# Integrate <msubsup> &#x222B;<!-- ∫ --> 0 <mrow class="MJX-TeXAtom-ORD"> <mi mat

Integrate ${\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}x}{\mathrm{cosh}x\phantom{\rule{mediummathspace}{0ex}}+\phantom{\rule{mediummathspace}{0ex}}\mathrm{cos}x}\frac{dx}{x}$
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I'm not sure how you showed the two integrals are equivalent, but the following is an evaluation of
${\int }_{0}^{\mathrm{\infty }}{e}^{-x}\mathrm{cos}\left(x\right)\mathrm{tanh}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}.$
For $\mathrm{\Re }\left(s\right)>0$, we have
$\begin{array}{rl}{\int }_{0}^{\mathrm{\infty }}\mathrm{tanh}\left(t\right){e}^{-st}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t& ={\int }_{0}^{\mathrm{\infty }}\frac{1-{e}^{-2t}}{1+{e}^{-2t}}\phantom{\rule{thinmathspace}{0ex}}{e}^{-st}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t\\ & ={\int }_{0}^{\mathrm{\infty }}\left(1-{e}^{-2t}\right){e}^{-st}\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}{e}^{-2tn}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t\\ & =\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}{\int }_{0}^{\mathrm{\infty }}\left({e}^{-\left(2n+s\right)t}-{e}^{-\left(2n+s+2\right)t}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{2n+s}-\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{2n+s+2}\\ & =\frac{1}{2}\left(\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{n+\frac{s}{2}}-\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{n+\frac{s}{2}+1}\right)\\ & \stackrel{\left(1\right)}{=}\frac{1}{4}\left(\psi \left(\frac{s}{4}+\frac{1}{2}\right)-\psi \left(\frac{s}{4}\right)-\psi \left(\frac{s}{4}+1\right)+\psi \left(\frac{s}{4}+\frac{1}{2}\right)\right)\\ & \stackrel{\left(2\right)}{=}\frac{1}{2}\left(\psi \left(\frac{s}{4}+\frac{1}{2}\right)-\psi \left(\frac{s}{4}\right)-\frac{2}{s}\right).\end{array}$
Therefore,
$\begin{array}{rl}{\int }_{0}^{\mathrm{\infty }}{e}^{-x}\mathrm{cos}\left(x\right)\mathrm{tanh}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}& =\mathrm{\Re }{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(1+i\right)x}\frac{\mathrm{tanh}\left(x\right)}{x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\\ & =\mathrm{\Re }{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(1+i\right)x}\mathrm{tanh}\left(x\right){\int }_{0}^{\mathrm{\infty }}{e}^{-xt}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\\ & =\mathrm{\Re }{\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\mathrm{tanh}\left(x\right){e}^{-\left(1+i+t\right)x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t\\ & =\mathrm{\Re }{\int }_{0}^{\mathrm{\infty }}\frac{1}{2}\left(\psi \left(\frac{1+i+t}{4}+\frac{1}{2}\right)-\psi \left(\frac{1+i+t}{4}\right)-\frac{2}{1+i+t}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t\\ & =\mathrm{\Re }\left(2\mathrm{ln}\mathrm{\Gamma }\left(\frac{3+i+t}{4}\right)-2\mathrm{ln}\mathrm{\Gamma }\left(\frac{1+i+t}{4}\right)-\mathrm{ln}\left(1+i+t\right)\right){|}_{0}^{\mathrm{\infty }}\\ & \stackrel{\left(3\right)}{=}\mathrm{\Re }\left(-2\mathrm{ln}\left(2\right)-2\mathrm{ln}\mathrm{\Gamma }\left(\frac{3+i}{4}\right)+2\mathrm{ln}\mathrm{\Gamma }\left(\frac{1+i}{4}\right)+\mathrm{ln}\left(1+i\right)\right)\\ & =-\frac{3\mathrm{ln}\left(2\right)}{2}+2\mathrm{\Re }\left(\mathrm{ln}\mathrm{\Gamma }\left(\frac{1+i}{4}\right)-\mathrm{ln}\mathrm{\Gamma }\left(\frac{3+i}{4}\right)\right).\end{array}$