How to do integral $\int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}dt$ and $\int_0^T \frac{1}{\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}dt$? I met these two integrals but don't know how to do them:

$$I_1 = \int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t$$

$$I_2 = \int_0^T \frac{1}{\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t$$

where $b&gt;0$, $T&gt;0$.

Please kindly help?



Thanks to hits from Fabien, for the first one, let $t=\frac{T}{u^2+1}$:

$$I_1 = \int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t = \frac{2}{T}e^{-\frac{b^2}{2T}} \int_0^{+\infty}  e^{-\frac{b^2}{2T}u^2} \text{d}u  = \frac{\sqrt{2\pi}}{b\sqrt{T}} e^{-\frac{b^2}{2T}} $$

seems this tally with Mhenni Benghorbal's result.

For the second one, let $t=\frac{T}{u^2+1}$, $a^2=\frac{b^2}{2T}$:

$$I_2  =2 e^{-\frac{b^2}{2T}} \int_0^{+\infty} \frac{1}{1+u^2}  e^{-\frac{b^2}{2T}u^2} \text{d}u  =2 e^{-a^2}  \int_0^{+\infty} \frac{1}{1+u^2}  e^{-a^2u^2} \text{d}u := 2 e^{-a^2} I_3 $$

Then

$$I_3 = \int_0^\infty \frac{1}{1+u^2} e^{-a^2u^2} \text{d}u = \int_0^\infty \text{d}u\, e^{-a^2u^2} \int_0^\infty e^{-x(1+u^2)} \text{d}x = \int_0^\infty \text{d}x\, e^{-x} \int_0^\infty e^{-(a^2+x)u^2} \text{d}u$$

So $$I_3 = \int_0^\infty \text{d}x e^{-x} \frac{1}{2} \sqrt{\frac{\pi}{a^2+x}} = \frac{\sqrt{\pi}}{2} \int_0^\infty \frac{e^{-x}}{\sqrt{a^2+x}} \text{d}x$$

Let $t=\sqrt{a^2+x}$, so $x=t^2-a^2$, $\text{d}x = 2t\text{d}t$,

$$I_3 = \sqrt{\pi} \int_a^\infty e^{-(t^2-a^2)} \text{d}t = \sqrt{\pi} e^{a^2} \frac{\sqrt{\pi}}{2}\, \text{erfc}(a)$$

So $$I_2 = 2e^{-a^2} I_3 =  \pi \, \text{erfc} \left(\frac{b}{\sqrt{2T}}\right)$$

Same as Mhenni Benghorbal's answer, $$I_2=  \, \pi- 
{{\rm erf}\left( {\frac {b}{\sqrt {2T}}}\right)}.$$

YEAH!
Hint :
Try the substitution $$t=\cfrac{T}{u^2+1}$$

The first integral has the shape of the gaussian.
The second one leads you to $$I(\beta) = \alpha \int_{\mathbb{R}^+} \cfrac{1}{u^2+1} \exp(-\beta(u^2+1)  ) \,du$$

Considering $$\begin{cases} I'(\beta)=\alpha \int_{\mathbb{R}^+} \exp(-\beta(u^2+1)  ) \,du=\alpha\exp(-\beta)\cfrac{\sqrt{\pi}}{2\sqrt{\beta}} \\ I(0) = \alpha \int_{\mathbb{R}^+} \cfrac{1}{u^2+1}  \,du=\cfrac{\alpha\pi}{2}\end{cases}$$

with the substitution $\gamma^2 = \beta$ ends the problem.
Here are closed forms for the first and second  integrals respectively


  $$ I_1 = {
\frac {\sqrt {2\pi }}{b\sqrt {T}}}{{\rm e}^{-\,{\frac {{b}^{2}}{2T}}}} $$
  
  $$I_2=  \, \pi- 
{{\rm erf}\left( {\frac {b}{\sqrt {2T}}}\right)}.$$


You can test them numerically.