Showing that $\int_0^{\infty} e^{-ax} \frac {\sin x}{x} dx = \frac {\pi}{2} - \arctan (a)$ for any $a >0$

Question

$\displaystyle \int_0^{\infty} e^{-ax} \dfrac {\sin x}{x} dx = \dfrac {\pi}{2} - \arctan (a)$ for any $a >0$

I am really not sure how to go about doing this. I checked Wolfram Alpha and the function does not have an elementary antidericative. I do not have that many special tools (mostly just Fubini's Theorem). My best guess is that we have to do a substitution of some kind, and maybe integration by parts. Then hopefully we will get a "known" integral. However, I don't really know in what direction to go.

Answer 1

You have $$ \frac{1}{2}\int_{-1}^1 e^{ixt} \, dt = \frac{\sin{x}}{x}. $$ Insert this and change the order of integration with Fubini and you just have to calculate $$ \frac{1}{2}\int_{-1}^{1} \frac{dt}{a-it}. $$ You can get rid of the imaginary part by noting that $$ \frac{1}{a-it} = \frac{a+it}{a^2+t^2} $$ and the latter term is odd, so the integral is equal to $$ \frac{1}{2} \int_{-1}^1 \frac{a \, dt}{a^2+t^2} = \int_0^1 \frac{a \, dt}{a^2+t^2} = \arctan{(1/a)} = \operatorname{arccot}{a} = \frac{\pi}{2}-\arctan{a}. $$

Answer 2

Let $$I(a)=\int_0^{\infty} e^{-ax} \frac{\sin x}{x}\,dx\qquad \text{for } a>0$$ Now differentiate with respect to $a$, \begin{align} I'(a)&=-\int_0^{\infty} e^{-ax}\sin x\,dx\\ &=-\frac{1}{a^2+1} \qquad (\text{integration by parts})\\ \implies I(a)=C-\arctan a \end{align} From the initial integral, we know that $\lim_{a\to\infty} I(a)=0$. We can thus find the constant $C=\pi/2$. Thus, $$\int_0^{\infty} e^{-ax} \frac{\sin x}{x}\,dx=\frac{\pi}{2}-\arctan a\qquad \text{for } a>0$$

Answer 3

By a fundamental property of the Laplace transform, if $\frac{f(x)}{x}$ is improperly Riemann-integrable over $\mathbb{R}^+$ we have $$ \int_{0}^{+\infty}\frac{f(x)}{x}\,dx = \int_{0}^{+\infty}(\mathcal{L} f)(s)\,ds.$$ On the other hand $$ \mathcal{L}\left(e^{-ax}\sin(x)\right)(s) = \frac{1}{1+(a+s)^2} $$ is trivial by integration by parts, hence the original integral equals $$ \int_{a}^{+\infty}\frac{ds}{1+s^2} = \frac{\pi}{2}-\arctan a $$ as wanted. Here the application of Fubini's theorem is hidden in the invoked property of $\mathcal{L}$.

Answer 4

Since $$I=\int_{0}^{\infty}\frac{e^{-ax}\sin\left(x\right)}{x}dx=\frac{1}{2i}\int_{0}^{\infty}\frac{e^{x\left(-a-i\right)}-e^{x\left(-a+i\right)}}{x}dx$$ then we can apply the complex version of the Frullani's theorem and get the result, recalling that $$\arctan\left(\frac{1}{x}\right)=\frac{\pi}{2}-\arctan\left(x\right),\, x>0.$$