Evaluate $$\displaystyle\int_0^\infty \frac {(\log x)^4dx}{(1+x)(1+x^2)}$$
This is a past final term exam problem of a complex analysis course at my university. I am studying for this year’s exam and I found this problem. The examiner assumes us to use residue calculus. Could you please give your valuable suggestions on how to proceed ?
We can use brute force.
First, notice that: $$ \int_0^\infty \frac{\log(x)}{(1+x)(1+x^2)}=\frac{-\pi^2}{16}, \int_0^\infty \frac{\log^2(x)}{(1+x)(1+x^2)}=\frac{\pi^3}{16} ,\int_0^\infty \frac{\log^3(x)}{(1+x)(1+x^2)}=\frac{-7\pi^4}{128}$$
Now, to calculate $\int_0^\infty \frac{\log^4(x)}{(1+x)(1+x^2)}$ we will consider the integral $\int_C\frac{\log^5(z)}{(1+z)(1+z^2)}$ and use the keyhole contour
Then
$\begin{align} \displaystyle \int_C \frac{\log^5(z)}{(1+z)(1+z^2)}dz = \int_{\epsilon}^{R} \frac{\log^5(x)}{(1+x)(1+x^2)}dx-\int_{\epsilon}^{R} \frac{(\log(x)+2\pi i)^5}{(1+x)(1+x^2)}dx+(\int_{\gamma_{\epsilon}}+\int_{\gamma_R})f(z)dz = 2\pi i(\operatorname{Res}_{z=i}f(z)+\operatorname{Res}_{z=-i}f(z)+\operatorname{Res}_{z=-1}f(z))\end{align}$
Where $f(z)= \frac{\log^5(x)}{(1+x)(1+x^2)}$
Notice that the branch of $\log(z)$ is with $0< \theta< 2\pi$ we have:
$\operatorname{Res}_{z=i}f(z)= \frac{\pi^5(1-i)}{128}$
$\operatorname{Res}_{z=-i}f(z)= \frac{243\pi^5(-1-i)}{128}$
$\operatorname{Res}_{z=-1}f(z)= \frac{i\pi^5}{2}$
Then: $2\pi i(\operatorname{Res}_{z=i}f(z)+\operatorname{Res}_{z=-i}f(z)+\operatorname{Res}_{z=-1}f(z))= \frac{-242\pi^6 i}{64}+\frac{180\pi^6}{64}$
When $\epsilon \rightarrow 0$ and $R \rightarrow \infty$:
$$\begin{align} \int_{\epsilon}^{R} \frac{\log^5(x)}{(1+x)(1+x^2)}dx-\int_{\epsilon}^{R} \frac{(\log(x)+2\pi i)^5}{(1+x)(1+x^2)}dx \end{align}$$
$$= -\int_{0}^{\infty}\frac{10\pi i \log^4(x)}{(1+x)(1+x^2)}+\int_{0}^{\infty}\frac{40\pi^2\log^3(x)}{(1+x)(1+x^2)}+\int_{0}^{\infty}\frac{80i\pi^3\log^2(x)}{(1+x)(1+x^2)}-\int_{0}^{\infty}\frac{80\pi^4\log(x)}{(1+x)(1+x^2)}-\int_{0}^{\infty}\frac{i32\pi^5}{(1+x)(1+x^2)}= -\int_{0}^{\infty}\frac{10\pi i \log^4(x)}{(1+x)(1+x^2)}- \frac{140\pi^6}{64}+\frac{320i\pi^6}{64}+\frac{320\pi^6}{64}-\frac{512i\pi^6}{64} = -\int_{0}^{\infty}\frac{10\pi i \log^4(x)}{(1+x)(1+x^2)}+ \frac{180\pi^6}{64}-\frac{192i\pi^6}{64}$$
And you must show that:
$|\int_{\gamma_{\epsilon}}f(z)dz| \rightarrow 0 , |\int_{\gamma_R}f(z)dz| \rightarrow 0$
So: $$-\int_{0}^{\infty}\frac{10\pi i \log^4(x)}{(1+x)(1+x^2)}+ \frac{180\pi^6}{64}-\frac{192i\pi^6}{64}=\frac{-242\pi^6 i}{64}+\frac{180\pi^6}{64}$$
Follows that:
$$\int_{0}^{\infty}\frac{\log^4(x)}{(1+x)(1+x^2)}=\frac{5\pi^5}{64}$$.
