Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
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.

share|cite|improve this question

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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 – Eric Angle Sep 5 '12 at 18:58
up vote 4 down vote accepted


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.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.