Integrate \(\int_{0}^{\infty} \frac{\mathrm{ln}^{3}(x)}{(1+x^{2})(1+x)^{2}} \mathrm{d} x \)

The following integral was a question on Mathematics Stack Exchange.

\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{3}(x)}{(1+x^{2})(1+x)^{2}} \mathrm{d} x

As usual, there are multiple clever solutions. However, a user noted that the integral could be evaluated via the Mellin transform but he did not provide any details so I will do it here.

Let us evaluate it via the Mellin transform
\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x

f(x) = \frac{1}{(1+x^{2})(1+x)^{2}} = -\frac{1}{2}\frac{x}{x^{2}+1} + \frac{1}{2}\frac{1}{x+1} + \frac{1}{2}\frac{1}{(x+1)^{2}}
via partial fraction expansion.

Applying the Mellin transform, yields
\mathcal{M}[f(x)](s) & = \int\limits_{0}^{\infty} \frac{x^{s-1}}{(1+x^{2})(1+x)^{2}} \\
& = -\frac{1}{2}\left[\frac{1}{2}\pi\sec\left(\frac{\pi}{2}s\right)\right] + \frac{1}{2}\pi\csc(\pi s) + \frac{1}{2} \mathrm{B}(s,2-s)

Taking the 3rd derivative of equation \eqref{eq:160813a4} with respect to s and then taking \(\lim s \to 1\) yields
\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{3}(x)}{(1+x^{2})(1+x)^{2}} \mathrm{d} x & = -\frac{1}{2}\left( -\frac{7}{960} \pi^{4} \right) + \frac{1}{2} \left( -\frac{7}{60} \pi^{4} \right) + \frac{1}{2}0 \\
& = -\frac{7}{128} \pi^{4}

Let us fill in the details. Handling the beta function first, we have
\mathrm{B}(s,2-s) = \frac{\Gamma(s)\Gamma(2-s)}{\Gamma(2)} = \Gamma(s)\Gamma(2-s)
To take derivatives, we note that
\frac{\mathrm d}{\mathrm d s} \Gamma(s) = \Gamma(s) \psi^{(0)}(s) \quad \mathrm{and} \quad \psi^{(n)}(s) = \frac{\mathrm{d}^{n}}{\mathrm{d} s^{n}} \psi^{(0)}(s)
Where \(\psi^{(n)}(s)\) is the polygamma function.

Taking the third derivative of equation \eqref{eq:160813a6} and letting \(\lim s \to 1\) equals 0. Here we used
\psi^{(0)}(1) = -\gamma \quad \mathrm{and} \quad \psi^{(1)}(1) = \frac{\pi^{2}}{6}
and fortunately \(\Gamma^{(3)}(s) = -\Gamma^{(3)}(2-s)\) which leads to some cancellations.

Doing the same for the first two terms on the right hand side of equation \eqref{eq:160813a4}, we have to be careful. Each of them individually goes to \(\infty\) as \(\lim s \to 1\) but the \((s-1)^{-4}\) terms in the Laurent expansions about \(s=1\) cancel. Here I used Wolfram Alpha. Doing it with the two terms combined yielded our final answer.

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