1. ## Invalid integral solution?

$\displaystyle \int^{\infty}_{0} \frac{\sin x}{x^{1-a}} \ dx$ converges for $\displaystyle -1<a<1$. But is the following solution only valid for $\displaystyle 0 < a <1$ ? Is it valid at all?

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x^{1-a}} \ dx = -\Im \int^{\infty}_{0} e^{-ix} x^{a-1} \ dx$

The issue is that $\displaystyle \int^{\infty}_{0} \frac{\cos x}{x^{1-a}} \ dx$ does not converge for $\displaystyle a \le 0$.

let $\displaystyle u = ix$

$\displaystyle = -\Im \int^{\infty}_{0} e^{-u} \big(\frac{u}{i}\big)^{a-1} \ \frac{du}{i}$

$\displaystyle = -\Im \ (-i)^{a} \int_{0}^{\infty} u^{a-1} e^{-u} \ du$

$\displaystyle = -\Im \ e^{\frac{-i a \pi}{2}} \ \Gamma(a) = \sin \Big( \frac{a \pi}{2} \Big) \Gamma(a)$

2. ## Re: Invalid integral solution?

The fact is that the definition of the Gamma function...

$\displaystyle \Gamma(z)= \int_{0}^{\infty} t^{z-1}\ e^{-t}\ dt$ (1)

... is valid only if $\displaystyle \Re (z) >0$, that is necessary condition for convergence of the integral in (1). For $\displaystyle \Re (z)<0$ the Gamma function can be computed using the relation...

$\displaystyle \Gamma(z)\ \Gamma(-z) = -\frac{\pi}{z\ \sin \pi z}$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: Invalid integral solution?

Can I invoke the gamma reflection formula for the other values of a?

4. ## Re: Invalid integral solution?

What Random Variable is asking for is probably: is the formula...

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

... valid for $\displaystyle -1 < a < 0$?...

All right!... a simple way is try to compute the (1) setting [for example...] $\displaystyle a=-\frac{1}{2}$ in some way and verify. If (1) is valid it should be...

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x^{\frac{3}{2}}}\ dx = -\sin \frac{\pi}{4}\ \Gamma (-\frac{1}{2})= \sqrt{2 \pi}$ (2)

Now we define...

$\displaystyle \varphi(t)= \int_{0}^{\infty} \frac{\sin t x}{x^{\frac{3}{2}}}\ dx$ (3)

... and then compute...

$\displaystyle \mathcal{L} \{\varphi(t)\} = \int_{0}^{\infty} e^{-s t}\ \int_{0}^{\infty} \frac{\sin t x}{x^{\frac{3}{2}}}\ dx\ dt = \int_{0}^{\infty} \frac{\mathcal{L} \{\sin t x \}}{x^{\frac{3}{2}}}\ dx =$

$\displaystyle = \int_{0}^{\infty} \frac{dx}{\sqrt{x}\ (s^{2}+x^{2})}$ (4)

The integral in (4) can be attacked with the usual complex analysis and we obtain...

$\displaystyle \mathcal {L} \{\varphi (t)\} = \frac{\pi}{\sqrt{2}\ s^{\frac{3}{2}}}$ (5)

... and from (5)...

$\displaystyle \varphi(t) = \frac{\pi}{\sqrt{2}}\ \frac{\sqrt{t}}{\Gamma (\frac{3}{2})} = \sqrt{2 \pi t}$ (6)

... so that is...

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x^{\frac{3}{2}}}\ dx = \varphi (1) = \sqrt{2 \pi}$ (7)

... and the (1) is verified for $\displaystyle a=- \frac{1}{2}$. Following the same procedure it is [probably...] possible to demonstrate that the same is for $\displaystyle -1<a<0$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$