I’m trying to evaluate the following integral through a contour method:
$$\int_{0}^{\infty} \frac{\ln^3 (1+x^2)}{1+2x^2} \, dx$$
Defining $$f(z)=\frac{\ln^3 (1+z^2)}{1+2z^2}$$
I’ve tried to use a semi-circular contour from $-R$ to $R$ as the integrand is even and have calculated the residue at $z=\frac{i}{\sqrt{2}}$ which gives:
$$\text{Res} \left(f(z),\frac{i}{\sqrt{2}}\right) = \frac{\ln^3 (2)}{2\sqrt{2}} i$$
However, I’m struggling because there are branch cuts that I’m not sure how to deal with here. I can evaluate the related integral:
$$\int_{0}^{\infty} \frac{\ln (1+x^2)}{1+2x^2} \, dx = -\frac{\pi\ln(6-4\sqrt{2})}{2\sqrt{2}}$$
Since $\ln (1+x^2)=\ln(1+i x)+\ln(2-i x)$ allowing the integral to be separated and each new integral to be integrated along a semi-circle without running into a branch cut. However, I’m not sure about the squared or cubed case.
EDIT
With the help of Mathematica, I can determine
$$\int_{0}^{\infty} \frac{\ln^3 (1+x^2)}{1+2x^2} \, dx = \pi\left(3\sqrt{2}\operatorname{Li}_3\left(\frac{1}{4}\left(2+\sqrt{2}\right)\right)-3\sqrt{2}\operatorname{Li}_3\left(\frac{1}{4+2\sqrt{2}}\right)-6\sqrt{2}\operatorname{Li}_2\left(\frac{1}{\sqrt{2}}\right)\ln(2)+\frac{3\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)\ln(2)}{\sqrt{2}}+3\sqrt{2}\operatorname{Li}_2\left(\frac{1}{4}\left(2+\sqrt{2}\right)\right)\ln\left(4-2\sqrt{2}\right)+\frac{3\operatorname{Li}_2\left(-2+2\sqrt{2}\right)\ln(2)}{\sqrt{2}}-3\sqrt{2}\operatorname{Li}_2\left(\frac{1}{4+2\sqrt{2}}\right)\ln(2)-3\sqrt{2}\operatorname{Li}_2\left(\frac{1}{4+2\sqrt{2}}\right)\ln\left(2+\sqrt{2}\right)+\frac{3\ln^3(2)}{8\sqrt{2}}-\frac{\ln^3\left(3-2\sqrt{2}\right)}{\sqrt{2}}-\frac{\ln^3\left(4-2\sqrt{2}\right)}{\sqrt{2}}-\sqrt{2}\ln^3\left(2-\sqrt{2}\right)-\sqrt{2}\ln^3\left(1+\sqrt{2}\right)-\sqrt{2}\ln^3\left(2+\sqrt{2}\right)-\frac{3\ln(2)\ln^2\left(3-2\sqrt{2}\right)}{\sqrt{2}}-\frac{9\ln^2(2)\ln\left(2-\sqrt{2}\right)}{4\sqrt{2}}+\frac{3\ln^2\left(4-2\sqrt{2}\right)\ln\left(2-\sqrt{2}\right)}{\sqrt{2}}+3\sqrt{2}\ln(2)\ln^2\left(2-\sqrt{2}\right)-\frac{75\ln^2(2)\ln\left(1+\sqrt{2}\right)}{4\sqrt{2}}+6\sqrt{2}\ln^2\left(2-\sqrt{2}\right)\ln\left(1+\sqrt{2}\right)-\frac{3\ln\left(\frac{577}{2}-204\sqrt{2}\right)\ln^2\left(1+\sqrt{2}\right)}{2\sqrt{2}}-\frac{3\ln(2)\ln^2\left(2+\sqrt{2}\right)}{\sqrt{2}}+3\sqrt{2}\ln(2)\ln^2\left(3+2\sqrt{2}\right)-\frac{3\ln\left(1-\frac{1}{\sqrt{2}}\right)\ln^2\left(3+2\sqrt{2}\right)}{2\sqrt{2}}+\frac{3\ln\left(1+\frac{1}{\sqrt{2}}\right)\ln^2\left(3+2\sqrt{2}\right)}{2\sqrt{2}}+\frac{3\ln^2(2)\ln\left(1393+985\sqrt{2}\right)}{2\sqrt{2}}-\frac{3\ln\left(12-8\sqrt{2}\right)\ln\left(3-2\sqrt{2}\right)\ln\left(1+\sqrt{2}\right)}{\sqrt{2}}+\frac{3\ln(2)\ln\left(1+\sqrt{2}\right)\ln\left(2+\sqrt{2}\right)}{\sqrt{2}}-\frac{3\ln\left(1+\sqrt{2}\right)\ln\left(3+2\sqrt{2}\right)\ln\left(136+96\sqrt{2}\right)}{\sqrt{2}}\right)-\frac{\pi^3\left(\ln\left(\frac{577}{8}-51\sqrt{2}\right)-2\ln\left(3-2\sqrt{2}\right)\right)}{4\sqrt{2}}$$
It may be possible to simplify this answer with polylogarithm identities, but I have not yet determined how to.
