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If you take the counter-clockwise contour of a closed semi-circle with radius $R$ in the complex plane with a contour along the real line from $-R$ to $+R$, we have
$$I_0=\int_{-\infty}^\infty\frac{1-x}{1-x^5}\ dx=P.V.\int_{-\infty}^\infty\frac{1-x}{1-x^5}\ dx$$
$$\lim_{R\to\infty}\oint_{\gamma...12
You may notice that:
$$ \frac{1}{n^2+1} = \int_{0}^{+\infty}\sin(x)e^{-nx}\,dx\tag{1}$$
from which$^{(*)}$:
$$ S=\sum_{n\geq 0}\frac{(-1)^n}{n^2+1}=1+\int_{0}^{+\infty}\sin(x)\sum_{n\geq 1}(-1)^n e^{-nx}\,dx \tag{2}$$
and:
$$ S = 1-\int_{0}^{+\infty}\frac{\sin(x)}{e^x+1}\,dx =\color{red}{\frac{1}...
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