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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.

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

3 Answers 3

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

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  • $\begingroup$ 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? $\endgroup$
    – inghans
    Apr 14, 2020 at 23:44
  • $\begingroup$ (Provided that $s>0$) $\endgroup$
    – Chappers
    Apr 15, 2020 at 0:30
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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.

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

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

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