the characteristic function of Levy Distribution The standard Levy distribution has the PDF:
$$f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2x}}\frac{1}{x^{3/2}},$$
where $x\geq0$.
My question is how to compute its characteristic function:
$$ \phi(t)=\int_{-\infty}^{\infty} e^{jtx}f(x)\mathrm{d}x.  $$
I tried to use the Residue theorem. 

I can't find the question this duplicated, so I'll write the answer here. Define $a:=(1-i)\sqrt{t}$ so $a^2=-2it$. Since $\phi(-t)=\phi^\ast(t)$, once we've proven that $t>0\implies \phi(t)=\exp -a$ we'll know that more generally $\phi(t)=\exp -(1-i\operatorname{sgn}t)\sqrt{|t|}$. For the $t>0$ case$$\phi(t)=\int_0^\infty\tfrac{\exp -a}{\sqrt{2\pi}}x^{-3/2}\exp\big[-\tfrac{1}{2}(x^{-1/2}-ax^{1/2})^2\big]dx.$$The substitution $x\mapsto a^{-2}x^{-1}$ obtains the alternative expression$$\phi(t)=\int_0^\infty\tfrac{\exp -a}{\sqrt{2\pi}}ax^{-1/2}\exp\big[-\tfrac{1}{2}(x^{-1/2}-ax^{1/2})^2\big]dx.$$Averaging the expressions and substituting $y=x^{-1/2}-ax^{1/2}$,$$\phi(t)=\int_\mathbb{R}\frac{\exp (-a-\tfrac{1}{2}y^2)}{\sqrt{2\pi}}dy=\exp -a.$$
