How does the answer of the Laplace transform $$\mathcal L \left\{ \frac{\sin t}{t} \right\}= \frac{\pi}{2}-\tan^{-1}(s)$$ solve the definite integral

$$\int_0^{\infty} \frac{\sin t}{t} dt = \frac{\pi}{2} $$

How are they related? why does this solve the definite integral?

Thank you.

  • $\begingroup$ Use Laplace of sint , then use.. division of t rule, which lets you integrate it from s to infty $\endgroup$ Commented Nov 28, 2016 at 5:36
  • $\begingroup$ I know how to do. I just don't know why it works?? $\endgroup$ Commented Nov 28, 2016 at 5:56
  • $\begingroup$ Go through the derivation of division of t rule. You will understand why it works $\endgroup$ Commented Nov 28, 2016 at 8:05
  • $\begingroup$ I understand the division of t rule and how it works. What i do not understand is by taking the Laplace transform ℒ{ sin(t)/t } it basically solve the definite integral? $\endgroup$ Commented Nov 28, 2016 at 8:43

2 Answers 2


Your statement is

$$\int_0^{\infty} dt \frac{\sin{t}}{t} e^{-s t} = \frac{\pi}{2} - \arctan{s} $$

Plug in $s=0$ to both sides.

There are lots of ways to prove the LT. One way to do it is to use the FT relation for the sinc term.

  • $\begingroup$ $s=0$ is a boundary point, hence you need an argument when taking the limit. $\endgroup$ Commented Dec 2, 2016 at 17:34
  • 1
    $\begingroup$ @AD. "hence?" If the integral converges (and the OP is not questioning that), then I do not need to elaborate. $\endgroup$
    – Ron Gordon
    Commented Dec 2, 2016 at 17:38

You need to use Abel's theorem to take the limit as s goes to zero. Tis is needed because 0 is the boudary of the LT.


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