Evaluation of $\int_{-\infty}^\infty \frac{\sin(at)}t\ dt$

Question

I just need to know if this integral converges. If it does, what is the value?

If not can anyone tell me what is the value of the $$ \int_{-\infty}^\infty 2\,\frac{\cos(w)\sin(w)}{w}\ dw ? $$

Answer 1

The improper integral $$\int_{-\infty}^{+\infty}\frac{\sin(at)}{t}\,dt$$ converges. Indeed, consider $$\int_{c}^{d}\frac{\sin(at)}{t}\,dt$$ If we substitue $x=at$, $dx=adt$, $$\int_{c}^{d}\frac{\sin(at)}{t}\,dt=\int_{ac}^{ad}\frac{\sin x}{x}\,dx$$ If $a>0$, as $c\to -\infty$ and $d\to +\infty$, $$\int_{-\infty}^{+\infty}\frac{\sin(at)}{t}\,dt=\int_{-\infty}^{+\infty}\frac{\sin x}{x}\,dx=\pi$$ I will leave the cases $a=0$, $a<0$ to the OP. We also have that $$ \int_{-\infty}^\infty 2\,\frac{\cos(w)\sin(w)}{w}\ dw= \int_{-\infty}^\infty \frac{\sin(2w)}{w}\ dw=...$$

Answer 2

In light of @Sasha's comment, $\int_0^{\infty}\frac{\sin x}{x}dx$ converges. Since $\int_0^{1}\frac{\sin x}{x}dx$ converges (because $\frac{\sin x}{x}$ is continous in $(0,1]$ and its limit is equal to $1$ when $x\to 0^+$) wee need just to show that $\int_1^{\infty}\frac{\sin x}{x}dx$ converges. Doing the integration by parts, so we have $$\int_1^{a}\frac{\sin x}{x}dx=\frac{-\cos x}{x}\big|_1^a+\int_1^{a}\frac{\cos x}{x^2}dx=\cos1-\frac{\cos a}{a}+\int_1^{a}\frac{\cos x}{x^2}dx$$ Now by taking the limit of both sides when $a\to \infty$ and the fact that $\lim_{a\to\infty}\frac{\cos a}{a}=0$ we have $$\int_1^{\infty}\frac{\sin x}{x}dx=\cos1+\int_1^{\infty}\frac{\cos x}{x^2}dx$$ But the last interal on the RHS converges(why?).

Answer 3

Setting $u= at$ then , since the integrand is an odd function we get

$$\int_{-\infty}^{+\infty}\frac{\sin(at)}{t}\,dt = 2sign(a)\int_{0}^{+\infty}\frac{\sin(at)}{t}\,dt=2sign(a)\int_{0}^{+\infty}\frac{\sin(t)}{t}\,dt= sign(a)\pi $$

See here, for the convergence, Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?