General solution of $ \frac{d^j}{d\sigma^j}(\exp(0.5(\alpha-\sigma)^2) $ How would you write the general solution, I'm assuming something like a sum, of:
$$ \frac{d^j}{d\sigma^j}\exp[0.5(\alpha-\sigma)^2] $$
Regards,
 A: You can have your solution in terms of The MeijerG function. It is a very general special function
$$ \left( -1 \right) ^{\frac{3}{2}\,j}{2}^{\frac{1}{2}\,j}
G^{1, 2}_{2, 3}\left(-\frac{\left( \alpha-\sigma \right) ^{2}}{2}\, \Big\vert\,^{-\frac{1}{2}\,j, -\frac{1}{2}\,j+\frac{1}{2}}_{-\frac{1}{2}\,j, \frac{1}{2}, 0}\right)
$$
Note that, the above formula is a unified formula. It gives derivatives of real order (including the integer order), if $j>0$, anti-derivatives of real order (including the integer order) if $j<0$, and the original function if $j=0$ of the function $ \mathrm{e}^{\frac{1}{2}(\alpha-\sigma)^2}\,.$ 
If you simplify the above Meijer G-function, you can get the answer in terms of the hypergeometric function
$$ {\frac { \left( \alpha-\sigma \right) ^{-j} \left( -1 \right) ^{j}
{\mathrm{F} (1/2,1;\,-1/2\,j+1,-1/2\,j+1/2;\,1/2\, \left( \alpha-\sigma \right) ^{2})}
}{\Gamma  \left( -j+1 \right) }}
\,.$$
But the above formula has a deficiency, since you need to deal with the poles of the gamma function in the denominator. This is due to the limitations of the hypergeometric function. 
