Prove
$$\int_{0}^{\infty}\frac{|\sin x|\sin x}{x}dx=1.$$
I know how to calculate $\int_{0}^{\infty}\frac{\sin x}{x}dx=\frac{\pi}{2}$, but the method cannot be applied here. So I am thinking
$$\sum_{k=0}^n(-1)^k\int_{k\pi}^{(k+1)\pi}\frac{\sin^2 x}{x}dx$$
but I don't know how to proceed.
Answer
By Lobachevsky integral formula: https://en.wikipedia.org/wiki/Lobachevsky_integral_formula
$$\int_{0}^{\infty}\frac{\sin x}{x}|\sin x|\,\mathrm{d}x=\int_0^{\pi/2}|\sin x|\,\mathrm{d}x=1.$$
No comments:
Post a Comment