一个关于sin的有趣的积分
Created at 2018-01-28 Updated at 2018-02-13 Category math / calculus
Solving the integral
$$\int_{0}^{+\infty}\frac{\sin x}{x}dx,$$
One easiest way to get this integral is to evaluate the following improper integral with parameter \(a\):
$$ I(a)=\int_0^\infty e^{-ax}\frac{\sin x}{x}dx, a\ge 0.$$
It is easy to see
$$I’(a)=-\int_0^\infty e^{-ax}\sin xdx=\frac{e^{-ax}}{a^2+1}(a\sin x+\cos x)\big|_0^\infty=-\frac{1}{a^2+1}.$$
Thus
$$I(\infty)-I(0)=-\int_0^\infty\frac{1}{a^2+1}da=-\frac{\pi}{2}.$$
Note \(I(\infty)=0\) and hence \(I(0)=\frac{\pi}{2}\).
$$I’(a) = \int_{0}^{\infty}-e^{-ax} \sin xdx $$
$$= \int_{0}^{\infty}e^{-ax} d\cos x $$
$$= e^{-ax}\cos x - -\int_{0}^{\infty}e^{-ax}\cos xdx$$
$$= \left[e^{-ax}\cos x + a\left[e^{-ax}\sin x + \int_{0}^{\infty}ae^{-ax}\sin xdx\right] \right]^{\infty}_0 $$
$$= \left[ e^{-ax}\cos x + ae^{-ax}\sin x - a^2 I’(a) \right]^{\infty}_0 $$
$$\therefore I’(a)= \frac{e^{-ax}}{1+a^2}(a\sin x + \cos x)\big|_0^\infty$$