Lobachevsky Integral Formula

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $f$ be a continuous function, periodic in $\pi$.


Then:

$\displaystyle \int_0^\infty \frac {\sin x} x f \left({x}\right) \rd x = \int_0^{\frac \pi 2} f \left({x}\right) \rd x$


Proof


Source of Name

This entry was named for Nikolai Ivanovich Lobachevsky.