3
$\begingroup$

I am calculating an integral $\displaystyle \int \limits_{-\infty}^{\infty}\dfrac{\sin(x)}{x}$ and I dont seem to be getting an answer.

When I integrate by parts twice, I get:
$$\displaystyle \int \limits _{-\infty}^{\infty}\frac{\sin(x)}{x}dx = \left[\frac{\sin(x)\ln(x) - \frac{\cos(x)}{x}}{2}\right ]_{-\infty}^{+\infty}$$

What will be the answer to that ?

$\endgroup$

marked as duplicate by Start wearing purple, Git Gud, M.H, Ayman Hourieh, Amzoti Jul 17 '13 at 20:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ You could use double integrals and switch limits. It is an improper integral. The answer is $\pi$. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jul 17 '13 at 20:09
  • $\begingroup$ @AnuragPallaprolu I looked it up on Wolfram Alpha and it said $\pi$. WHat is an improper integral ? :) $\endgroup$ – Little Child Jul 17 '13 at 20:11
  • 2
    $\begingroup$ @LittleChild I think we could use a bit of wiki here. :) Its an integral whose limits reach either infinities. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jul 17 '13 at 20:12
  • $\begingroup$ @AnuragPallaprolu Double integral = split the integral into two ?? $\endgroup$ – Little Child Jul 17 '13 at 20:13
  • 1
    $\begingroup$ @LittleChild Accepting answers is (of course) entirely up to the OP. However, in the present case, since you admitted in a comment that the posted answer did not help you, I wonder why you accepted this answer (... 7 minutes after it got posted!). A consequence is that it makes other, perhaps more satisfying, answers less likely to be posted. $\endgroup$ – Did Jul 29 '13 at 15:58
3
$\begingroup$

Hint: From the viewpoint of improper Lebesgue integrals or in sense of Cauchy principal value is integral is legitimate. Integration by parts. \begin{align} \int \limits_{-\infty}^{\infty}\dfrac{\sin(x)}{x} \mbox{d} x = & \lim_{t\to\infty}\int \limits_{-t}^{\frac{1}{t}}\dfrac{\sin(x)}{x} \mbox{d} x + \lim_{t\to\infty}\int \limits_{\frac{1}{t}}^{t}\dfrac{\sin(x)}{x} \mbox{d} x \\ = & \lim_{t\to\infty}\int \limits_{-t}^{\frac{1}{t}}\sin(x)(\log x)^\prime \mbox{d} x + \lim_{t\to\infty}\int \limits_{\frac{1}{t}}^{t}\sin(x)(\log x)^\prime\mbox{d} x \\ \end{align}

$\endgroup$
  • $\begingroup$ I don't think you're allowed to take the limit that way. You can't take to 0 and to infinity in the same limit. Perhaps there's a rule that says when that's okay? $\endgroup$ – Adar Hefer Jul 17 '13 at 20:13
  • 1
    $\begingroup$ Yeah, it is bad form to compute these as a single limit. Once you know the integral converges, you can get the answer this way; but, without proving convergence, this can give you unfortunate results in some cases. $\endgroup$ – Nick Peterson Jul 17 '13 at 20:16
  • $\begingroup$ Oh, if I already know convergence then this will reproduce the result had I taken the limit using two limits? $\endgroup$ – Adar Hefer Jul 17 '13 at 20:20
  • 2
    $\begingroup$ How does this help to go toward the answer? $\endgroup$ – Sungjin Kim Jul 17 '13 at 20:38
  • 3
    $\begingroup$ @LittleChild: Well.. if this answer helped you very little, then why would you accept it? Using integration by parts on this integral only makes it more complicated. So, I even doubt that this answer helped you "very little". I recommend following the link and read answers there. $\endgroup$ – Sungjin Kim Jul 18 '13 at 7:22

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