integral with gaussian function I am trying to evaluate the following integral:
$$
\int_0^\infty{z^{m-1}\over\left[1+\left(\eta z\right)^n\right]^p}e^{-(z-b)^2\over c}\,{\rm d}z,
$$
where the integration is w.r.t. to $z$, and the other parameters are real positive.
Any idea?
 A: This answer is far from complete and will be deleted if a more complete answer is submitted.
The value of the definite integral $\int_0^{\infty}\frac{z^{m-1}}{\left(1+\left(\eta z\right)^n\right)^p} \exp{\left(-\frac{\left(z-b\right)^2}{c}\right)}\ dz$ will be a function of the six parameters $\eta,m,n,p,b,\text{and}\ c$, all of which are assumed to be positive and real. Call this function $f$, so that:
$$f\left(\eta,m,n,p,b,c\right)=\int_0^{\infty}\frac{z^{m-1}}{\left(1+\left(\eta z\right)^n\right)^p} \exp{\left(-\frac{\left(z-b\right)^2}{c}\right)}\ dz.$$
A good place to begin would be to see if any of the parameter dependencies can be immediately simplified by rescaling. First note that since $c>0$, $c$ will have a positive real square-root $\sqrt{c}$, and since $b>0$ we can rewrite $b$ as the product of $\sqrt{c}$ and some positive scalar $\tilde{b}$: $b=\sqrt{c}\ \tilde{b}$. Also, make the substitution $z=\sqrt{c}\ \tilde{z}$, $dz=\sqrt{c}\ d\tilde{z}$, in the integral above to get:
$$f\left(\eta,m,n,p,\sqrt{c}\ \tilde{b},c\right) = \int_0^{\infty}\frac{\left(\sqrt{c}\ \tilde{z}\right)^{m-1}}{\left(1+\left(\eta \left(\sqrt{c}\ \tilde{z}\right)\right)^n\right)^p} \exp{\left(-\frac{\left(\sqrt{c}\ \tilde{z}-\sqrt{c}\ \tilde{b}\right)^2}{c}\right)}\ \sqrt{c}\ d\tilde{z}\\
 = \sqrt{c}\int_0^{\infty}\frac{\left(\sqrt{c}\right)^{m-1}\tilde{z}^{m-1}}{\left(1+\left(\left(\eta\sqrt{c}\right)\tilde{z}\right)^n\right)^p} \exp{\left(-\frac{\left(\sqrt{c}\right)^2\left(\tilde{z}-\tilde{b}\right)^2}{c}\right)}\ d\tilde{z}\\
 = \left(\sqrt{c}\right)^{m}\int_0^{\infty}\frac{\tilde{z}^{m-1}}{\left(1+\left(\tilde{\eta}\tilde{z}\right)^n\right)^p} \exp{\left(-\left(\tilde{z}-\tilde{b}\right)^2\right)}\ d\tilde{z},$$
where in the last line we introduced that rescaled parameter $\tilde{\eta}=\eta\sqrt{c}$ for convenience. Now we have the following relation:
$$f\left(\eta,m,n,p,b,c\right)=f\left(\frac{\tilde{\eta}}{\sqrt{c}},m,n,p,\sqrt{c}\ \tilde{b},\sqrt{c}^2\right)\\
=c^{\frac{m}{2}}\int_0^{\infty}\frac{\tilde{z}^{m-1}}{\left(1+\left(\tilde{\eta}\tilde{z}\right)^n\right)^p} \exp{\left(-\left(\tilde{z}-\tilde{b}\right)^2\right)}\ d\tilde{z}\\
=c^{\frac{m}{2}}\ f\left(\tilde{\eta},m,n,p,\tilde{b},1\right).$$
I.e., if we can can solve the special case of the five-parameter integral with $c=1$, $g\left(\eta,m,n,p,b\right)=f\left(\eta,m,n,p,b,1\right)$, then we've automatically solved the general case as well.
Further useful reduction of the integral $g\left(\eta,m,n,p,b\right)=\int_0^{\infty}\frac{z^{m-1}}{\left(1+\left(\eta z\right)^n\right)^p} \exp{\left(-\left(z-b\right)\right)}\ dz$ may be possible using Feynman's trick. In particular, an expression for the value of the integral integer values of $m$ and $p$ can probably be built up iteratively from the $m=p=1$ case through partial differentiation of the integrand w.r.t. $b$ and $\eta$.
