Integral $\int_{0}^{\infty}\frac{\ln\left(x\right)\left(e^{-x}-\cos x\right)}{x\left(1+x^{4}\right)}dx$

Question

$$I=\int_{0}^{\infty}\frac{\ln\left(x\right)\left(e^{-x}-\cos x\right)}{x\left(1+x^{4}\right)}dx=\frac{\pi^{2}}{8}\left(1-e^{-\frac{1}{\sqrt{2}}}\cos\left(\frac{1}{\sqrt{2}}\right)\right)$$

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!

Answer 1

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 $$ \int_{0}^{\infty} \frac{e^{-x}-\cos(x)}{x (1+x^{4})} \, \ln(x) \, \mathrm dx = -I(1) = \frac{\pi^{2}}{8} \left(1- e^{-1 / \sqrt{2}} \cos \left(\frac{1}{\sqrt{2}} \right)\right). $$

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$(1)$ See here for a way to show that the integral vanishes without using complex analysis or special functions.

Answer 2

If complex integration is acceptable, we can consider the integrand $\displaystyle f(z)=\frac{\ln z(e^{-z}-e^{iz})z^3}{1+z^4}$ and a closed contour: $r\to R\to iR \,(\,\text{along an arch of the radius R})\to ir\to r\,(\,\text{along an arch of the radius r})$.

The integrals along the arches tend to $0$ - as $r\to 0$ and $R\to\infty$. We also have a single simple pole inside the contour (at $z=e^{\frac{\pi i}4}$).

Hence, $$\oint \frac{\ln z(e^{-z}-e^{iz})z^3}{1+z^4}dz=$$ $$=\int_0^\infty\frac{\ln x(e^{-x}-e^{ix})x^3}{1+x^4}dx+\int_\infty^0\frac{\ln(ix)(e^{-ix}-e^{-x})x^3}{1+x^4}dx=2\pi i\underset{z=e^{\frac{\pi i}4}}{\operatorname{Res}}\frac{\ln z(e^{-z}-e^{iz})z^3}{1+z^4}$$ Taking the real part $$2\Re\int_0^\infty\frac{\ln x(e^{-x}-e^{ix})x^3}{1+x^4}dx=\Re\frac{\pi i}2\int_0^\infty\frac{e^{-ix}-e^{-x}}{1+x^4}x^3dx+\Re2\pi i\underset{z=e^{\frac{\pi i}4}}{\operatorname{Res}}\frac{\ln z(e^{-z}-e^{iz})z^3}{1+z^4}$$ $$2\int_0^\infty\frac{\ln x(e^{-x}-e^{ix})x^3}{1+x^4}dx=2I_2=\frac\pi2\int_0^\infty\frac{x^3\sin x}{1+x^4}dx-\frac{\pi^2}2\Re\underset{z=e^{\frac{\pi i}4}}{\operatorname{Res}}\frac{(e^{-z}-e^{iz})z^3}{1+z^4}$$ $$I_2=\frac\pi8\Im\int_{-\infty}^\infty\frac{x^3e^{ix}}{1+x^4}dx-\frac{\pi^2}4\Re\underset{z=e^{\frac{\pi i}4}}{\operatorname{Res}}\frac{(e^{-z}-e^{iz})z^3}{1+z^4}$$ $$=\frac{\pi^2}4\Re\underset{\binom{z=e^{\frac{\pi i}4}}{z=e^{\frac{3\pi i}4}}}{\operatorname{Res}}\frac{e^{iz}z^3}{1+z^4}-\frac{\pi^2}4\Re\underset{z=e^{\frac{\pi i}4}}{\operatorname{Res}}\frac{(e^{-z}-e^{iz})z^3}{1+z^4}$$ $$=\frac{\pi^2}8e^{-\frac1{\sqrt 2}}\cos\frac1{\sqrt2}-0=\frac{\pi^2}8e^{-\frac1{\sqrt 2}}\cos\frac1{\sqrt2}$$

Using the same approach for $I_1$ we get (as we do not have poles inside the contour): $$\oint\frac{\ln z(e^{-z}-e^{iz})}{z}dz=2I_1-\frac{\pi i}2\int_0^\infty\frac{e^{-ix}-e^{-x}}xdx=0$$ $$\Rightarrow\,\,I_1=\frac{\pi i}4\int_0^\infty\frac{e^{-ix}-e^{-x}}xdx=-\frac\pi4\Im\int_0^\infty\frac{e^{-ix}}xdx=\frac{\pi^2}8$$