SolAug2706

Sketch: Using the reflection formula for the gamma function

$\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$

we can rewrite the integral as

$\frac12 \int_0^1 \log \left(\frac{\pi}{\sin(\pi x)}\right) \, dx$

which evaluates to $log(2π) / 2$

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