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

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

• Try differentiating the LHS with respect to $a$. – Starfall Apr 19 '18 at 14:11
• Using Fubini, write $\frac{\sin x}{x}$ as $\int_0^1 f(t,x)\,dt$. – Daniel Fischer Apr 19 '18 at 14:12
• I suspect, the Laplace transform might be helpful here. Integrals from $0$ to infinity with an $x$ in a denominator can often be done that way, as the denominator's $x$ goes away after performing Laplace appropriately through $L(xf(x))=-F'(s)$ – imranfat Apr 19 '18 at 14:12
• @Starfall Unfortunately we did not go over interchanging the derivative and integral. – Helix Apr 19 '18 at 14:12
• Actually, I don't suspect, I KNOW it can be done through Laplace; looking over my notes, but I have class coming up now... – imranfat Apr 19 '18 at 14:14

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, $$\int_0^{\infty} e^{-ax} \frac{\sin x}{x}\,dx=\frac{\pi}{2}-\arctan a\qquad \text{for } a>0$$
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 $$\frac{1}{2} \int_{-1}^1 \frac{a \, dt}{a^2+t^2} = \int_0^1 \frac{a \, dt}{a^2+t^2} = \arctan{(1/a)} = \operatorname{arccot}{a} = \frac{\pi}{2}-\arctan{a}.$$
• Alternatively, use $${1\over2}\int_{-1}^1\cos(xt)\,dt={\sin x\over x}$$ and then (after Fubinating) integrate $$\int_0^\infty e^{-ax}\cos(xt)\,dx$$ (by parts, twice). – Barry Cipra Apr 19 '18 at 14:28
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 $\mathcal{L}$.
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 $$\arctan\left(\frac{1}{x}\right)=\frac{\pi}{2}-\arctan\left(x\right),\, x>0.$$