-2\pi\ln\!\Gamma(1/4)\approx-0.5208856126\!\dots$$ I haven't really tried much of anything worth mentioning; I've had basically no experience with $\ln\!\Gamma$.


$$ \begin{align} \int_0^\infty\frac{2\log(x)}{e^x+e^{-x}}\,\mathrm{d}x &=\frac{\partial}{\partial t}\int_0^\infty\frac{2x^t}{e^x+e^{-x}}\,\mathrm{d}x \end{align} $$ $$ \begin{align} \int_0^\infty\frac{2x^t}{e^x+e^{-x}}\,\mathrm{d}x &=\int_0^\infty2x^te^{-x}\left(1-e^{-2x}+e^{4x}-\dots\right)\,\mathrm{d}x\\ &=2\Gamma(t+1)\left(1-\frac1{3^{t+1}}+\frac1{5^{t+1}}-\dots\right)\\[6pt] &=2\Gamma(t+1)\beta(t+1) \end{align} $$ where $\beta(s)$ is the Dirichlet beta function.

Now we need to compute $\frac{\mathrm{d}}{\mathrm{d}x}\Gamma(x)\beta(x)$ at $x=1$.

$\Gamma(1)=1$ and $\Gamma'(1)=-\gamma$ as shown in this answer.

$\beta(1)=\frac\pi4$ using Gregory's Series. Finally, we need to compute $$ \beta'(1)=\sum_{k=0}^\infty(-1)^k\frac{\log(2k+3)}{2k+3} $$ which I am attempting to do in a manner similar to this answer. I will try to finish this in a bit.

  • $\begingroup$ Keep in mind that the answer involves $\ln\!\Gamma(\frac{1}{4})$. This might make that sum very hard to solve. $\endgroup$ – Akiva Weinberger Aug 27 '14 at 2:00

EDIT: I'm in the process of trying to make my answer more rigorous.

Let's start with the integral $$I(a) =\int_{0}^{\infty} \frac{\ln (a^{2}+x^{2})}{\cosh x} \, dx, \quad a>0.$$

Differentiating with respect to $a$, we get $$ \begin{align} I'(a) &= \int_{0}^{\infty} \frac{2a}{(a^{2}+x^{2})\cosh x} \, dx \\ &= \int_{0}^{\infty} \frac{2}{(1+u^{2})\cosh (au)} \, du. \end{align}$$

For $0 < b < a <c < \infty$, the function $\frac{2a}{a^{2}+x^{2}}$ is dominated on $(0, \infty)$ by the integrable function $\frac{2c}{(b^{2}+x^{2})\cosh x}$.

This permits to differentiate under the integral sign for all $a>0$.

From the answers to this question, we know that $$\int_{0}^{\infty} \frac{2}{(1+u^{2}) \cosh (au)} \, du= \psi\left(\frac{3}{4}+ \frac{a}{2 \pi} \right) - \psi \left(\frac{1}{4} + \frac{a}{2 \pi} \right). $$

Therefore, $$I(a) = 2 \pi \left[\ln \Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right) - \ln \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right) \right] + C.$$

To find the constant of integration, notice that $$ \begin{align} &\int_{0}^{\infty} \frac{\ln(a^{2}+u^{2})}{\cosh u}\, du - \int_{0}^{\infty}\frac{\ln(a^{2})}{\cosh u} \, du \\ &= \int_{0}^{\infty} \frac{\ln(1+ \frac{u^{2}}{a^{2}})}{\cosh u} \, du \\ &= 2 \pi \left[\ln \Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right) - \ln \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right) \right]+C - \ln(a^{2})\frac{\pi}{2} \\ &= 2 \pi \ln \left[\frac{\Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right)}{\sqrt{a} \ \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right)} \right] +C . \end{align}$$

Letting $ a \to \infty$, we get $$ 0 = 2 \pi \lim_{a \to \infty} \log \left[\frac{\Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right)}{\sqrt{a} \ \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right)} \right] + C.$$

From Stirling's approximation formula for the gamma function, we know that $$ \frac{\Gamma(x+\frac{1}{2})}{\Gamma(x)} \sim \frac{\sqrt{\frac{2\pi}{x+1/2}} \left(\frac{x+1/2}{e} \right)^{x+1/2}}{\sqrt{\frac{2\pi}{x}} \left(\frac{x}{e} \right)^{x}} =\sqrt{x} \left(1+ \frac{1}{2x} \right)^{x} e^{-1/2} \sim \sqrt{x} .$$