As @Random Variable suggested, we can use a keyhole contour to determine
$$\int_{-\infty}^{\infty} \frac{\ln^3 (1+x^2)}{1+2x^2} \, dx = \frac{\pi^3 \ln \left(3-2\sqrt{2}\right)}{\sqrt{2}} - \frac{\pi \ln^3 (2)}{\sqrt{2}} - 6 \pi \underbrace{\int_{1}^{\infty} \frac{\ln^2 (t^2-1)}{1-2t^2} \, dt}_{I}$$
A method of evaluating $I$ is as follows: $$I = \int_{0}^{1} \frac{\ln^2 (1-x)}{x^2-2} \, dx + 2\int_{0}^{1}\frac{\ln(1-x) \ln(1+x)}{x^2-2} \, dx + \int_{0}^{1}\frac{\ln^2 (1+x)}{x^2-2} \, dx + 4 \int_{0}^{1}\frac{\ln^2 (x)}{x^2-2} \, dx - 4 \int_{0}^{1} \frac{\ln (1-x^2) \ln (x)}{x^2-2} \, dx$$ for which each of these integrals, Mathematica is able to evaluate in terms of polylogarithms.
Are there any further simplifications, rigorous proofs of the integrals Mathematica is able to determine, or any alternate methods at arriving at the answer?
The answer below is addressed to the part of the OP asking "...rigorous proofs of the integrals Mathematica is able to determine...". Well, believe it or not, you can actually do all extremely elegantly and even get much more.
A solution in large steps by Cornel Ioan Valean
Key steps: $1)$ Derive and prepare to use that $\displaystyle \int_0^{\pi/2}\frac{\cos(2n\theta)}{\cos^2(\theta)+y^2 \sin^2(\theta)}\textrm{d}\theta=\frac{\pi}{2} \frac{1}{y}\left(\frac{y-1}{y+1}\right)^n, \ y>0$; $2)$ then use the Fourier series of $\log^3(\cos(\theta))$; and finally $3)$ employ all necessary generating functions, which happily are known.
Your desired result is: $$\int_0^{\pi/2}\frac{\log^3(\cos(\theta))}{\cos^2(\theta)+y^2 \sin^2(\theta)}\textrm{d}\theta=-\frac{1}{8}\int_0^{\infty} \frac{\log^3(1+x^2)}{1+y^2 x^2}\textrm{d}x$$ $$= \frac{\pi}{8}\left(6\zeta(3)-\log(2)\pi^2-4\log^3(2)\right)\frac{1}{y}+\frac{3}{8}\pi(4\log^2(2)+\pi^2)\frac{1}{y}\log\left(\frac{2y}{1+y}\right)$$ $$-\frac{3}{4}\log(2)\pi\frac{1}{y}\log^2\left(\frac{2y}{1+y}\right)-\frac{\pi}{4}\frac{1}{y}\log^3\left(\frac{2y}{1+y}\right)+\frac{3}{4}\pi\frac{\log(y)}{y}\log^2\left(\frac{2y}{1+y}\right)$$ $$-\frac{3}{4}\pi\frac{\log(1-y)}{y}\log^2\left(\frac{2y}{1+y}\right)-\frac{3}{2}\log(2)\pi \frac{1}{y}\operatorname{Li}_2\left(\frac{1-y}{1+y}\right)-\frac{3}{4}\pi \frac{1}{y}\operatorname{Li}_3\left(\frac{1-y}{1+y}\right)$$ $$-\frac{3}{2}\pi \frac{1}{y}\operatorname{Li}_3\left(\frac{2y}{1+y}\right), \ y>0;$$
