Lobachevsky Integral Formula

Theorem

Let $f$ be a continuous real function, periodic with period $\pi$.

Then:

$\ds \int_0^\infty \frac {\sin x} x \map f x \rd x = \int_0^{\frac \pi 2} \map f x \rd x$

Proof

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Source of Name

This entry was named for Nikolai Ivanovich Lobachevsky.