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

[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.

 

[fredoniahead]

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

 

[tim_lou]

this trick works all the time!

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

 

[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.