Tuesday, 8 March 2016

calculus - Evaluating the integral: $intlimits_{0}^{infty}left(frac{sin(ax)}{x}right)^2 dx , a neq 0 $




Integrate from 0 to infinity
$$\int\limits_{0}^{\infty}\left(\frac{\sin(ax)}{x}\right)^2 dx , a \neq 0 $$





I tried evaluating the indefinite integral of that function using Sine integral. But I failed to do it. I am having no idea to evaluate the definite integral. I seek for help


Answer



First from here https://math.stackexchange.com/q/2508827 it is easy to prove using the changes $u=2y$ that $$\int_0^\infty \frac{\sin(u)}{u}dx= \int_0^\infty\left(\frac{\sin(u)}{u}\right)^2dx$$



Inserting the change of variables $u=ax$ one gets



$$\int_0^\infty\left(\frac{\sin(ax)}{x}\right)^2dx = a\int_0^\infty\left(\frac{\sin(ax)}{ax}\right)^2d(ax)\\= a \operatorname{sign}(a)\int_0^\infty\left(\frac{\sin(u)}{u}\right)^2dx =|a|\int_0^\infty \frac{\sin(u)}{u}dx =\color{blue}{\frac{|a|π}{2} }$$




this comes from the Dirichlet integral Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...