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Gamma function

Beta function

For $0<a<2$,
$$
I(a)=\int_0^\infty\frac{x^{a-1}}{1+x^2}\,dx
=\frac\pi2\csc\frac{\pi a}{2}.
$$
Differentiating at $a=1$ and expanding the secant gives
$$
\int_0^\infty\frac{(\log x)^2}{1+x^2}\,dx=\frac{\pi^3}{8},
\qquad
\int_0^\infty\frac{(\log x)^4}{1+x^2}\,dx=\frac{5\pi^5}{32}.
$$

Logarithmic moments of the Cauchy kernel

Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 12B / Solution

Logarithmic moments of the Cauchy kernel

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Logarithmic moments of the Cauchy kernel

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