And if we want more ...: $$\int_0^{\pi/2}\frac{\log^4(\cos(\theta))}{\cos^2(\theta)+y^2 \sin^2(\theta)}\textrm{d}\theta=\frac{1}{16}\int_0^{\infty} \frac{\log^4(1+x^2)}{1+y^2 x^2}\textrm{d}x$$ $$= \frac{\pi}{96}(7\pi^4+24 \log^2(2)\pi^2+48\log^4(2)-288\log(2)\zeta(3))\frac{1}{y}$$ $$-\frac{1}{2}\log(2)\pi(4\log^2(2)+3\pi^2)\frac{1}{y}\log\left(\frac{2y}{1+y}\right)+\frac{\pi^3}{2}\frac{1}{y}\log^2\left(\frac{2y}{1+y}\right)$$ $$-3\log(2)\pi \frac{\log(y)}{y}\log^2\left(\frac{2y}{1+y}\right)+3\log(2)\pi \frac{\log(1-y)}{y}\log^2\left(\frac{2y}{1+y}\right)$$ $$+\frac{3}{2}\log(2)\pi \frac{1}{y}\log^3\left(\frac{2y}{1+y}\right)+\frac{\pi}{2}\frac{\log(y)}{y}\log^3\left(\frac{2y}{1+y}\right)-\frac{\pi}{2}\frac{\log(1-y)}{y}\log^3\left(\frac{2y}{1+y}\right)$$ $$ -\frac{\pi}{8}\frac{1}{y}\log^4\left(\frac{2y}{1+y}\right)+\frac{\pi}{4}(12\log^2(2)+\pi^2)\frac{1}{y}\operatorname{Li}_2\left(\frac{1-y}{1+y}\right)+3\log(2)\pi \frac{1}{y}\operatorname{Li}_3\left(\frac{1-y}{1+y}\right)$$ $$+6\log(2)\pi \frac{1}{y}\operatorname{Li}_3\left(\frac{2y}{1+y}\right)-\frac{3}{2}\pi\frac{1}{y} \operatorname{Li}_4\left(\frac{1-y}{1+y}\right)-3\pi\frac{1}{y} \operatorname{Li}_4\left(\frac{2y}{1+y}\right)$$ $$ -3\pi \frac{1}{y} \operatorname{Li}_4\left(\frac{y-1}{2y}\right), \ y>0.$$
The Fourier series used above are:
$$ \log^3(\cos(x))$$ $$=-\log^3(2)-\frac{1}{4}\log(2)\pi^2-\frac{3}{2}\zeta(3)+\sum_{n=1}^{\infty}(-1)^{n-1} \biggr(\left(3\log^2(2)+\frac{\pi^2}{4}\right)\frac{1}{n}$$ $$-3\log(2)\frac{1}{n^2}+\frac{3}{2}\frac{1}{n^3}+6\log(2)\frac{H_n}{n}-3\frac{H_n}{n^2}+3\frac{H_n^2}{n}\biggr) \cos (2 n x),\ -\frac{\pi}{2}<x<\frac{\pi}{2}$$ and $$\log^4(\cos(x))$$ $$=\log^4(2)+\frac{1}{2}\log^2(2)\pi^2+6\log(2)\zeta(3)+\frac{19}{240}\pi^4$$ $$-\sum_{n=1}^{\infty}(-1)^{n-1}\biggr((4\log^3(2)+\log(2)\pi^2+6\zeta(3))\frac{1}{n}-\left(6\log^2(2)+\frac{\pi^2}{2}\right)\frac{1}{n^2}$$ $$+6\log(2)\frac{1}{n^3}-3\frac{1}{n^4}+(12\log^2(2)+\pi^2)\frac{H_n}{n}-12\log(2)\frac{H_n}{n^2}+6\frac{H_n}{n^3}+12\log(2)\frac{H_n^2}{n}$$ $$-6\frac{H_n^2}{n^2}+4\frac{H_n^3}{n}+2\frac{H_n^{(3)}}{n}\biggr) \cos (2 n x),\ -\frac{\pi}{2}<x<\frac{\pi}{2}.$$
A final note: All the details of the results above with full solutions will be found in More (Almost) Impossible Integrals, Sums, and Series (2023), the sequel of (Almost) Impossible Integrals, Sums, and Series (2019).