Therefore, $$ \lim_{a \to \infty} \ln \left[\frac{\Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right)}{\sqrt{a} \ \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right)} \right] = \lim_{a \to \infty} \ln \ \frac{\sqrt{\frac{1}{4} + \frac{a}{2 \pi}}}{\sqrt{a}} = \lim_{a \to \infty} \ln \sqrt{\frac{1}{4a}+\frac{1}{2 \pi}} = - \frac{\ln (2 \pi)}{2}, $$

which implies $$C= \pi \ln (2 \pi).$$

So we have $$\int_{0}^{\infty} \frac{\ln(a^{2}+u^{2})}{\cosh u} \, du = 2 \pi \left[\ln \Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right) - \ln \Gamma \left(\frac{1}{4} + \frac{a}{2\pi} \right) \right] + \pi \ln (2 \pi), \quad a>0.$$

For $0 \le a < 1$, $\frac{\left|\ln(a^{2}+u^{2})\right|}{\cosh u} $ is dominated on $(0, \sqrt{1-a^{2}})$ by the integrable function $\frac{2\left|\ln u\right|}{\cosh u}$.

And on $(\sqrt{1-a^{2}}, \infty)$, $\left|\frac{\ln(a^{2}+u^{2})}{\cosh u} \right|$ is dominated by the integrable function $\frac{2u}{\cosh u}$.

Therefore, $$\lim_{a \to 0^{+}} \int_{0}^{\infty} \frac{\ln(a^{2}+u^{2})}{\cosh u}\, du = 2 \int_{0}^{\infty} \frac{\ln u}{\cosh u} \, du,$$ and the result follows after applying the reflection formula for the gamma function.

  • $\begingroup$ Wow. This approach is really interesting... I suppose one way to becoming a great integrator is to memorize a lot of results. (By the way, since some partial answers on this thread note that this integral is equal to $-\dfrac{\pi\gamma}4+\beta'(1)$, this answer can be used to find $\beta'(1)$.) $\endgroup$ – Akiva Weinberger Aug 31 '14 at 0:26

Just a partial answer for now.

We have to compute: $$ I = \int_{0}^{+\infty}\frac{2\log x}{e^{x}+e^{-x}}\,dx = 2\frac{\partial}{\partial\alpha}\left.\left(\int_{0}^{+\infty}\frac{z^\alpha}{e^z+e^{-z}}\,dz\,\right)\right|_{\alpha=0^+}$$ Since: $$\int_{0}^{+\infty}\frac{z^\alpha}{e^z+e^{-z}}\,dz=\int_{1}^{+\infty}\frac{\log^\alpha t}{t^2+1}\,dt=\int_{0}^{1}\frac{(-\log t)^\alpha}{1+t^2}\,dt$$ and $$\int_{0}^{1}(-\log t)^\alpha\, t^{2n}\,dt = \frac{\Gamma(\alpha+1)}{(2n+1)^{\alpha+1}},$$ it follows that: $$\int_{0}^{+\infty}\frac{z^\alpha}{e^z+e^{-z}}\,dz = \Gamma(\alpha+1)\cdot\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^{\alpha+1}}=\Gamma(\alpha+1)\cdot\beta(\alpha+1).\tag{1}$$

Now we "just" need to differentiate the RHS of $(1)$ with respect to $\alpha$ and take the limit as $\alpha\to 0^+$. With the aid of Mathematica (see here for the relation between the Dirichlet beta function and the Hurwitz zeta function) for computing the Taylor series of the $\Gamma$ and $\beta$ functions I got:

$$I=\frac{1}{2} \left(-\pi (\gamma+\log 4)-\text{StieltjesGamma}\left[1,\frac{1}{4}\right]+\text{StieltjesGamma}\left[1,\frac{3}{4}\right]\right).$$

  • $\begingroup$ Are those generalized Stieltjes constants? I had to look them up... I hadn't seen them before. I suppose they came out of replacing the beta with the Hurwitz zetas and differentiating them. $\endgroup$ – Akiva Weinberger Aug 27 '14 at 1:02
  • $\begingroup$ A notation note: The generalized Stieltjes constants can be written as $\gamma_{1}(\frac{1}{4})$ and $\gamma_{1}(\frac{3}{4})$, I think. $\endgroup$ – Akiva Weinberger Aug 27 '14 at 1:06

If several change of variables are made, it can be shown that this integral is equivalent to the famous Vardi integral, $\displaystyle \int_{\pi/4}^{\pi/2}\ln(\ln(\tan(x)))dx=\frac{\pi}{2}\ln\left(\sqrt{2\pi}\frac{\Gamma(3/4)}{\Gamma(1/4)}\right)$, which has been done on the site. Well, it's twice the Vardi integral.

