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 $b>0$, $T>0$.
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.
