Lobachevsky integral formula

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

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

• Hardy, G. H., The Integral ${\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x}}\,dx={\frac {\pi }{2}},}$ The Mathematical Gazette, Vol. 5, No. 80 (June–July 1909), pp. 98–103 JSTOR 3602798
• Dixon, A. C., Proof That ${\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x}}\,dx={\frac {\pi }{2}},}$ The Mathematical Gazette, Vol. 6, No. 96 (January 1912), pp. 223–224. JSTOR 3604314