INTEGRAL ln x/cosh^2 x \,dx

In this post We will prove the following integral


\int_0^\infty \frac{ \ln x}{\cosh^2 (x)}\,dx=\ln \left( \frac{\pi}{4}\right)-\gamma


Lets first evaluate the integral


\begin{aligned} \int_0^\infty \frac{x^{s}}{\cosh^2 (x)}dx&=4\int_0^\infty \frac{x^{s}}{(e^x+e^{-x})^2}dx\\ &=4\int_0^\infty \frac{x^{s}e^{-2x}}{e^{-2x}(e^x+e^{-x})^2}dx\\ &=4\int_0^\infty \frac{x^{s}e^{-2x}}{(1+e^{-2x})^2}dx\\ &=-4\sum_{k=1}^\infty (-1)^{k-1}k\int_0^\infty x^{s}e^{-2kx}dx \qquad 2kx \mapsto x\\ &=\frac{4}{2^{s+1}}\sum_{k=1}^\infty \frac{(-1)^{k-1}k}{k^{s+1}}\int_0^\infty x^{s}e^{-x}dx \\ &=\frac{2}{2^{s}}\Gamma(s+1)\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^{s}} \\ &=2^{1-s}\Gamma(s+1)\eta(s) \\ \end{aligned}


\boxed{\int_0^\infty \frac{x^{s}}{\cosh^2 (x)}dx=2^{1-s}\Gamma(s+1)\eta(s)}(1)


Differentiating (1) w.r. to s


\begin{aligned} \int_0^\infty \frac{x^{s} \ln x}{\cosh^2 (x)}\,dx&=\frac{d}{ds}\left(2^{1-s}\Gamma(s+1)\eta(s) \right)\\ &=-2^{1-s}\,\ln 2\,\Gamma(s+1)\eta(s) + 2^{1-s}\left(\Gamma^{\prime}(s+1)\eta(s)+\Gamma(s+1)\eta^{\prime}(s) \right)\\ \end{aligned}


Now, letting s \rightarrow 0


\begin{aligned} \int_0^\infty \frac{ \ln x}{\cosh^2 (x)}\,dx&=-2\,\ln 2\,\Gamma(1)\eta(0) + 2\left(\Gamma^{\prime}(1)\eta(0)+\Gamma(1)\eta^{\prime}(0) \right)\\ &=-2\left(\ln 2+\gamma \right)\eta(0)+2\eta^{\prime}(0)\\ \end{aligned}(2)


Now recall that

\eta(s) = (1-2^{1-s})\zeta(s)(3)


And consequently


\eta^{\prime}(s) = 2^{1-s} \, \ln 2 \, \zeta(s)+(1-2^{1-s})\zeta^{\prime}(s)(4)


We also have the following relations (proved here)

\zeta(0) = -\frac{1}{2}

and

\zeta'(0) = - \frac{1}{2} \log 2 \pi

Setting s=0 in (3) and (4) we get

\eta(0) = \frac{1}{2}(5)


\eta^{\prime}(0) =\frac{1}{2}\ln \left( \frac{\pi}{2}\right)
(6)

Plugging (5) and (6) in (2)


\begin{aligned} \int_0^\infty \frac{ \ln x}{\cosh^2 (x)}\,dx&=\left(-\ln 2-\gamma \right)+\ln \left( \frac{\pi}{2}\right)\\ &=\ln \left( \frac{\pi}{4}\right)-\gamma \qquad \blacksquare \end{aligned}

Appendix

Recall the geometric series

\begin{aligned} \sum_{k=0}^\infty (-1)^kx^k&=\frac{1}{1+x} \qquad \text{differentiating w.r. to x}\\ \sum_{k=1}^\infty (-1)^k k \,x^{k-1}&=-\frac{1}{(1+x)^2}\\ \sum_{k=1}^\infty (-1)^k k \,x^{k}&=-\frac{x}{(1+x)^2} \qquad \text{switch}\,x \mapsto e^{-2x} \\ \sum_{k=1}^\infty (-1)^k k \,e^{-2kx}&=-\frac{e^{-2x}}{(1+e^{-2x})^2} \end{aligned}


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