# Integrability of sin(x)*x^(-a)

1. Jan 10, 2009

### Preno

Hello, question: does the integral $$\int_0^\infty \frac{\sin(x)}{x^a}$$ converge (in the sense of Lebesgue principal value) for all $$a \in (0;2)$$? For a=1/2, it's the Fresnel integral, but other than that, I'm not sure how to approach this.

2. Jan 11, 2009

It can be shown with some clever maneuvering and the use of the Gamma function that:

$$\int_{0}^{\infty}\frac{sin(x)}{x^{a}}dx=\frac{\sqrt{\pi}{\Gamma}(1-\frac{a}{2})}{{\Gamma}(\frac{a}{2}+\frac{1}{2})}$$

Gamma is undefined at 0, so one can see that a=2 leads to Gamma(0) and a=0 gives 1.

Of course, if a=1/2, then we have $$\sqrt{\pi}$$, which is the solution of the Fresnel integral.

Remember that $${\Gamma}(\frac{1}{2})=\sqrt{\pi}$$.

3. Jan 11, 2009

### tim_lou

this trick works all the time!!!

$$\int_0^\infty \frac{\sin(x)}{x^a}dx=\int_0^\infty \int_0^\infty \frac{t^{a-1}}{\Gamma(a)} \sin(x)e^{-xt} dxdt$$

4. Jan 12, 2009

### Preno

Thanks for the replies.

Fredoniahead: thanks for the formula (Maple says it needs an extra $$2^{-a}$$ factor). It's curious/didactic that the actual integral is undefined for a=0, while the formula is perfectly well-behaved there.

tim_lou: neat trick, I'll remember it.