$$I=\int_{0}^{\infty}\frac{\ln\left(x\right)\left(e^{-x}-\cos x\right)}{x\left(1+x^{4}\right)}dx=\frac{\pi^{2}}{8}\left(1-e^{-\frac{1}{\sqrt{2}}}\cos\left(\frac{1}{\sqrt{2}}\right)\right)$$
I got this Integral from one of my friends, he used Wolfram to get the closed form of the Integral.
Well, here is my attempt:
Using Partial Fraction Decomposition:
$$I=\displaystyle{\underbrace{\int_{0}^{\infty}\frac{\ln\left(x\right)\left(e^{-x}-\cos x\right)}{x}dx}_{I_1}-\underbrace{\int_{0}^{\infty}\frac{\ln\left(x\right)\left(e^{-x}-\cos x\right)x^{3}}{1+x^{4}}dx}_{I_2}}$$
For $I_1$ Wolfram Gives:
$$I_1=
\frac{\pi^2}{8}$$
Wolfram can not evaluate $I_2$ but as we already know $I$, this gives:
$$I_2=\frac{\pi^{2}}{8}e^{-\frac{1}{\sqrt{2}}}\cos\left(\frac{1}{\sqrt{2}}\right)$$
How can we evaluate these? Preferably without Complex Analysis if possible!
Best Answer
Assume that $a>0$.
Very similar to my answer here, let $$I(a) = \int_{0}^{\infty} \frac{\cos(ax)-e^{-{ax}}}{x(1+x^{4})} \, \ln(x) \, \mathrm dx.$$
Then $$ \begin{align} I^{(4)}(a) + I(a) &= \int_{0}^{\infty} \frac{\cos(ax)-e^{-ax}}{x} \, \ln(x) \, \mathrm dx \\ &= \int_{0}^{\infty} \frac{\cos(u) -e^{-u}}{u} \, \ln(u) \, \mathrm du - \int_{0}^{\infty} \frac{\cos(u) -e^{-u}}{u} \, \ln(a) \, \mathrm du \\ &\overset{(1)}= \int_{0}^{\infty} \frac{\cos(u) -e^{-u}}{u} \, \ln(u) \, \mathrm du -0\\ &=\lim_{s \to 0^{+}} \frac{\mathrm d}{\mathrm ds}\int_{0}^{\infty} \left(\cos(u) - e^{-u} \right) u^{s-1} \, \mathrm du \\ &= \lim_{s \to 0^{+}}\frac{\mathrm d}{\mathrm ds} \, \Gamma(s) \left( \cos \left(\frac{\pi s}{2} \right)-1\right) \\ &= \lim_{s \to 0^{+}} \left(\Gamma'(s) \left(\cos \left(\frac{\pi s}{2}\right)-1 \right)- \Gamma(s)\frac{\pi}{2} \sin \left(\frac{\pi s}{2} \right) \, \right) \\ &= \lim_{s \to 0^{+}} \left(\left(- \frac{1}{s^{2}} + O(1) \right)\left(\cos \left(\frac{\pi s}{2}\right)-1 \right) - \frac{\pi}{2}\left(\frac{1}{s} +O(1) \right)\sin \left(\frac{\pi s}{2} \right)\right) \\ &= \frac{\pi^{2}}{8} - \frac{\pi^{2}}{4} \\ &= - \frac{\pi^{2}}{8}. \end{align}$$
The general solution of the above linear differential equation with constant coefficients is $$ \small I(a) = C_{1}e^{a / \sqrt{2}} \cos \left(\frac{a}{\sqrt{2}} \right) +C_{2}e^{a / \sqrt{2}} \sin \left(\frac{a}{\sqrt{2}} \right) +C_{3} e^{-a / \sqrt{2}} \cos \left(\frac{a}{\sqrt{2}} \right) +C_{4} e^{-a / \sqrt{2}} \sin \left(\frac{a}{\sqrt{2}} \right) - \frac{\pi^{2}}{8}.$$
The initial conditions are $\lim_{a \to 0^{+}} I(a) = 0$ and $$\lim_{a \to 0^{+}} I'(a) = \int_{0}^{\infty} \frac{\ln(x)}{1+x^{4}} \, \mathrm dx = \lim_{s \to 1} \frac{\mathrm d}{\mathrm ds} \int_{0}^{\infty} \frac{x^{s-1}}{1+x^{4}} \, \mathrm dx =\lim_{s \to 1} \frac{\mathrm d}{\mathrm ds} \frac{\pi}{4} \, \csc \left(\frac{\pi s}{4} \right) = -\frac{\pi^{2}}{8 \sqrt{2}}.$$
Since $I(a)$ remains finite as $a \to + \infty$, $C_{1}$ and $C_{2}$ must be zero.
And the initial condition $\lim_{a \to 0^{+}} I(a) =0$ means that $C_{3}= \frac{\pi^{2}}{8}$.
Finally, to satisfy the initial condition $\lim_{a \to 0^{+}} I'(a) = -\frac{\pi^{2}}{8 \sqrt{2}}$, $C_{4}$ must be zero.
Therefore, $$I(a) = \frac{\pi^{2}}{8} e^{-a / \sqrt{2}} \cos \left(\frac{a}{\sqrt{2}} \right) - \frac{\pi^{2}}{8},$$
and $$ \int_{0}^{\infty} \frac{e^{-x}-\cos(x)}{x (1+x^{4})} \, \ln(x) \, \mathrm dx = -I(1) = \frac{\pi^{2}}{8} \left(1- e^{-1 / \sqrt{2}} \cos \left(\frac{1}{\sqrt{2}} \right)\right). $$
$(1)$ See here for a way to show that the integral vanishes without using complex analysis or special functions.