# How to prove $\operatorname{si}(0) = -\pi/2$ without contour [duplicate]

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

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## marked as duplicate by David Speyer, Sasha, Fabian, William, Thomas AndrewsSep 7 '12 at 21:17

Isn't $\operatorname{Si}(0)=\int_0^0\frac{\sin x}{x}\ dx=0$? –  axblount Sep 5 '12 at 18:55
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? –  Cameron Buie Sep 5 '12 at 18:56
I think you mean $Si(\infty)$ or $-si(0)$. See en.wikipedia.org/wiki/Sine_integral#Sine_integral –  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.