# A 'complicated' integral: $\int \limits_{-\infty}^{\infty}\frac{\sin(x)}{x}$ [duplicate]

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 ?

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You could use double integrals and switch limits. It is an improper integral. The answer is $\pi$. –  Torsten Hĕrculĕ Cärlemän Jul 17 '13 at 20:09
@AnuragPallaprolu I looked it up on Wolfram Alpha and it said $\pi$. WHat is an improper integral ? :) –  Little Child Jul 17 '13 at 20:11
@LittleChild I think we could use a bit of wiki here. :) Its an integral whose limits reach either infinities. –  Torsten Hĕrculĕ Cärlemän Jul 17 '13 at 20:12
@AnuragPallaprolu Double integral = split the integral into two ?? –  Little Child Jul 17 '13 at 20:13
@O.L. Perhaps you could explain briefly why this follows from your link? –  user1729 Jul 17 '13 at 20:39