I am trying to evaluate the following $$I = \int_0^s u^{-3n-1} \exp{\left[-\left(l u^n + \frac{m}{u^{n}}\right)^2\right]}\,du,$$ where $l, m$ and $n$ are positive constants.
I tried to substitute $v = u^{-n}$, and got the following integral $$I =\frac{1}{n} \int_{s^{-1/n}}^{\infty} v^{2} \exp{\left[-\left(\frac{l}{v} + m v\right)^2\right]}\,dv.$$ I was trying to follow in Oliver's footsteps in this Question, but I am not sure if that is completely possible with this integrand.
Edit
Upon integration by parts once we get \begin{eqnarray} I &=& \frac{v^2\sqrt{\pi } e^{-4 l m} \left(\text{erfc}\left(\frac{l}{v}-m v\right)-e^{4 l m} \text{erfc}\left(\frac{l}{v}+m v\right)\right)}{4 m}|_{s^\frac{-1}{n}}^{\infty} - \frac{1}{n} \int_{s^\frac{1}{n}}^{\infty} v \exp{\left[-\left(\frac{l}{v} + m v\right)^2\right]}\,dv\\ \end{eqnarray} Any help will be greatly appreciated. Thank you.
Hint. One may start with the following evaluation, for $a>0$, $b>0$, $$ \begin{align} &\int_a^\infty e^{\large-bx^2-\frac1{x^2}}\,dx \\\\&=\frac{\sqrt{\pi}\left(e^{2\sqrt{b}}+e^{-2\sqrt{b}} \right)}{4\sqrt{b}}+\frac{\sqrt{\pi}e^{-2\sqrt{b}}}{4\sqrt{b}}\text{erf}\left(\frac1{a}-a\sqrt{b}\right)-\frac{\sqrt{\pi}e^{2\sqrt{b}}}{4\sqrt{b}}\text{erf}\left(\frac1{a}+a\sqrt{b}\right). \tag1 \end{align} $$ By differentiating both sides with respect to $b$ one gets the closed form : $$ \int_a^\infty x^2e^{\large-bx^2-\frac1{x^2}}\,dx=-\partial_b\left( \text{RHS}\right). \tag2 $$ Then, by the change of variable $v=lx$, one obtains $$ \int_{\large s^{-1/n}}^{\infty} v^{2} \exp{\left[-\left(\frac{l}{v} + m v\right)^2\right]}\,dv=l^3e^{\large-2ml}\int_{\large \frac{s^{-1/n}}{l}}^\infty x^2e^{\large-(ml)^2x^2-\frac1{x^2}}\,dx \tag3 $$ and one may conclude with $(2)$.