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!
 A: 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.
A: I just briefly want to mention that these integrals are examples of convolutions and may be handled by the convolution theorem for Laplace transforms.
The convolution theorem for Laplace transforms states that, for two real-valued functions $f$ and $g$ having Laplace transforms $F$ and $G$, respectively, the following relation holds:
$$\int_0^T dt \, f(t)g(T-t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, F(s) G(s) e^{s T} $$  
where $c$ is greater than the real part of each pole of $FG$.  For example, when $f(t) = t^{-1/2} e^{-b^2/(4 t)}$ and $g(t) = t^{-1/2}$, we have $F(s) = \sqrt{\pi/s} \, e^{-b \sqrt{s}}$ and $G(s) =  \sqrt{\pi/s}$.  Accordingly, the integral is then the inverse Laplace transform of $(\pi/s) e^{-b \sqrt{s}}$, which one may show is 
$$\pi \, \operatorname{erfc}{\left ( \frac{b}{2 \sqrt{T}} \right )} $$
A: 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. 
