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

How to prove $\operatorname{si}(0) = -\pi/2$ without contour integration ? Where $\operatorname{si}(x)$ is the sine integral.


marked as duplicate by David E Speyer, Sasha, Fabian, William, Thomas Andrews Sep 7 '12 at 21:17

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    $\begingroup$ Isn't $\operatorname{Si}(0)=\int_0^0\frac{\sin x}{x}\ dx=0$? $\endgroup$ – axblount Sep 5 '12 at 18:55
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    $\begingroup$ Well if you're talking about $$\text{Si}(z):=\int_0^z\frac{\sin t}t\,dt,$$ then $\text{Si}(0)=0$. Did you mean something else, perhaps? $\endgroup$ – Cameron Buie Sep 5 '12 at 18:56
  • $\begingroup$ I think you mean $Si(\infty)$ or $-si(0)$. See en.wikipedia.org/wiki/Sine_integral#Sine_integral $\endgroup$ – Eric Angle Sep 5 '12 at 18:58


Note that our integral may be rewritten as $$\int_{0}^{\infty} \int_{0}^{\infty} e^{-xy} \sin x \ dy \ dx = \int_{0}^{\infty} \frac{\sin x}{x} \ dx$$ but integrating with respect to x we get that $$\int_{0}^{\infty} \int_{0}^{\infty} e^{-xy} \sin x \ dx \ dy = \int_{0}^{\infty} \frac{1}{1+y^2} \ dy$$ Hence I hope you can handle it on your own.


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