The closed form of $\int_0^{\infty} \frac{\log(\cosh(x))}{x} e^{-x} \ dx$

Question

An integral I discussed last days in a chat, and it looks like a hard nut since after some manipulations of the initial form we reach an integral where the integrand is expressed in
terms of digamma function that can be further reduced to some very resistent series.
I conjecture that the integral has a closed form expressed in terms of G-Barnes function
(and newer constants?) Well, I might be wrong, but here we have many integration gurus
that could probably answer that.

$$\int_0^{\infty} \frac{\log(\cosh(x))}{x} e^{-x} \ dx$$

Answer 1

The integral to evaluate is \begin{align} I(1) = \int_{0}^{\infty} e^{-x} \, \ln(\cosh(x)) \, \frac{dx}{x}. \end{align} The evaluation is as follows. Consider \begin{align} \int_{1}^{\infty} e^{-\alpha x} d\alpha = \left[ - \frac{1}{x} e^{-\alpha x} \right]_{1}^{\infty} = \frac{e^{-x}}{x}. \end{align} Let \begin{align} I(\alpha) = \int_{0}^{\infty} e^{-\alpha x} \, \ln(\cosh(x)) \, \frac{dx}{x}, \end{align} where $I(1)$ is the integral to evaluate and $I(\infty) = 0$, for which differentiation with respect to $\alpha$ leads to \begin{align} \partial_{\alpha} I(\alpha) = - \int_{0}^{\infty} e^{- \alpha x} \ln(\cosh(x)) dx. \end{align} Using integration by parts it is seen that \begin{align} \partial_{\alpha} I(\alpha) &= \left[ \frac{1}{\alpha} \, e^{- \alpha x} \ln(\cosh(x)) \right]_{0}^{\infty} - \frac{1}{\alpha} \int_{0}^{\infty} e^{- \alpha x} \tanh(x) dx \\ &= - \frac{1}{\alpha} \int_{0}^{\infty} e^{- \alpha x} \tanh(x) dx \\ &= - \frac{1}{2 \alpha} \left[ \beta\left( \frac{\alpha}{2} \right) - \beta\left( \frac{\alpha + 2}{2} \right) \right] \\ &= - \frac{1}{4\alpha} \left[ 2 \psi\left( \frac{\alpha + 2}{4} \right) - \psi\left( \frac{\alpha}{4} \right) - \psi\left( \frac{\alpha}{4} + 1 \right) \right] \\ &= \frac{1}{\alpha^{2}} - \frac{1}{\alpha} \beta\left(\frac{\alpha}{2}\right), \end{align} where $2 \beta(x) = \psi((x+1)/2) - \psi(x/2)$. Since \begin{align} \beta\left(\frac{x}{2}\right) = 2 \partial_{x} \ln\left( \frac{\Gamma((x+2)/4)}{\Gamma(x/4)} \right) \end{align} then \begin{align} \partial_{\alpha} I(\alpha) &= \frac{1}{\alpha^{2}} - \frac{2}{\alpha} \partial_{\alpha} \ln\left( \frac{\Gamma((\alpha+2)/4)}{\Gamma(\alpha/4)} \right) \end{align} and \begin{align} \int_{1}^{\infty} \partial_{\alpha} I(\alpha) \, d\alpha &= 1 - 2 \int_{1}^{\infty} \frac{1}{\alpha} \partial_{\alpha} \ln\left( \frac{\Gamma((\alpha+2)/4)}{\Gamma(\alpha/4)} \right) \\ I(\infty) - I(1) &= 1 - 2 \left[ \frac{1}{\alpha} \ln\left( \frac{\Gamma((\alpha+2)/4)}{\Gamma(\alpha/4)} \right) \right]_{1}^{\infty} - 2 \int_{1}^{\infty} \ln\left( \frac{\Gamma((\alpha+2)/4)}{\Gamma(\alpha/4)} \right) \, \frac{d\alpha}{\alpha^{2}} \\ - I(1) &= 1 + 2 \ln\left( \frac{\Gamma(3/4)}{\Gamma(1/4)} \right) - \frac{1}{2} \int_{1/4}^{\infty} \ln\left( \frac{\Gamma(u+1/2)}{\Gamma(u)} \right) \, \frac{du}{u^{2}}. \end{align} This is \begin{align} I(1) &= \ln\left( \frac{\Gamma^{4}(1/4)}{2 \pi^{2} e} \right) - \frac{1}{2} \int_{1/4}^{\infty} \ln\left( \frac{\Gamma(u)}{\Gamma(u + 1/2)} \right) \, \frac{du}{u^{2}}. \end{align} It is of note that \begin{align} \ln\left( \frac{\Gamma^{4}(1/4)}{2 \pi^{2} e} \right) < \int_{1/4}^{\infty} \ln\left( \frac{\Gamma(u)}{\Gamma(u + 1/2)} \right) \, \frac{du}{u^{2}} < \ln\left( \frac{\Gamma^{4}(1/4)}{4 \pi^{2} e} \right). \end{align}

