Post #4520182

Lobachevsky's integral formula \[\displaystyle \begin{aligned} \int_{0}^{\infty} \frac{\sin^{2}x}{x^{2}}\,f(x)\,dx \\[0.5em] = \int_{0}^{\infty} \frac{\sin x}{x}\,f(x)\,dx \\[0.5em] = \int_{0}^{\pi/2} f(x)\,dx \end{aligned} \]