<s>[INST] Closed form of $ \int_0^\infty x e^{-b \pi x^2}(1+x^{-a})^{-n}\mathrm{d}x$ 
Is there an exact closed-form expression or a closed-form upper bound for the following integral?
  $$I = \int_{0}^{\infty}\frac{x e^{-b \pi x^2}}{(1+x^{-a})^{n}}\mathrm{d}x, \tag1$$
  where $a > 2$, $b, n > 0$.
  If yes, how can we obtain it?

My attempt: If we multiply $I$ in $(1)$ by $2\pi b$, we get the probability density function of Rayleigh random variable, i.e, $f_X(x) = 2\pi b xe^{-b \pi x^2}$. So we can write $I$ as
$$I = \mathbb{E}\left[\left(\frac{1}{1+X^{-a}}\right)^{n}\right].$$
I could not procced in this way.
 [/INST] WolframAlpha ist able to compute the anti-derivative of your integrand for $a=2$ and small fixed integer $n$. From these solutions I came up with a conjecture for $a=2$ and general integer $n$. 
Some notation:
$$ I_{a,n}(c)= \int_{0}^{\infty}\frac{x e^{-c x^2}}{(1+x^{-a})^{n}}\mathrm{d}x, \tag1$$
where $a > 2$, $c, n > 0$.
The conjecture for $a=2$:
$$ F_{2,n}(x,c)= \int\frac{x e^{-c x^2}}{(1+x^{-2})^{n}}\mathrm{d}x = p_1(c) ~ \mathrm{e}^c ~ \mathrm{Ei}(-c(x^2+1)) - \frac{\mathrm{e}^{-cx^2}~p_2(c,x)}{c(x^2+1)^{n-1}}$$
$$ I_{2,n}(c) = -F_{2,n}(0,c) = -p_1(c) ~ \mathrm{e}^c ~ \mathrm{Ei}(-c) + \frac{p_2(c,0)}{c}$$
Where $p_1(c)$ is a polynomial in $c$ of degree $n-1$ and $p_2(x,c)$ is a polynomial in $c$ of degree $n-1$ and in $x$ of degree $2(n-1)$ with only even terms in $x$.
I don't have a general form for $p_1$ and $p_2$, but using the above restrictions it's possible to solve for their coefficients given some fixed $n$ by differentiating $F_{2,n}$ and comparing with the integrand.
Some solutions:
$$I_{2,1} = \left(\frac{1}{2} \right)\mathrm{e}^c \mathrm{Ei}(-c) +
\frac{1}{c}\left( \frac{1}{2} \right) \text{(already given by JJacquelin)}$$
$$I_{2,2} = \left(\frac{1}{2} \, c + 1 \right)\mathrm{e}^c
\mathrm{Ei}(-c) + \frac{1}{c}\left( \frac{1}{2} \, c + \frac{1}{2}
\right)$$
$$I_{2,3} = \left(\frac{1}{4} \, c^{2} + \frac{3}{2} \, c + \frac{3}{2}
\right)\mathrm{e}^c \mathrm{Ei}(-c) + \frac{1}{c}\left( \frac{1}{4} \,
c^{2} + \frac{5}{4} \, c + \frac{1}{2} \right)$$
$$I_{2,4} = \left(\frac{1}{12} \, c^{3} + c^{2} + 3 \, c + 2
\right)\mathrm{e}^c \mathrm{Ei}(-c) + \frac{1}{c}\left( \frac{1}{12} \,
c^{3} + \frac{11}{12} \, c^{2} + \frac{13}{6} \, c + \frac{1}{2}
\right)$$
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