Lobachevsky integral formula

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Let ${\displaystyle f(x)}$ be a continuous function satisfying the ${\displaystyle \pi }$-periodic assumption ${\displaystyle f(x+\pi )=f(x)}$, and ${\displaystyle f(\pi -x)=f(x)}$, for ${\displaystyle 0\leq x<\infty }$. If the integral ${\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x}}f(x)\,dx}$ is taken to be an improper Riemann integral, we have Lobachevsky's Dirichlet integral formula

${\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}x}{x^{2}}}f(x)\,dx=\int _{0}^{\infty }{\frac {\sin x}{x}}f(x)\,dx=\int _{0}^{\pi /2}f(x)\,dx}$

Moreover, we have the following identity as an extension of the Lobachevsky Dirichlet integral formula^[1]

${\displaystyle \int _{0}^{\infty }{\frac {\sin ^{4}x}{x^{4}}}f(x)\,dx=\int _{0}^{\pi /2}f(t)\,dt-{\frac {2}{3}}\int _{0}^{\pi /2}\sin ^{2}tf(t)\,dt.}$

As an application, take ${\displaystyle f(x)=1}$. Then

${\displaystyle \int _{0}^{\infty }{\frac {\sin ^{4}x}{x^{4}}}\,dx={\frac {\pi }{3}}.}$