# how to integrate $\int_{-\infty}^{+\infty} \frac{\sin(x)}{x} \,dx$? [duplicate]

How can I do this integration using only calculus? (not laplace transforms or complex analysis)

$$\int_{-\infty}^{+\infty} \frac{\sin(x)}{x} \,dx$$

I searched for solutions not involving laplace transforms or complex analysis but I could not find.

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## marked as duplicate by Bruno Joyal, Chandrasekhar, anon, JavaMan, Davide GiraudoFeb 6 '12 at 19:58

It is impossible to express the indefinite integral $$\int \frac{\sin x}{x}\, dx$$ in terms of elementary functions, so you cannot find its antiderivative and then take the appropriate limits. – Clive Newstead Feb 6 '12 at 19:39
Please see: math.stackexchange.com/questions/5248/… – user9413 Feb 6 '12 at 19:41
That link is really fascinating. – Sangchul Lee Feb 6 '12 at 19:57
I found the solution by the link – alice Feb 6 '12 at 20:01
It's also possible to compute the integral using a contour integral, if you have some knowledge of complex analysis. – user170231 Sep 22 '14 at 21:55

Putting rigor aside, we may do like this: \begin{align*} \int_{-\infty}^{\infty} \frac{\sin x}{x} \; dx &= 2 \int_{0}^{\infty} \frac{\sin x}{x} \; dx \\ &= 2 \int_{0}^{\infty} \sin x \left( \int_{0}^{\infty} e^{-xt} \; dt \right) \; dx \\ &= 2 \int_{0}^{\infty} \int_{0}^{\infty} \sin x \, e^{-tx} \; dx dt \\ &= 2 \int_{0}^{\infty} \frac{dt}{t^2 + 1} \\ &= \vphantom{\int}2 \cdot \frac{\pi}{2} = \pi. \end{align*} The defects of this approach are as follows: