How to compute: $I_x =\int_0^1 t^x e^{-\frac{1}{1-t^2}} dt~~~x>0$?

Question

$$I_x =\int_0^1 t^x e^{-\frac{1}{1-t^2}} dt~~~x>0$$

I first tried to see if I could solve the integer case.

$$I_n =\int_0^1 t^ne^{-\frac{1}{1-t^2}} dt $$

I have tried to find a possible relationship between $I_n $ and $I_{n+1}$ I could not get it after integration by parts the result does not look nice.

At some point I thought it is related to a Gamma function.But after the change of variables I did get a convincing out come.

Please help me to to evaluate $I_x$ or $I_n$

Answer 1

Set $1/(1-t^2)=\tau$. Therefore $$ I_x=(1/2)\int_1^\infty d\tau (\tau -1)^{-1/2} \tau ^{-3/2}\left(\frac{\tau-1}{\tau}\right)^{x/2}e^{-\tau}=(1/2)\int_1^\infty d\tau\ e^{-\tau} (\tau-1)^{\frac{x-1}{2}} \tau^{\frac{1}{2} (-x-3)}\ , $$ which Mathematica can solve in terms of a Meijer G-function (see above for equivalent Gradshteyn formula 3.384.4) $$ I_x=\frac{1}{2} \Gamma \left(\frac{x+1}{2}\right) G_{1,2}^{2,0}\left(1\left| \begin{array}{c} \frac{x+3}{2} \\ 0,1 \\ \end{array} \right.\right)\ . $$