Question

Evaluate ${\int }_0^{\infty }\sin xdx$

Answer 1

Let us first evaluate;
$I=\int e^{-sx}\sin xdx$ and $J=\int e^{-sx}\cos xdx$
Using integer by parts, we get
$I=-e^{-sx}\cos x-sJ$ (i)
$J=e^{-sx}\sin x+sI$ (ii)
Subtracting equations (i) and (ii), we get
$I=-e^{-sx}\left|\frac{\cos x+s\sin x}{1+s^2}\right|$
$⟹J=e^{-sx}|\sin x-\frac{s^2}{s^2+1}\sin x-\frac{s}{s^2+1}\cos x|$
$e^{-sx}\left|\frac{\sin x-s\cos x}{1+s^2}\right|$
Thus, ${\int }_0^{\infty }{e}^{-sx}\sin xdx=\frac{1}{s^2+1}$
${\int }_0^{\infty }{e}^{-sx}\cos xdx=\frac{s}{s^2+1}$
Now, ${\int }_0^{\infty }\sin xdx=\underset{s\to 0}{lim}{\int }_0^{\infty }{e}^{-sx}\sin xdx=\underset{s\to 0}{lim}\frac{1}{s^2+1}=1$