# How to solve or approximate this special integral related to inverse gamma distribution

I encountered a problem to get the expectation with respect to an inverse gamma distribution. My problem can be simplified as to solve the following integral: $$\int_{0}^{\infty}x^{-a}\exp\left(-\left[\frac{b}{x} + \frac{c}{\,\sqrt{\, x\,}\,}\right]\right)\,\mathrm{d}x, \quad\mbox{where}\quad a,b,c \in \mathbb{R}\quad\mbox{and}\quad a > 1\,,\ b > 0.$$ It is basically something like $E\left[c_{0}/\sqrt{\, X\,}\right]$ if $X$ follows inverse gamma distribution.

Is there an analytical form or approximation for above the integral? Thanks in advance!

Not a complete answer, but for what it's worth: via Mathematica I found that your integral $I$ can be expressed in the terms of the Kummer confluent hypergeometric function, ${}_1F_1$: $$I=b^{-a} \left(b \Gamma (a-1) \, _1F_1\left(a-1;\frac{1}{2};\frac{c^2}{4 b}\right)-\sqrt{b} c \Gamma \left(a-\frac{1}{2}\right) \, _1F_1\left(a-\frac{1}{2};\frac{3}{2};\frac{c^2}{4 b}\right)\right)\,.$$ For certain specific values of $a$ and $b$ (integer, half-integer, quarter-integer), this can be simplified considerably, in terms of error and Bessel fn.'s especially. E.g., for $a=3/2$ and $b=2$ the expression simplifies to $\sqrt{\frac{\pi }{2}} e^{\frac{c^2}{8}} \text{erfc}\left(\frac{c}{2 \sqrt{2}}\right)$.