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 $f(\pi-x)=f(x)$, $0\leq x<\infty$. If the integral $\int_0^\infty \frac{\sin^4x}{x^4}f(x)dx$ defined in the sense of the improper Riemann integral, then we show the following equality $$\int_0^\infty \frac{\sin^4x}{x^4}f(x)dx=\int_0^{\frac{\pi}{2} }f(t)dt-\frac{2}{3}\int_0^{\frac{\pi}{2} }\sin^2tf(t)dt$$ hence if we take $f(x)=1$, then we have $$\int_0^\infty \frac{\sin^4x}{x^4}dx=\frac{\pi}{3}$$ Moreover, we give a method for computing $\int_0^\infty \frac{\sin^{2n}x}{x^{2n}}f(x)dx$ for $n\in \mathbb N$Comment: Paper is re-written with new presentation and new results adde