<s>[INST] Closed form of $\int_0^{\infty} \frac{\log(x)}{\cosh(x) \sec(x)- \tan(x)} \ dx$ What real analysis tools would you recommend me for getting the closed form of the integral below?
$$\int_0^{\infty} \frac{\log(x)}{\cosh(x) \sec(x)- \tan(x)} \ dx$$
 [/INST] The place I would start is the nifty result, proven here, that
$$\frac{\sin{x}}{\cosh{t} - \cos{x}} = 2 \sum_{k=1}^{\infty} e^{-k t} \sin{k x} $$
Of course, the integral actually looks like
$$\int_0^{\infty} dx \frac{\cos{x}}{\cosh{x} - \sin{x}} \log{x} $$
so we need to map $x \mapsto \pi/2 - x$ and we have that the integral is actually
$$2 \sum_{m=0}^{\infty} (-1)^m \int_0^{\infty} dx \, e^{-(2 m+1) x} \cos{(2 m+1) x}\, \log{x} + 2 \sum_{m=1}^{\infty} (-1)^{m+1} \int_0^{\infty} dx \, e^{-2 m x} \sin{2 m x}\, \log{x} $$
We then note that
$$-k \int_0^{\infty} dx \, e^{-(1-i) k x} \log{x} = \frac{\log (k)}{2}+\frac{\gamma }{2}+\frac{\log (2)}{4}+\frac{\pi }{8}+i \left(\frac{\log (k)}{2}+\frac{\gamma }{2}+\frac{\log (2)}{4}-\frac{\pi }{8}\right)$$
So the integral is
$$-\sum_{m=0}^{\infty} (-1)^m \frac{\log{(2 m+1)}}{2 m+1}  - \sum_{m=1}^{\infty} (-1)^{m+1} \frac{\log{(2 m)}}{2 m}\\ - \left (\gamma + \frac12 \log{2} + \frac{\pi}{4} \right )\sum_{m=0}^{\infty} \frac{(-1)^m}{2 m+1} -   \left (\gamma + \frac12 \log{2} - \frac{\pi}{4} \right )\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{2 m}$$
The third and fourth sums are well known.  The second is less-well known, but may be shown to be
$$\frac{1}{4} \left(3 \log ^2(2)-2 \gamma  \log (2)\right)$$
(See, for example, Williams & Hardy, The Red Book of Mathematical Problems, problem 71.)
The first sum, however, is less well known.  It may be evaluated by considering the derivative of a generalized zeta function (or a Lerch transcendent).  The result is 
$$\sum_{m=0}^{\infty} (-1)^m \frac{\log{(2 m+1)}}{2 m+1} = \frac12 \log{2} \left [\psi{\left ( \frac{3}{4} \right )} - \psi{\left ( \frac{1}{4} \right )}  \right ] - \frac14 \left [ \gamma_1{\left ( \frac{3}{4} \right )}-\gamma_1{\left ( \frac{1}{4} \right )}\right ]$$
where $\gamma_1(a)$ is a generalized Stieltjes constant.  
Put this all together and the result agrees with a numerical evaluation performed with Mathematica ($ \approx -1.35775$).
ADDENUDUM
Apparently I forgot that I had encountered that first sum some time ago here, and the result may be expressed in simpler terms:
$$ \sum_{m=1}^{\infty} (-1)^{m} \frac{\log{(2 m+1)}}{2 m+1} = -\frac{\pi}{4} \gamma - \frac{\pi}{4} \log{\frac{\Gamma{\left ( \frac{3}{4} \right )}^4}{\pi}} $$
so we may now put this all together as follows:
$$\frac{\pi}{4} \gamma + \frac{\pi}{4} \log{\frac{\Gamma{\left ( \frac{3}{4} \right )}^4}{\pi}} - \frac{3}{4} \log^2{2} + \frac12 \gamma \log{2} - \left (\gamma + \frac12 \log{2} + \frac{\pi}{4} \right )\frac{\pi}{4} -   \left (\gamma + \frac12 \log{2} - \frac{\pi}{4} \right )\frac12 \log{2}$$
or
$$\int_0^{\infty} dx \frac{\cos{x}}{\cosh{x} - \sin{x}} \log{x}  = \frac{\pi}{4} \log{\frac{\Gamma{\left ( \frac{3}{4} \right )}^4}{\pi}}  - \log^2{2} - \frac{\pi^2}{16}$$
ADDENDUM II
The methodology outlined above may be applied to other, similar integrals.  For example, it is a simpler matter to show that
$$\int_0^{\infty} dx \frac{x \sin{x}}{\cosh{x}-\cos{x}} \log{x} = \frac{1}{6} \pi ^2 \left(-12 \log{A}+1+\frac{\log{2}}{2}+\log{\pi} \right) $$
where $A$ is Glaisher's constant.
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