An extension of Lobachevsky formula

In this paper we extend the Dirichlet integral formula of Lobachevsky. Let $f(x)$ be a continuous function and satisfy in the $\pi$-periodic assumption $f(x+\pi )=f(x)$, and $f(\pi -x)=f(x)$, $0\le x<\mathrm{\infty }$. If the integral ${\int }_0^{\mathrm{\infty }}\frac{{\sin }^{4}x}{x^4}f(x)dx$ defined in the sense of the improper Riemann integral, then we show the following equality $${\int }_0^{\mathrm{\infty }}\frac{{\sin }^{4}x}{x^4}f(x)dx={\int }_0^{\frac{\pi }{2}}f(t)dt-\frac{2}{3}{\int }_0^{\frac{\pi }{2}}{\sin }^{2}tf(t)dt$$ hence if we take $f(x)=1$, then we have $${\int }_0^{\mathrm{\infty }}\frac{{\sin }^{4}x}{x^4}dx=\frac{\pi }{3}$$ Moreover, we give a method for computing ${\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2n}x}{x^{2n}}f(x)dx$ for $n\in \mathbb{N}$