An interesting integral!

[Hamza Mahmood]

\[ \int_{0}^{\infty }\frac{\ln^{2}x}{x^2+2x\cos\theta+1 }dx=?\ where\ \theta\in [0,\pi ] \]

[mathofusva]

Notice that
\begin{eqnarray*}
\int_0^\infty\,\frac{\ln^2x}{1 + 2x\cos\theta + x^2}\,dx & = & \int_0^1\,\frac{\ln^2x}{1 + 2x\cos\theta + x^2}\,dx + \int_1^\infty\,\frac{\ln^2x}{1 + 2x\cos\theta + x^2}\,dx\\
& = & 2\,\int_0^1\,\frac{\ln^2x}{1 + 2x\cos\theta + x^2}\,dx,
\end{eqnarray*}
where the substitution $x = 1/t$ has been used in the integral over $[1, \infty)$.

Recall that
$$\sum_{k=1}^\infty\,x^k\sin k\alpha = \frac{x\sin\alpha}{1 - 2x\cos\alpha + x^2}.$$
Replacing $\alpha$ by $\pi + \theta$ yields
$$\sum_{k=1}^\infty\,(-1)^{k-1}x^k\sin k\theta = \frac{x\sin\theta}{1 + 2x\cos\theta + x^2}.$$
Thus,
\begin{eqnarray*}
\int_0^1\,\frac{\ln^2x}{1 + 2x\cos\theta + x^2}\,dx & = & \sum_{k=1}^\infty\,(-1)^{k-1}\frac{\sin k\theta}{\sin\theta}\,\int_0^1\,x^{k-1}\ln^2x\,dx\\
& = & \frac{2}{\sin\theta} \sum_{k=1}^\infty\,(-1)^{k-1}\frac{\sin k\theta}{k^3}.
\end{eqnarray*}
Since
$$\sum_{k=1}^\infty\,(-1)^{k-1}\frac{\cos k\theta}{k^2} = \frac{\pi^2}{12} - \frac{1}{4}\theta^2,$$
integrating with respect to $\theta$ gives
$$\sum_{k=1}^\infty\,(-1)^{k-1}\frac{\sin k\theta}{k^3} = \frac{\theta}{12}(\pi^2- \theta^2).$$
Combining the results above asserts that
$$\int_0^\infty\,\frac{\ln^2x}{1 + 2x\cos\theta + x^2}\,dx = \frac{\theta}{3\sin\theta}(\pi^2- \theta^2).$$

Remark. If $p$ is a positive integer, along the same lines, one can establish that
$$\int_0^\infty\,\frac{\ln^{p}(1/x)}{1 + 2x\cos\theta + x^2}\,dx = \frac{2\,p!}{\sin\theta}\, \sum_{k=1}^\infty\,(-1)^{k-1}\frac{\sin k\theta}{k^{p+1}}.$$

[Hamza Mahmood]

Thank you!!