Answer 2

I have come up with two ways to show that $$ \int_0^\infty e^{-x}\log(\cosh(x))\,\frac{\mathrm{d}x}{x} =1+\sum_{k=1}^n(-1)^k\frac{\log(2k+1)}{k}\tag{1} $$

---

Approach 1: Use the series $$ \log(1+x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{x^n}{n}\tag{2} $$ to get $$ \begin{align} \frac{\log(\cosh(x))}{x}e^{-x} &=\frac{x-\log(2)+\log(1+e^{-2x})}{x}e^{-x}\\ &=\left(1+\frac1x\sum_{k=1}^\infty(-1)^k\frac{1-e^{-2kx}}{k}\right)e^{-x}\\ &=e^{-x}+\sum_{k=1}^\infty(-1)^k\frac{e^{-x}-e^{-(2k+1)x}}{kx}\tag{3} \end{align} $$ and then use the integral $$ \int_0^\infty(e^{-x}-e^{-kx})\frac{\mathrm{d}x}{x}=\log(k)\tag{4} $$ to get $$ \int_0^\infty\frac{\log(\cosh(x))}{x}e^{-x}\,\mathrm{d}x =1+\sum_{k=1}^\infty(-1)^k\frac{\log(2k+1)}{k}\tag{5} $$

---

Approach 2: Use the integral $$ \frac{\log(\cosh(x))}{x}=1+\frac{\log\left(1-{\large\frac{1-e^{-2x}}2}\right)}{x}\tag{6} $$ the sum $$ \sum_{n=k}^\infty2^{-n-1}\binom{n}{k}=1\tag{7} $$ and $(2)$ to get $$ \begin{align} \int_0^\infty e^{-x}\log(\cosh(x))\,\frac{\mathrm{d}x}{x} &=1-\sum_{n=1}^\infty\frac1{n2^n}\int_0^\infty e^{-x}\left(1-e^{-2x}\right)^{\large n}\,\frac{\mathrm{d}x}{x}\\ &=1+\sum_{n=1}^\infty\frac1{n2^n}\sum_{k=1}^n(-1)^k\binom{n}{k}\log(2k+1)\\ &=1+\sum_{n=1}^\infty\frac1{2^n}\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}\frac{\log(2k+1)}{k}\\ &=1+\sum_{k=1}^\infty(-1)^k\frac{\log(2k+1)}{k}\tag{8} \end{align} $$

---

I haven't yet found a closed form for this sum, but we can use a similar sum to help accelerate the convergence of the series in $(8)$.

In equation $(8)$ of this answer it is shown that $$ \sum_{n=1}^\infty(-1)^n\frac{\log(n)}{n} =\gamma\log(2)-\frac{\log(2)^2}{2}\tag{9} $$ We can use this to accelerate the convergence of $$ \begin{align} &\sum_{k=1}^\infty(-1)^k\frac{\log(2k+1)}{k}\\ &=\sum_{k=1}^\infty(-1)^k\frac{\log(2k)}{k} +\sum_{k=1}^\infty(-1)^k\frac{\log\left(1+\frac1{2k}\right)}{k}\\ &=\gamma\log(2)-\frac{3\log(2)^2}{2} +\sum_{k=1}^\infty\frac{(-1)^k}k\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n(2k)^n}\\ &=\gamma\log(2)-\frac{3\log(2)^2}{2} +\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{k+n+1}}{n2^nk^{n+1}}\\ &=\gamma\log(2)-\frac{3\log(2)^2}{2} +\sum_{n=1}^\infty\frac{(-1)^n}{n2^n}\zeta(n+1)(1-2^{-n})\\ &=\gamma\log(2)-\frac{3\log(2)^2}{2}+\log(5/6) +\sum_{n=1}^\infty\frac{(-1)^n(2^n-1)}{n4^n}(\zeta(n+1)-1)\tag{10} \end{align} $$ Thus, we get $$ \begin{align} \int_0^\infty e^{-x}\log(\cosh(x))\,\frac{\mathrm{d}x}{x} &=1+\gamma\log(2)-\frac{3\log(2)^2}{2}+\log(5/6)\\ &+\sum_{n=1}^\infty\frac{(-1)^n(2^n-1)}{n4^n}(\zeta(n+1)-1)\tag{11} \end{align} $$ The series in $(11)$ converges about $0.6$ digits per term.

