Integral of $\ln(x)\operatorname{sech}(x)$

Question

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 $\ln\!\Gamma$.

Answer 1

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:

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

Answer 2

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

Answer 3

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:

$$\int_{0}^{1}\frac{\ln(\ln(1/t))}{t^{2}+1}dt$$

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

$$\int_{1}^{\infty}\frac{\ln(\ln(u)))}{u^{2}+1}du$$

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

$$1/2\int_{0}^{\infty}\frac{\ln(w)}{\cosh(w)}dw$$

Answer 4

You can get the value of the integral you're interested in from the integral $$I(a) =\int_{0}^{\infty} \frac{\ln (1+\frac{x^{2}}{a^{2}})}{\cosh x} \, dx, \quad a>0.$$

Notice that $\lim_{a \to \infty} I(a) = 0$.

Differentiating under the integral sign, we get $$ \begin{align} I'(a) &= \int_{0}^{\infty} \frac{2a}{(a^{2}+x^{2})\cosh x} \, dx - \frac{2}{a} \int_{0}^{\infty} \frac{dx}{\cosh x} \, dx \\ &= \int_{0}^{\infty} \frac{2}{(1+u^{2})\cosh (au)} \, du - \frac{\pi}{a}. \end{align}$$

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

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

Using Stirling's approximation formula for the gamma function, we see 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 $$I(a) = 2 \pi \left[\ln \Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right) - \ln \Gamma \left(\frac{1}{4} + \frac{a}{2\pi} \right) \right] - \pi \ln (a) + \pi \ln (2 \pi).$$

But since $$ \int_{0}^{\infty} \frac{\ln (a^{2}+x^{2})}{\cosh x} \, dx = I(a) + \ln(a^{2}) \int_{0}^{\infty} \frac{dx}{\cosh x} \, dx = I(a) + \pi \ln (a),$$ it follows that $$2 \int_{0}^{\infty} \frac{\ln x}{\cosh x} \, dx = \lim_{a \to 0^{+}} 2 \pi \left[\ln \Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right) - \ln \Gamma \left(\frac{1}{4} + \frac{a}{2\pi} \right) \right] + \pi \ln (2 \pi). $$

The final step is to apply the reflection formula for the gamma function.

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If we had started with the integral $\int_{0}^{\infty} \frac{\ln(a^{2}+x^{2})}{\cosh x} \, dx $, we wouldn't have had a known initial condition.

Answer 5

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

\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}

Answer 6

By expanding $\operatorname {sech x}$, we have $$ \begin{aligned} I& =\int_0 \ln x \operatorname{sech} x d x\\&=2 \int_0^{\infty} \frac{\ln x}{e^x+e^{-x}} d \\ & =2 \int_0^{\infty} \frac{e^{-x} \ln x}{1+e^{-2 x}} \\ & =2 \sum_{n=0}^{\infty}(-1)^n \underbrace{\int_0^{\infty} e^{-(2 n+1) x} \ln x d x}_{J’_ n(1)} \end{aligned} $$ where $$ J_n(a)=\int_0^{\infty} x^{a-1} e^{-(2 n+1) x} d x=\frac{\Gamma(a)}{(2 n+1)^a} $$ and $$ J_n^{\prime}(1)=-\frac{\ln (2 n+1)+\gamma}{2 n+1} $$ Hence

$$ I=-2 \sum_{n=0}^{\infty} \frac{(-1)^n \ln (2 n+1)}{2 n+1}-2 \gamma \sum_{n=0}^{\infty} \frac{(-1)^n}{2 n+1} $$

By the post, we have $$ \sum_{n=0}^{\infty} \frac{(-1)^n \ln (2 n+1)}{2 n+1}=\frac{\pi}{4}\left(4 \ln \Gamma\left(\frac{1}{4}\right)-\gamma-2 \ln 2-3 \ln \pi\right) $$ Hence we can conclude that $$ \begin{aligned} & I=-2 \sum_{n=0}^{\infty} \frac{(-1)^n \ln (2 n+1)}{2 n+1}-2 \gamma \sum_{n=0}^{\infty} \frac{(-1)^n}{2 n+1} \\ & =-2 \cdot \frac{\pi}{4}\left(4 \ln \Gamma\left(\frac{1}{4}\right)-\gamma-2 \ln 2-3 \ln \pi\right)-2 \gamma \cdot \frac{\pi}{4} \\ & =\pi \ln 2+\frac{3 \pi}{2} \ln \pi-2 \pi \ln \Gamma\left(\frac{1}{4}\right)\blacksquare \end{aligned} $$