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8

It is a consequence of the $\Gamma$ reflection formula: $$\Gamma(z)\,\Gamma(1-z) = \frac{\pi}{\sin(\pi z)} \tag{1}$$ and the Cantarini's trick (aka the Laplace transform of the sine function): $$\int_{0}^{+\infty} \sin(a t)\,e^{-bt} = \frac{a}{a^2+b^2}\tag{2}$$ from which:  \frac{1}{\pi^2+\log^2(x)} = \int_{0}^{+\infty} \frac{\sin(\pi t)}{\pi} ...

0

Hint : Apply Liouville theorem to $\frac{1}{f}$.

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Show that $g(z):=\frac1{f(z)}$ is constant.

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