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Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$?

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.

  • $\begingroup$ 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. $\endgroup$ – Clive Newstead Feb 6 '12 at 19:39
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    $\begingroup$ Please see: math.stackexchange.com/questions/5248/… $\endgroup$ – user9413 Feb 6 '12 at 19:41
  • $\begingroup$ That link is really fascinating. $\endgroup$ – Sangchul Lee Feb 6 '12 at 19:57
  • $\begingroup$ I found the solution by the link $\endgroup$ – alice Feb 6 '12 at 20:01
  • $\begingroup$ It's also possible to compute the integral using a contour integral, if you have some knowledge of complex analysis. $\endgroup$ – 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:

  1. Interchanging the order of two integral needs justification, and in fact this is the hardest step in this proof. (There are several ways to resolve this problem, though not easy.)
  2. It is nothing but a disguise of Laplace transform method. So this calculation contains no new information on the integral.
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