Vardi's Integral: $\int_{\pi/4}^{\pi/2} \ln (\ln(\tan x))dx $

Begin with $\int_{\pi/4}^{\pi/2}\ln(\ln(\tan(x)))dx$ and let $1/t=\tan(x)$.

It then becomes:


Let $u=1/t$ and it becomes:


Now, let $w=\ln(u)$ and it becomes:


  • $\begingroup$ Following the links, I must say that the best evaluation I've found is here. (As noted on this question, all that's needed is an evaluation of $\beta'(1)$, which - after using estimates and Sterling approximations in an amazing way to get $\beta'(0)$ - follows from the beta reflection formula.) $\endgroup$ – Akiva Weinberger Aug 29 '14 at 12:08

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{\infty}\ln\pars{x}\sech\pars{x}\,\dd x =\pi\ln\pars{2} + {3 \over 2}\,\pi\ln\pars{\pi} -2\pi\ln\pars{\Gamma\pars{1 \over 4}}:\ {\large ?}}$

\begin{align}&\color{#c00000}{\int_{0}^{\infty}\ln\pars{x}\sech\pars{x}\,\dd x} =\ \overbrace{2\int_{0}^{\infty}{\expo{x}\ln\pars{x} \over \expo{2x} + 1}\,\dd x} ^{\ds{\mbox{Set}\ \expo{x} \equiv t\ \imp\ x = \ln\pars{t}}}\ =\ \overbrace{2\int_{1}^{\infty}{t\ln\pars{\ln\pars{t}} \over t^{2} + 1}\,{\dd t \over t}} ^{\ds{t\ \mapsto\ {1 \over t}}} \\[3mm]&=2\int_{1}^{0}{\ln\pars{\ln\pars{1/t}} \over 1/t^{2} + 1}\, \pars{-\,{\dd t \over t^{2}}} \end{align}

Then, $$\begin{array}{|c|}\hline\\ \quad\color{#c00000}{\int_{0}^{\infty}\ln\pars{x}\sech\pars{x}\,\dd x} =2\int_{0}^{1}{\ln\pars{\ln\pars{1/t}} \over 1 + t^{2}}\,\dd t\quad \\ \\ \hline \end{array} $$

I already evaluated this integral. The result is: $$ \color{#c00000}{\int_{0}^{\infty}\ln\pars{x}\sech\pars{x}\,\dd x} =\pi\ln\pars{\Gamma^{2}\pars{3/4} \over \root{\pi}} $$

Also, $$ \Gamma\pars{3 \over 4} = {\pi \over \Gamma\pars{1/4}\sin\pars{\pi/4}} ={\root{2}\pi \over \Gamma\pars{1/4}} $$ such that

\begin{align}&\color{#66f}{\large\int_{0}^{\infty}\ln\pars{x}\sech\pars{x}\,\dd x} =\pi\ln\pars{{2\pi^{2} \over \Gamma^{2}\pars{1/4}}\,{1 \over \root{\pi}}} =-\pi\ln\pars{\Gamma^{2}\pars{1/4} \over 2\pi^{3/2}} \\[3mm]&=\color{#66f}{\large\pi\ln\pars{2} + {3 \over 2}\,\pi\ln\pars{\pi} -2\pi\ln\pars{\Gamma\pars{1 \over 4}}} \end{align}

  • $\begingroup$ By the way, in the link you gave, you began your solution by immediately putting it in this form:$$\int_{0}^{\infty}{\ln\pars{t} \over 1 + \expo{2t}}\,\expo{t}\dd t$$which is how the problem was posed on this thread. $\endgroup$ – Akiva Weinberger Sep 4 '14 at 22:45
  • $\begingroup$ Thank you for taking the time to answer the question!\begin{align}\\ \end{align}I must say that the main part of your solution is the use of the obscure 1846 Carl Malmsten identity. Up until then, it is essentially the same as Jack D'Azzurio's partial solution. $\endgroup$ – Akiva Weinberger Sep 4 '14 at 22:46
  • $\begingroup$ @columbus8myhw That's true but I did't in march. I wrote this one in this way just to show the equivalence between both of my answers. We don't need "to reinvent the wheel". I appreciated it was useful for you. Thanks. $\endgroup$ – Felix Marin Sep 5 '14 at 1:54

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