I already posted this in the Applied section but I don't think thats actually the right place so here it is again!

I think I'm pretty close to getting this but can't quite figure out what do next!

From earlier in the question...

$\displaystyle \textrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2} dt \sim 1 - \frac{e^{-x^2}}{\pi} \sum^{\infty}_{n=0} \Gamma \Big{(}n + \frac{1}{2}\Big{)}\frac{(-1)^n}{x^{2n+1}}$ (*)

Show that erf(x) can be expressed in terms of the incomplete gamma function, and that the above asymptotic expansion agrees with the result obtained in Problem 6. [The formula $\displaystyle \Gamma\Big{(}\frac{1}{2} + n)\Gamma\Big{(}\frac{1}{2} - n\Big{)} = (-1)^n \pi$ (**) for integer n may be useful.]

So I think I have to show the relation... $\displaystyle \gamma\Big{(}\frac{1}{2}, x\Big{)} = \sqrt{\pi} \textrm{erf}(\sqrt{\pi})$ (***) as that's the relation I've seen elsewhere.

The result obtained in question 6 was...

$\displaystyle \gamma(\alpha,x) \sim \Gamma(\alpha) - e^{-x}x^{\alpha - 1} \sum^{\infty}_{n=0}\frac{\Gamma(\alpha)}{\Gamma(\a lpha - n)}\frac{1}{x^n}$

And also...

$\displaystyle \gamma(\alpha,x) \sim \Gamma(\alpha) - e^{-x}x^{\alpha - 1} \sum^{\infty}_{n=0}\frac{(\alpha-1)(\alpha-2)\ldots(\alpha-n)}{x^n}$

Right so intro done!

Now what I've done...

From (*)

$\displaystyle \sqrt{\pi} \textrm{erf}(x) = 2 \int^x_0 e^{-t^2} dt \sim \sqrt{\pi} - \frac{e^{-x^2}}{\sqrt{\pi}} \sum^{\infty}_{n=0} \Gamma\Big{(}n + \frac{1}{2}\Big{)}\frac{(-1)^n}{x^{2n+1}}$

$\displaystyle = \Gamma\Big{(}\frac{1}{2}\Big{)} - \frac{e^{-x^2}}{\sqrt{\pi}} \sum^{\infty}_{n=0} \Gamma\Big{(}n + \frac{1}{2}\Big{)}\frac{(-1)^n}{x^{2n+1}}$

(This step seems to be showing that $\displaystyle \alpha = \frac{1}{2}$ which corresponds to (***))

Using (**)

$\displaystyle = \Gamma\Big{(}\frac{1}{2}\Big{)} - \sqrt{\pi} e^{-x^2} $$\displaystyle \sum^{\infty}_{n=0} \frac{1}{\Gamma\Big{(}\frac{1}{2} - n\Big{)}}\frac{1}{x^{2n+1}}$

$\displaystyle = \Gamma\Big{(}\frac{1}{2}\Big{)} - \sqrt{\pi} e^{-x^2} \sum^{\infty}_{n=0} \frac{x^{-(n+1)}}{\Gamma\Big{(}\frac{1}{2} - n\Big{)}}\frac{1}{x^n}$

$\displaystyle = \Gamma\Big{(}\frac{1}{2}\Big{)} - e^{-x^2} \sum^{\infty}_{n=0} \frac{\Gamma\Big{(}\frac{1}{2}\Big{)}}{\Gamma\Big{ (}\frac{1}{2} - n\Big{)}}\frac{x^{-(n+1)}}{x^n}$

But I'm stuck now. I need to do something to the $\displaystyle e^{-x^2} \textrm{ and the } x^{-(n+1)}$ but I'm not sure what... I don't even know if what I've done so far is right to be honest... Any help please!