Improper integral of $\frac{\sin x}{x}e^{-ax}$

Question

For $a>0$ define $$I(a)=\int_0^\infty \frac{\sin x}{x}e^{-ax} \, dx,$$ I can show it is continuous at $0$, but by differentiating in $a$, I can't see why $$I(a)=\frac{\pi}{2}-\arctan(a).$$ Thanks for any hints!

Answer 1

Note that

\begin{align} {dI \over da} & = {d \over da}\left( \int_0^\infty {\sin x \over x} e^{-ax} \, dx \right) = \int_0^\infty {\sin x \over x}{d \over {da}} (e^{-ax}) \, dx = \int_0^\infty {\sin x \over x} (-x e^{-ax}) \, dx \\[10pt] & = - \int_0^\infty (\sin x) e^{ - ax} \, dx \\[10pt] I'(a) & = - \int_0^\infty (\sin x) e^{-ax} \, dx = \left. {e^{-ax} (a\sin x + \cos x) \over 1 + a^2} \right|_0^\infty = 0 - {1 \over 1 + a^2} \\[10pt] I'(a) & = - {1 \over 1 + a^2} \\[10pt] I(a) & = - \arctan (a) + C \end{align}

Also we have

$$\mathop {\lim }\limits_{a \to \infty } I(a) = 0$$

and hence

$$ - \lim_{a \to \infty } \arctan (a) + C = 0\,\,\, \to \,\,\,\, - {\pi \over 2} + C = 0\,\,\,\, \to \,\,\,\,C = {\pi \over 2}$$

Answer 2

Notice, we know the Laplace Transform of $\sin x$ $$L[\sin x]=\int_{0}^{\infty}e^{-st}\sin t\ dt=\frac{1}{s^2+1}$$

Using property of Laplace transform as follows $$I(a)=\int_{0}^{\infty}e^{-ax}\ \frac{\sin x}{x}\ dx$$$$=\int_{a}^{\infty}L[\sin x]\ ds$$ $$=\int_{a}^{\infty}\frac{1}{s^2+1}\ ds$$ $$=[\tan^{-1}(s)]_{a}^{\infty}$$ $$=\tan^{-1}(\infty)-\tan^{-1}(a)$$ $$=\frac{\pi}{2}-\tan^{-1}(a)$$