Answer 3

The result given in attachment is not a closed form : The aim is only to transform the integral into a series. The Laplace transform appearing in the calculus comes from H. Bateman,Tables of Integral Transforms, p.49 , Eq.17. Mc.Graw-hill Edit. (1954)

Numerical computations show that the convergence of the series given by Robjohn is faster than the convergence of my series. So, I advice to use the Robjohn's series instead of mine.

Answer 4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\ln\pars{\cosh\pars{x}} \over x}\,\expo{-x}\,\dd x}$

\begin{align}&\color{#c00000}{% \int_{0}^{\infty}{\ln\pars{\cosh\pars{x}} \over x}\,\expo{-x}\,\dd x} =\int_{0}^{\infty}{x + \ln\pars{1 + \expo{-2x}} - \ln\pars{2} \over x}\, \expo{-x}\,\dd x \\[3mm]&=1 + \color{#00f}{\int_{0}^{\infty} {\ln\pars{1 + \expo{-2x}} - \ln\pars{2} \over x}\,\expo{-x}\,\dd x} \end{align}

\begin{align}&\color{#00f}{\int_{0}^{\infty} {\ln\pars{1 + \expo{-2x}} - \ln\pars{2} \over x}\,\expo{-x}\,\dd x} =\int_{0}^{\infty}\bracks{\sum_{n = 1}^{\infty} {\pars{-1}^{n + 1} \over n}\expo{-2nx} -\ln\pars{2}}\expo{-x}\,{\dd x \over x} \\[3mm]&=-\int_{0}^{\infty}\ln\pars{x}\bracks{\sum_{n = 1}^{\infty} {\pars{-1}^{n + 1} \over n}\pars{-2n - 1}\expo{-\pars{2n + 1}x} +\ln\pars{2}\expo{-x}}\,\dd x \end{align}

However, $$ \int_{0}^{\infty}\ln\pars{x}\expo{-\alpha x}\,\dd x =-\,{\ln\pars{\alpha} + \gamma \over \alpha}\,,\qquad\Re\pars{\alpha} > 0 $$

Then, \begin{align}&\color{#00f}{\int_{0}^{\infty} {\ln\pars{1 + \expo{-2x}} - \ln\pars{2} \over x}\,\expo{-x}\,\dd x} =-\braces{\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n}\, \bracks{\ln\pars{2n + 1} + \gamma}} + \gamma\ln\pars{2} \\[3mm]&=\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n}\,\ln\pars{2n + 1} =\int_{0}^{2}\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over \xi n + 1}\,\dd\xi =\int_{0}^{2} \sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n + 1/\xi}\,{\dd\xi \over \xi} \\[3mm]&=\int_{0}^{2}{\Phi\pars{-1,1,1/\xi} \over \xi}\,\dd\xi \end{align} where $\ds{\Phi\pars{z,s,\alpha}}$ is the Lerch Trascendent Function.

$$\color{#00f}{\large\int_{0}^{\infty} {\ln\pars{\cosh\pars{x}} \over x}\,\expo{-x}\,\dd x} =\color{#00f}{\large 1 + \int_{0}^{2}{\Phi\pars{-1,1,1/\xi} \over \xi}\,\dd\xi} $$

So far, I was unable to evaluate or/and find the 'logarithm involved sums' and we didn't find anything related to $\ds{\Phi}$ integration. Besides the usual links, this one seems to be very interesting.