How to Evaluate proper Integral. Recently I stumbled upon an integral and its solution in a physics article but I couldn't understand how it was evaluated.I have plotted the function and it indicates that the value of the integral should be finite.
I want to ensure the answer is correct.Does anybody know how to do it?
$$\int_0^T d t'  \frac{1}{(t'(T-t')) ^{3/2}}\exp \left[ -\frac{A}{t'}- \frac{B}{T-t'}\right]= \sqrt{  \frac{\pi}{T^{3}} }  \frac{ \sqrt{A}+ \sqrt{B} }{ \sqrt{AB} }\exp\left[-\frac{(\sqrt{A}+\sqrt{B})^{2}}{T}\right] $$ 
 A: Assuming $A,B,T>0$,
$$\begin{align}
I=&\int_{0}^{T}\frac{1}{(t(T-t))^{3/2}}\exp\left[-\frac{A}{t}-\frac{B}{T-t}\right]\,dt\\
\stackrel{x=1/t}{=}&\int_{1/T}^{\infty}\frac{x}{(Tx-1)^{3/2}}\exp\left[-Ax-\frac{Bx}{Tx-1}\right]\,dx\\
\stackrel{t=Tx}{=}&\;\frac{1}{T^{2}}\int_{1}^{\infty}\frac{t}{(t-1)^{3/2}}\exp\left[-\frac{1}{T}\left(At+\frac{Bt}{t-1}\right)\right]\,dt\\
\stackrel{x=t-1}{=}&\;\frac{e^{-(A+B)/T}}{T^{2}}\int_{0}^{\infty}\frac{x+1}{x^{3/2}}\exp\left[-\frac{1}{T}\left(Ax+\frac{B}{x}\right)\right]\,dx\\
\stackrel{t=\sqrt{x}}{=}&\;\frac{2e^{-(A+B)/T}}{T^{2}}\int_{0}^{\infty}\left(1+\frac{1}{t^{2}}\right)\exp\left[-\frac{1}{T}\left(At^{2}+\frac{B}{t^{2}}\right)\right]\,dt.
\end{align}$$
Then using
$$\int_{0}^{\infty}\exp\left[-ax^{2}-\frac{b}{x^{2}}\right]\,dx=\int_{0}^{\infty}\exp\left[-bx^{2}-\frac{a}{x^{2}}\right]\frac{1}{x^{2}}\,dx=\frac{\sqrt{\pi}e^{-2\sqrt{ab}}}{2\sqrt{a}} \tag{1}$$
we have
$$\int_{0}^{\infty}\left(1+\frac{1}{t^{2}}\right)\exp\left[-\frac{1}{T}\left(At^{2}+\frac{B}{t^{2}}\right)\right]\,dt=\frac{\sqrt{\pi T}e^{-2\sqrt{AB}/T}}{2}\left(\frac{1}{\sqrt{A}}+\frac{1}{\sqrt{B}}\right).$$
Thus
$$\begin{align}
I&=\frac{2e^{-(A+B)/T}}{T^{2}}\frac{\sqrt{\pi T}e^{-2\sqrt{AB}/T}}{2}\left(\frac{1}{\sqrt{A}}+\frac{1}{\sqrt{B}}\right)\\
&=\boxed{\sqrt{\frac{\pi}{T^{3}}}\left(\frac{1}{\sqrt{A}}+\frac{1}{\sqrt{B}}\right)\exp\left[-\frac{(\sqrt{A}+\sqrt{B})^{2}}{T}\right]}
\end{align}$$

Proof of $(1)$:
$$\begin{align}
\int_{0}^{\infty}\exp\left[-ax^{2}-\frac{b}{x^{2}}\right]\,dx
\stackrel{t=\sqrt{a}x}{=}&\;\frac{1}{\sqrt{a}}\int_{0}^{\infty}\exp\left[-t^{2}-\frac{ab}{t^{2}}\right]\,dt\\
\stackrel{x=\sqrt{ab}/t}{=}&\;\frac{\sqrt{ab}}{\sqrt{a}}\int_{0}^{\infty}\exp\left[-\frac{ab}{x^{2}}-x^{2}\right]\frac{1}{x^{2}}\,dx\\
={}&\;\frac{1}{2\sqrt{a}}\int_{0}^{\infty}\left(\frac{\sqrt{ab}}{x^{2}}+1\right)\exp\left[-\frac{ab}{x^{2}}-x^{2}\right]\,dx \tag{2}\\
\stackrel{t=x-\sqrt{ab}/x}{=}&\;\frac{1}{2\sqrt{a}}\int_{-\infty}^{\infty}\exp\left[-t^{2}-2\sqrt{ab}\right]\,dt\\
={}&\;\frac{e^{-2\sqrt{ab}}}{2\sqrt{a}}\int_{-\infty}^{\infty}e^{-t^{2}}\,dt\\
={}&\;\frac{\sqrt{\pi}e^{-2\sqrt{ab}}}{2\sqrt{a}},
\end{align}$$
where $(2)$ was obtained by averaging the previous two representations of the integral.
