# 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 ?

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

• 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
• @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. – Did Jul 29 '13 at 15:58