1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Use Laplace of sint , then use.. division of t rule, which lets you integrate it from s to infty $\endgroup$ Nov 28, 2016 at 5:36
  • $\begingroup$ I know how to do. I just don't know why it works?? $\endgroup$ Nov 28, 2016 at 5:56
  • $\begingroup$ Go through the derivation of division of t rule. You will understand why it works $\endgroup$ 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? $\endgroup$ Nov 28, 2016 at 8:43

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ $s=0$ is a boundary point, hence you need an argument when taking the limit. $\endgroup$ 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
    Dec 2, 2016 at 17:38
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.