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I am trying to evaluate the following $$I = \int_0^s u^{-n-1} \exp{\left[-\left(l u^n + \frac{m}{u^{n}}\right)^2\right]}\,du,$$ where $l, m$ and $n$ are non-negative constants. I tried to substitute $v = u^n$, and got the following integral $$I = \int_0^{s^\frac{1}{n}} v^{-2} \exp{\left[-\left(l v + \frac{m}{v}\right)^2\right]}\,dv,$$ which I tried to integrate by parts but it kept unravelling. Any help will be greatly appreciated. Thanks you.

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Hint. One may observe that, for $a>0$, $b>0$, on has $$ \int_a^\infty \exp{\left[-b\left(x-\frac1x\right)^2\right]}\,dx=\frac{\sqrt{\pi}}{4\sqrt{b}}\text{erfc}\left(a\sqrt{b}-\frac{\sqrt{b}}{a}\right)+\frac{\sqrt{\pi}e^{4b}}{4\sqrt{b}}\text{erfc}\left(a\sqrt{b}+\frac{\sqrt{b}}{a}\right). \tag1 $$ Starting from the initial integral making two successive changes of variable $$v=\frac1{u^n},\quad x=\sqrt{\frac{m}{l}}\:v,$$ one gets $$ \begin{align} \int_0^s u^{-n-1} e^{-\left(l u^n + \frac{m}{u^{n}}\right)^2}du =\frac1n\int_{1/s^{1/n}}^\infty e^{\large -\left(mv+\frac{l}v\right)^2}dv =\frac{\sqrt{m}e^{\large-4ml}}{n\sqrt{l}}\int_a^\infty e^{-\large ml\left(x-\frac1x\right)^2}dv \tag2 \end{align} $$ with $\displaystyle a=\sqrt{\frac{m}{l}}\frac1{s^{1/n}}$, then one may apply $(1)$ to evaluate $(2)$.

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  • $\begingroup$ @Oliver Oloa, Firstly thank you very much for answering, I know I already accepted this answer last night, but I am working through the calculation now, I believe there are 3 mistakes in equation 2, but one of them is that the sign of the exponent term became minus instead of plus, I do n't understand how this happened, so I am guessing this is a typo, but if this is corrected, we can not apply (1). Do you know of an equal identity in the plus sign case, if there is one ? $\endgroup$ – Comic Book Guy Jul 16 '16 at 21:47
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    $\begingroup$ @Comic Book Guy You are welcome. The minus sign is not a typo. It is correct. from $\left(mv+\frac{l}v\right)^2$, you get $ml\left(x+\frac{1}x\right)^2$, then observe that $\left(x+\frac{1}x\right)^2=\left(x-\frac{1}x\right)^2+4$ (do you see why?), which gives the factor $e^{\large-4ml}$ in front of the integral. $\endgroup$ – Olivier Oloa Jul 16 '16 at 22:45
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    $\begingroup$ @Oliver Oloa, that makes a lot of sense, thank you. $\endgroup$ – Comic Book Guy Jul 17 '16 at 0:12

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