# find the integral $\int ^\infty_0e^{-xu}\frac{\sin u}{u}du$

Find the integral $$\int ^\infty_0e^{-xu}\frac{\sin u}{u}du$$

my idea

$L\left(\frac{1}{t}f(t)\right)=\int_{s}^{\infty}L(f(t))dt$

$L(\sin bt)=t=\frac{1}{1+s^2}$

$\int ^\infty_0e^{-xu}\frac{\sin u}{u}du$ ?? how we processed

• Does this involve complex numbers? – Yip Jung Hon May 1 '18 at 5:35


In honour of Cleo:

$$\tan^{-1}\Big(\frac{1}{x}\Big)$$

• Out of curiosity, has anybody seen Cleo of late. Haven't seen her in MSE for the last two years ... – Nilotpal Kanti Sinha May 1 '18 at 6:39

In honor of Feynman:

Let us define (with $x>0$) $$I(x) = \int_0^\infty e^{-x u} \frac{\sin u}{u} du\;.$$

Differentiating (under the integral sign), we obtain $$I'(x) = -\int_0^\infty e^{-x u} \sin u\;du = -\operatorname{Im}\int_0^\infty e^{-xu +i u}\;du =\operatorname{Im} \frac1{i-x} = - \frac{1}{1+x^2}\;.$$

We thus obtain $$I(x) = -\arctan(x) + c \;.$$ It is easy to see that $I(u) \to 0$ for $x \to \infty$, this sets $c=\pi/2$ and thus $$I(x) = \frac{\pi}{2} -\arctan(x) = \arctan(1/x)\;.$$

• Good day, I've not seen integrating techniques like these before. May I ask what are these integrating techniques you've used here called? – Yip Jung Hon May 1 '18 at 6:13
• @YipJungHon: it is called Differentiation Under the Integral Sign. – Fabian May 1 '18 at 12:53

Sure your idea works: Let $f(t)=\sin(t)$, then \begin{align} \int_0^{\infty}e^{-st} \frac{\sin(t)}{t} &= \mathcal{L}\left\{\tfrac{1}{t} f(t)\right\}(s) = \int_s^{\infty} \mathcal{L}\left\{f(t)\right\}(s')\;ds'\\ &= \int_s^{\infty} \frac{ds'}{1+s'^2} = \left[\tan^{-1}(s')\right]_s^{\infty}\\ &= \tfrac{\pi}{2} - \tan^{-1}(s) \end{align}