# Integral by parts $\int_0^\infty e^{-st}\frac{\sin(t)}{t} dt$

I want to solve the following integral by parts:

$$\int_0^\infty e^{-st}\frac{\sin(t)}{t} dt$$

I have been trying but I don't know what else to do. The result should be $$\frac{\pi}{2}-\arctan\left(s\right)$$. I took $$\frac{\sin(t)}{t}$$ as u and $$e^{-st}$$ as dv, obtaining this:

$$du = \bigg[\frac{\cos(t)}{t}-\frac{\sin(t)}{t^2}\bigg]dt$$

$$v = -\frac{1}{s} e^{-st}$$

Applying the formula for definite integration by parts:

$$uv\bigg|_0^\infty -\int_0^\infty vdu$$

$$-\frac{1}{s}e^{-st}\frac{\sin(t)}{t}\bigg|_0^\infty +\frac{1}{s}\int_0^\infty e^{-st}\bigg[\frac{\cos(t)}{t}-\frac{\sin(t)}{t^2}\bigg]dt$$

From that point onwards, I am stuck. I would be most grateful if you may help me.

Thanks in advance.

• @EricTowers It is not constant, though. It is $\frac{\pi}{2}-\arctan\left(s\right)$. Sorry, my bad. Apr 14, 2020 at 23:24

## 3 Answers

Per double integral \begin{align} \int_0^\infty \frac{e^{-st}}t \sin t \>dt =\int_0^\infty \int_s^\infty e^{-xt}\sin t \>dx \>dt = \int_s^\infty \frac1{1+x^2}dx = \tan^{-1}\frac1s\\ \end{align}

• Thanks a lot! I considered the possibility of a double integral but didn't know how to proceed. By the way, it looks like you applied the Fubini theorem. Am I right? Apr 14, 2020 at 23:44
• (Provided that $s>0$) Apr 15, 2020 at 0:30

Your method doesn't work. Here is small trick which makes the evaluation of the integral easy: Let $$F(u)=\int_0^{\infty} e^{-st} \frac {\sin (ut)} t dt$$. Then $$F'(u)=\int_0^{\infty} e^{-st}\cos (ut) dt$$. This is a standard integral. I will leave the evaluation (by two integration by parts) to you. Once you write down $$F'(u)$$ and observe that $$F(0)=0$$ you can write down $$F(u)$$ (in terms of $$\arctan$$). Put $$u=1$$ to finish.

• I haven't thought about that before. I will take a look to your proposed solution. Thanks a lot. Apr 14, 2020 at 23:31
• @inghans This is a standard methodology that is called by some "Feynman's Trick," named after the Nobel Prize winning theoretical physicist Richard Feynam. Apr 14, 2020 at 23:33
• @MarkViola Thanks for the information. I didn't know the name for this trick. Apr 14, 2020 at 23:39
• Although some of us prefer calling it the more descriptive name differentiation under the integral sign, or Leibniz's integral rule. Apr 15, 2020 at 0:29

So, your integral is the Laplace transform of $$\sin(t)/t$$. Per an identity on this reference sheet,

$$\mathcal L \left\{ \frac{f(t)}{t} \right\} = \int_s^\infty F(u) du$$

where $$F$$ is the Laplace transform of $$f$$. Take $$f(t) = \sin(t)$$. Then by this identity and formula $$(7)$$ from the sheet with $$a=1$$,

$$\int_0^\infty e^{-st} \frac{\sin(t)}{t} dt = \int_s^\infty \frac{1}{u^2 + 1} du$$

The latter is a fairly standard integral.

$$\int_s^\infty \frac{1}{u^2 + 1} du = \lim_{u \to \infty} \arctan(u) - \arctan(s) = \frac \pi 2 - \arctan(s)$$