Solution to $\int_{0}^{y} x^{-a} \exp \left[- \frac{(b - cx^{-d})^2}{2} \right] dx$ Is there a solution to this integral?
$$\int_{0}^{y} x^{-a} \exp \left[- \frac{(b - cx^{-d})^2}{2} \right] dx,$$
where $a > 0$ and $d > 0$.
 A: Maple does not know a symbolic answer for this.  The very special case $a=2,b=0,c=1,d=2$ is evaluated by Maple in terms of the Whittaker M function or a $\;{}_1F_1$ hypergeometric function:
$$
\int_{0}^{y}\!{\frac {1}{{x}^{2}}{{\rm e}^{-1/2\,{x}^{-4}}}}\,{\rm d}x
=-4/5\,{\frac {\sqrt [8]{2}}{\sqrt [4]{{{\rm e}^{{y}^{-4}}}}\sqrt {y}}
{{\rm M}_{1/8,\,5/8}\left(1/2\,{y}^{-4}\right)}}+1/4\,{\frac {{2}^{3/4
}\pi}{\Gamma \left( 3/4 \right) }}-{\frac {1}{\sqrt {{{\rm e}^{{y}^{-4
}}}}y}}
\\
=
2/5\,{\frac {1}{\sqrt [4]{{{\rm e}^{{y}^{-4}}}}\sqrt {y}} \left( {y}
^{-4} \right) ^{{\frac{9}{8}}}
{\mbox{$_1$F$_1$}(1;\,9/4;\,1/2\,{y}^{-4})} \left( {{\rm e}^{1/4\,{y}^
{-4}}} \right) ^{-1}}+1/4\,{\frac {{2}^{3/4}\pi}{\Gamma \left( 3/4
 \right) }}-{\frac {1}{\sqrt {{{\rm e}^{{y}^{-4}}}}y}}
$$
A: A closed form expression in terms of a special function exists for $b=0$, when
$$\int_0^y x^{-a} \exp \left(-\tfrac{1}{2} c^2 x^{-2d} \right) \,dx=\frac{1}{2 d}y^{1-a} E_{1-\frac{a-1}{2 d}}\left(\tfrac{1}{2} c^2 y^{-2 d}\right),\;\;c,d>0,$$
with $E_n(x)$ the exponential integral function.
A: By using series expansion, change of variable and Eq. (3.381.9) of the book: "I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, 8th ed. Burlington, MA, USA: Academic Press, 2015", I was able to find this solution:
$$\int_{0}^{y} x^{-a} \exp \left[- \frac{(b - cx^{-d})^2}{2} \right] dx 
\\= \exp(-b^2/2)  \sum_{k=0}^{\infty} \frac{(bc)^k}{k!} \Gamma\left(\frac{dk+a-1}{2d}, \frac{c^2 y^{-2d}}{2} \right)\frac{1}{2d(c^2/2)^{\frac{dk+a-1}{2d}}}.$$
I've run some simulations and it works.However, there might be some convergence problem for some values, I think so.
