اگر $f$ یک تابع و برای هر عدد حقیقی $x$ و هر عدد حسابی $n$ داشته باشیم $f(x^+_-n\pi)=f(x)$ آنگاه،
اگر $f$ تابعی زوج باشد:
$$ 1)\int_0^ \infty\frac{sin(x)}{x}f(x)=\int_0^ \infty\frac{sin^2(x)}{x^2}f(x)= \int_0^ \frac{\pi}{2}f(x)dx $$
واگر $f$ تابعی فرد باشد:
$$2)\int_0^ \infty\frac{sin(x)}{x}f(x)dx= \int_0^ \frac{\pi}{2}cos(x)f(x)dx$$
اثبات1):
$$I:=2\int_0^ \infty\frac{sin(x)}{x}f(x)dx=\int_{- \infty}^ \infty\frac{sin(x)}{x}f(x)dx$$
$$=\int_{- \infty }^0\frac{sin(x)}{x}f(x)dx+\int_0^ \infty\frac{sin(x)}{x}f(x)dx$$
$$\int_{\frac{-\pi}{2}}^0\frac{sin(x)}{x}f(x)dx+\int_\frac{-3\pi}{2}^{\frac{-\pi}{2}}\frac{sin(x)}{x}f(x)dx+\int_ \frac{-5\pi}{2} ^{\frac{-3\pi}{2}}\frac{sin(x)}{x}f(x)dx+...$$
$$+\int_0^{\frac{\pi}{2}}\frac{sin(x)}{x}f(x)dx+\int_\frac{\pi}{2}^{\frac{3\pi}{2}}\frac{sin(x)}{x}f(x)dx+\int_ \frac{3\pi}{2} ^{\frac{5\pi}{2}}\frac{sin(x)}{x}f(x)dx+...$$
$$=\int_ \frac{-\pi}{2} ^{\frac{\pi}{2}}\frac{sin(x)}{x}f(x)dx+\int_\frac{-3\pi}{2}^{\frac{3\pi}{2}}\frac{sin(x)}{x}f(x)dx+\int_ \frac{-5\pi}{2} ^{\frac{5\pi}{2}}\frac{sin(x)}{x}f(x)dx+...$$
$$ \sum_{n=0}^ \infty \int_ {(k-\frac{1}{2})\pi} ^{(k+\frac{1}{2})\pi} \frac{sin(x)}{x}f(x)dx=\int_ {(k-\frac{1}{2})\pi} ^{(k+\frac{1}{2})\pi}\sum_{n=0}^ \infty \frac{sin(x)}{x}f(x)dx $$
حالا با تغییر متغیر $t:=x-n\pi$ داریم:
$$2I=\int_ \frac{-\pi}{2}^\frac{\pi}{2}\sum_{n=0}^ \infty \frac{sin(t)(-1)^n}{t+n \pi}f(t)dt=\int_ \frac{-\pi}{2}^\frac{\pi}{2}sin(t)f(t)(\sum_{n=0}^ \infty \frac{(-1)^n}{t+n \pi})dt$$
$$=\int_ \frac{-\pi}{2}^\frac{\pi}{2}sin(t).csc(t).f(t)dt=\int_ \frac{-\pi}{2}^\frac{\pi}{2}f(t)dt=2\int_0^\frac{\pi}{2}f(t)dt$$
$$ \Rightarrow \int_0^ \infty\frac{sin(x)}{x}f(x)= \int_0^ \frac{\pi}{2}f(x)dx$$
اثبات قسمت دوم $1$ و $2$ مشابه است.
مثال:
$$2)\int_0^ \infty\frac{sin(x)}{x}dx=\int_0^ \frac{\pi}{2}dx=\frac{\pi}{2}$$
$\Box$
