Scary looking integral from a movie background I watched recently this movie: https://www.imdb.com/title/tt3149038/mediaviewer/rm261224704
and saw on the poster background, (top-left) the following integral
$$\int_{\large\frac{v}{\sqrt{t}}}^{+\infty}\frac{e^{\large-\frac{u^2}{2}}du}{(1-\tfrac{y^2}{u^2})^{3/2}}$$
I also addded a picture below just in case from the first link it's unclear.
I do not know how to approach such a scary looking integral, elementary methods that I know fails, but I am genuinely curious if it's actually something meaningful or just something trown randomly. Also I  learnt normal distribution recently and thought that it might be related, but the denominator doesn't match in any way. How can this integral be approached in order to obtain a closed form and does it have any meaning in mathematics, statistics or physics? 

 A: For the sake of completion, here's a solution to the integral.
We'll analyze the following auxiliary integral:
\begin{align*}
 \frac{1}{\sqrt{\pi}}\int_{\sqrt{a^2 +1}}^{\infty} \frac{x e^{- (bx)^2}}{(x^2 -1)^{\frac32}} \mathrm{d}x
& \overset{\color{blue}{t = b^2(x^2 -1)}}{=} \frac{be^{-b^2}}{2\sqrt{\pi}}\int_{(ab)^2}^{\infty} \frac{e^{-t}}{t^{\frac{3}{2}}}\mathrm{d}t\\
& \overset{\text{I.B.P.}}{=}  \frac{be^{-b^2}}{\sqrt{\pi}}\left[\frac{e^{-(ab)^2}}{ab} -\int_{(ab)^2}^{\infty} \frac{e^{-t}}{t^{\frac{1}{2}}}\mathrm{d}t\right]\\
&  \overset{\color{blue}{s = \sqrt{t}}}{=}  e^{-b^2}\left[\frac{e^{-(ab)^2}}{a\sqrt{\pi}} - b\frac{2}{\sqrt{\pi}}\int_{ab}^{\infty} e^{-s^2}\mathrm{d}s\right]\\
& =e^{-b^2}\left[\frac{e^{-(ab)^2}}{a\sqrt{\pi}} - b\,\mathrm{erfc}\left(ab\right)\right]
\end{align*}
Now, using Feynman's trick we can differentiate both sides of the previous equation with respect to $b$ to obtain:
$$
-\frac{2b}{\sqrt{\pi}}\int_{\sqrt{a^2 +1}}^{\infty} \frac{x^3 e^{- (bx)^2}}{(x^2 -1)^{\frac32}} \mathrm{d}x = \left(2b^2 -1 \right)e^{-b^2}\mathrm{erfc}\left(ab\right) -\frac{2be^{-(a^2+1)b^2}}{a\sqrt{\pi}} \tag{1}
$$
Finally, since the integral of interest can be re-written as
$$
\int_{\large\frac{v}{\sqrt{t}}}^{\infty}\frac{e^{-\frac{u^2}{2}}}{\left(1-\frac{y^2}{u^2}\right)^{\frac32}}\mathrm{d}u \overset{\color{blue}{xy =u }}{=} y\int_{\large\color{purple}{\frac{v}{y\sqrt{t}}}}^{\infty} \frac{x^3 e^{-\left(\color{purple}{\frac{y}{\sqrt{2}}} x\right)^2}}{(x^2 -1)^{\frac{3}{2}}}\mathrm{d}x
$$
we can substitute $a = \sqrt{\frac{v^2}{y^2t}-1}$ and $b = \frac{y}{\sqrt{2}}$ into equation $(1)$ to get the final result
$$
\boxed{\int_{\large\frac{v}{\sqrt{t}}}^{\infty}\frac{e^{-\frac{u^2}{2}}}{\left(1-\frac{y^2}{u^2}\right)^{\frac32}}\mathrm{d}u  =y^2 e^{-\frac{v^2}{2t}} \sqrt{\frac{t}{v^2 -y^2t}}+\sqrt{\frac{\pi}{2}}\left(1 -y^2\right)e^{-\frac{y^2}{2}}\mathrm{erfc}\left(\sqrt{\frac{v^2}{2t} -\frac{y^2}{2}}\right)}
$$
valid for $v> |y|\sqrt{t}$.
