Calculating an Exponential Integral 
Calculate $\int_0^{\infty} \frac{x^{2N+1}}{a+x^{-b}} e^{-c x^2} dx$ where $a,c > 0$ and $b>1$.

The best I could do about this integral is to find an upperbound for it:
$\int_0^{\infty} \frac{x^{2N+1}}{a+x^{-b}} e^{-c x^2} dx \leq \int_0^{\infty} \frac{x^{2N+1}}{x^{-b}} e^{-c x^2} dx= \int_0^{\infty} x^{2N+b+1} e^{-c x^2} dx$
Then I can use $\int_{0}^{\infty} x^{n} e^{-ax^2}\,\mathrm{d}x = 
\begin{cases}
       \frac{1}{2}\Gamma \left(\frac{n+1}{2}\right)/a^{\frac{n+1}{2}} & (n>-1,a>0) \\
       \frac{(2k-1)!!}{2^{k+1}a^k}\sqrt{\frac{\pi}{a}} & (n=2k, k \;\text{integer}, a>0) \\
       \frac{k!}{2a^{k+1}} & (n=2k+1,k \;\text{integer}, a>0)
\end{cases} $
Any idea that how I can calculate the integral, not jut an upper bound?
 A: I don't know an answer in the general case but concerning the specific case $b=2$ we may start by defining :
\begin{align}
\tag{1}f(a,c)&:=\int_0^{\infty} \frac{x^{-1}}{a+x^{-2}} e^{-c x^2} dx\\
&=\int_0^{\infty} \frac{x}{1+ax^2} e^{-c x^2} dx\\
&=e^{\,\large{\frac ca}}\int_0^{\infty} \frac{c\;x}{c\;(1+ax^2)} e^{\,\large{-\frac ca (1+ax^2)}} dx\\
\text{for}\ \ u&:=\frac ca (1+ax^2)\quad \text{this becomes :}\\
f(a,c)&=\frac{e^{\,\large{\frac ca}}}{2\;a}\int_{\large{\frac ca}}^{\infty} \frac{e^{-u}}{u} \;du\\
\tag{2}f(a,c)&=-\frac{e^{\,\large{\frac ca}}}{2\;a}\operatorname{Ei}\left(-\frac ca\right)\\
\end{align}
With $\;\displaystyle \operatorname{Ei}(x):=-\int_{-x}^\infty\frac{e^{-t}}t\,dt\;$ the exponential integral (so that $\;\displaystyle \operatorname{Ei}(x)'=\dfrac {e^x}x$).
Successive derivatives of $f(a,c)$ relatively to $c$ (under the integral sign) will give you :
$$\left(\frac d{dc}\right)^n f(a,c)=(-1)^n\int_0^{\infty} \frac{x^{2n-1}}{a+x^{-2}} e^{-c x^2} dx$$
that is :
$$\tag{3}\int_0^{\infty} \frac{x^{2n-1}}{a+x^{-2}} e^{-c x^2} dx=-\frac{1}{2\;a}\left(-\frac d{dc}\right)^n \left[e^{\,\large{\frac ca}}\;\operatorname{Ei}\left(-\frac ca\right)\right]$$
that should simply be (for $n> 0$) :
$$\tag{4}\int_0^{\infty} \frac{x^{2n-1}}{a+x^{-2}} e^{-c x^2} dx=-\frac{(-1)^n}{2\;a}\left[\frac{e^{\,\large{\frac ca}}}{a^n}\;\operatorname{Ei}\left(-\frac ca\right)-\sum_{k=1}^n\frac {(k-1)!\,(-1)^k}{a^{n-k}\,x^k}\right]$$
I'll let you prove this last part and reverify all this...
Hoping this helped any way,
A: $\int_0^\infty\dfrac{x^{2N+1}e^{-cx^2}}{a+x^{-b}}~dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^nx^{2n+b+2N+1}}{n!(ax^b+1)}~dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^nx^{\frac{2n+2N+1}{b}+1}}{n!(ax+1)}~d\left(x^\frac{1}{b}\right)$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^nx^\frac{2(n+N+1)}{b}}{bn!(ax+1)}~dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^n\left(\dfrac{x}{a}\right)^\frac{2(n+N+1)}{b}}{bn!(x+1)}~d\left(\dfrac{x}{a}\right)$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^nx^\frac{2(n+N+1)}{b}}{a^{\frac{2(n+N+1)}{b}+1}bn!(x+1)}~dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^n\pi}{a^{\frac{2(n+N+1)}{b}+1}bn!\sin\left(\pi+\frac{2(n+N+1)\pi}{b}\right)}$ (according to https://en.wikipedia.org/wiki/Beta_function#Properties)
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}c^n\pi}{a^{\frac{2(n+N+1)}{b}+1}bn!\sin\frac{2(n+N+1)\pi}{b}}$