Then the question is how to calculate $ \int_0^\infty \frac{\log(x)}{(1+x)(1+x^2)}=\frac{-\pi^2}{16}, \int_0^\infty \frac{\log^2(x)}{(1+x)(1+x^2)}=\frac{\pi^3}{16} ,\int_0^\infty \frac{\log^3(x)}{(1+x)(1+x^2)}=\frac{-7\pi^4}{128}$
To calculate these integrals you should proceed in the same way. That is, to calculate $ \int_0^\infty \frac{\log^{n}(x)}{(1+x)(1+x^2)}$ you should consider $\int_C\frac{\log^{n+1}(z)}{(1+z)(1+z^2)}$. In our case, $n=1,2,3$.
For $-1\lt a\lt0$, $$ \begin{align} &\int_0^\infty\frac{x^a}{(1+x)(1+x^2)}\,\mathrm{d}x\\ &=\int_0^\infty\frac{x^a}2\left(\frac1{1+x}+\frac{1-x}{1+x^2}\right)\mathrm{d}x\\ &=\frac12\int_0^\infty\frac{x^a}{1+x}\,\mathrm{d}x+\frac14\int_0^\infty\frac{x^{\frac{a-1}2}}{1+x}\,\mathrm{d}x-\frac14\int_0^\infty\frac{x^{\frac{a}2}}{1+x}\,\mathrm{d}x\\ &=\frac12\Gamma(1+a)\Gamma(-a)+\frac14\Gamma\!\left(\frac{1+a}2\right)\Gamma\!\left(\frac{1-a}2\right)-\frac14\Gamma\!\left(\frac{2+a}2\right)\Gamma\!\left(-\frac{a}2\right)\\ &=-\frac12\frac\pi{\sin(\pi a)}+\frac14\frac\pi{\cos\left(\frac\pi2a\right)}+\frac14\frac\pi{\sin\left(\frac\pi2a\right)}\\ &=\frac\pi4\frac{1+\cos\left(\frac\pi2a\right)-\sin\left(\frac\pi2a\right)}{\left(1+\cos\left(\frac\pi2a\right)\right)\cos\left(\frac\pi2a\right)}\\ &=\frac\pi2\frac1{1+\cos\left(\frac\pi2a\right)+\sin\left(\frac\pi2a\right)}\\ \end{align} $$ This can be analytically continued to $-1\lt a\lt 2$.
Taking $4$ derivatives gives $$ \begin{align} &\int_0^\infty\frac{\log(x)^4x^a}{(1+x)(1+x^2)}\,\mathrm{d}x\\ &=\frac{\pi^5}{32}\frac{\scriptsize105+45\left(\sin\left(\frac\pi2a\right)+\cos\left(\frac\pi2a\right)\right)-54\sin(\pi a)-11\left(\sin\left(\frac{3\pi}2a\right)-\cos\left(\frac{3\pi}2a\right)\right)-\cos(2\pi a)}{2\left(1+\cos\left(\frac\pi2a\right)+\sin\left(\frac\pi2a\right)\right)^5} \end{align} $$ and evaluating at $a=0$ gives $$ \int_0^\infty\frac{\log(x)^4}{(1+x)(1+x^2)}\,\mathrm{d}x =\frac{5\pi^5}{64} $$
$$\int_{0}^{+\infty}\frac{(1-x)\log^4 x}{1-x^4}\,dx=\int_{0}^{1}\frac{(1-x)\log^4 x}{1-x^4}\,dx+\int_{0}^{1}\frac{\left(1-\frac{1}{x}\right)\log^4 x}{x^2\left(1-\frac{1}{x^4}\right)}\,dx $$ can be written as $$ \int_{0}^{1}\frac{\log^4(x)}{1+x^2}\,dx = \sum_{k\geq 0}(-1)^k \int_{0}^{1}x^{2k}\log^4(x)\,dx=\sum_{k\geq 0}\frac{24(-1)^k}{(2k+1)^5} $$ and the RHS is well-known to be related to Euler numbers. The final outcome is $$ \int_{0}^{+\infty}\frac{\log^4 x}{(1+x)(1+x^2)}\,dx = \color{red}{\frac{5\pi^5}{64}}.$$
