Grigorios Kostakos wrote:$1.\displaystyle \int_{0}^{+\infty}\frac{\log^n{x}}{1+x^2}\,{\rm{d}}x$
Well, for the first one for the moment... We have successively:
\begin{align*} \int_{0}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x &= \int_{0}^{1} \frac{\log^n x}{1+x^2}\, {\rm d}x + \int_{1}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x\\ &= \int_{0}^{1} \frac{\log^n x}{1+x^2}\, {\rm d}x - \int_{1}^{0} \frac{\log^n \left ( \frac{1}{x} \right )}{1+ \left ( \frac{1}{x} \right )^2} \frac{1}{x^2} \, {\rm d}x\\ &= \int_{0}^{1} \frac{\log^n x}{1+x^2} \, {\rm d}x + \int_{0}^{1} \frac{\log^n \left ( \frac{1}{x} \right )}{1+x^2}\, {\rm d}x\\ &= \int_{0}^{1} \frac{\log^n x}{1+x^2}\, {\rm d}x+ \int_{0}^{1} \frac{(-1)^n \log^n x}{1+x^2} \, {\rm d}x \\ &= \left ( 1+(-1)^n \right ) \int_{0}^{1} \frac{\log^n x}{1+x^2} \, {\rm d}x\\ &=\left ( 1+(-1)^n \right )\int_{0}^{1} \log^n x \sum_{k=0}^{\infty} (-1)^k x^{2k} \, {\rm d}x \\ &= \left ( 1+(-1)^n \right ) \sum_{k=0}^{\infty} (-1)^k \int_{0}^{1}x^{2k} \log^n x \, {\rm d}x\\ &= \left ( 1+(-1)^n \right )(-1)^n n! \sum_{k=0}^{\infty} \frac{(-1)^k}{\left ( 2k+1 \right )^{n+1}} \\ &= \left ( 1+(-1)^n \right )(-1)^n n! \sum_{k=0}^{\infty} \left [ \frac{1}{\left ( 4k+1 \right )^{n+1}} - \frac{1}{\left ( 4k+3 \right )^{n+1}} \right ] \\ &=\left ( 1+(-1)^n \right )(-1)^n n! \frac{1}{4^{n+1}} \sum_{k=0}^{\infty} \left [ \frac{1}{\left ( k+ \frac{1}{4} \right )^{n+1}} - \frac{1}{\left ( k+\frac{3}{4} \right )^{n+1}} \right ] \\ &= \frac{1}{4^{n+1}}\left ( 1+(-1)^n \right ) \left [ \psi^{(n)} \left ( \frac{3}{4} \right ) - \psi^{(n)} \left ( \frac{1}{4} \right )\right ] \end{align*}
Thus, distinguishing cases for $n$ (either it is odd or even) we get the following
beautiful closed form:
$$\int_{0}^{\infty}\frac{\log^n x}{1+x^2}\, {\rm d}x = \left\{\begin{matrix} 0 &, &n \; {\rm odd} \\\\ \displaystyle \frac{2}{4^{n+1}}\left [ \psi^{(n)} \left ( \frac{3}{4} \right ) - \psi^{(n)} \left ( \frac{1}{4} \right ) \right ]& ,& n \; {\rm even} \end{matrix}\right.$$
Recalling also that , if $n$ is even:
$$\psi^{(n)}(1-z)-\psi^{(n)} (z)= \pi \; \frac{\mathrm{d}^{n} }{\mathrm{d} x^{n}} \cot \pi z \tag{ polygamma reflection formula}$$
holds, the above formula reduces down to - the not so beautiful -
$$\int_{0}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x = \left[ \frac{2\pi}{4^{n+1}} \frac{\mathrm{d}^n }{\mathrm{d} x^n} \cot \pi z \right]_{z=1/4} \;\; n \; \; {\rm even}$$