Calculate: $F(x)=\int_{0}^{+\infty}\frac{e^{i xt}}{t^{\alpha}}dt\quad \text{avec}~x\in \mathbb{R}~\text{ et }~0<\alpha<1$ I would like to calculate this integral:
$F(x)=\int_{0}^{+\infty}\frac{e^{i 
xt}}{t^{\alpha}}dt\quad \text{avec}~x\in \mathbb{R}~\text{ et }~0<\alpha<1$
I calculated : $\displaystyle F(ix)=\int_{0}^{+\infty}\frac{e^{-xt}}{t^{\alpha}}dt=x^{\alpha-1}\Gamma(1-\alpha)$ but not $F(x)$
I would be interest for any replies or any comments.
 A: For $0 < \varepsilon < R < \infty$, consider the contour $C_{\varepsilon,R}$ consisting of


*

*the interval $[\varepsilon,R]$ on the real axis,

*the quarter circle from $R$ to $iR$,

*the interval $[iR,i\varepsilon]$ on the imaginary axis, and

*the quarter circle from $i\varepsilon$ to $\varepsilon$.


Since the integrand $z^{-\alpha} e^{ixz}$ is holomorphic in - for example - $\mathbb{C}\setminus (-\infty,0]$, we have
$$\int_{C_{\varepsilon,R}} z^{-\alpha} e^{ixz}\,dz = 0.$$
By Jordan's lemma, the integral over the large quarter circle tends to $0$ as $R\to\infty$. By the standard estimate (ML inequality), the integral over the small semicircle tends to $0$ as $\varepsilon \to 0$.
It follows that
$$\int_0^\infty \frac{e^{ixt}}{t^\alpha}\,dt = e^{-\alpha\pi i/2} \int_0^\infty \frac{e^{-xt}}{t^\alpha}\,dt = e^{-\alpha\pi i/2} x^{\alpha-1}\Gamma(1-\alpha).$$
A: 1)-for the case  $x>0$ , suppose that :$u=(xt)^{1-\alpha}$ , we get :
$F(x)=\displaystyle\int_0^{+\infty}\frac{e^{i xt}}{t^{\alpha}}dt=\frac1{x^{1-\alpha}(1-\alpha)}\int_0^{+\infty}e^{\displaystyle i u^\beta}du$ with $\beta=\dfrac1{1-\alpha}>1$
the last integral is classic (generalisation of  FRESNEL integral) to be:
$\displaystyle\frac1{\beta}\Gamma(\frac1{\beta})e^{\displaystyle i\frac{\pi}{2\beta}}$
we get finally :$F(x)=\dfrac{\Gamma(1-\alpha) e^{\displaystyle i\frac{\pi}2(1-\alpha)}}{x^{1-\alpha}}$
2)- for $x<0$ we get with conjugate $F(x)=\dfrac{\Gamma(1-\alpha) e^{\displaystyle -i\frac{\pi}2(1-\alpha)}}{(-x)^{1-\alpha}}$
3)-for $x=0$ the integral is diverge 
