Are you serious?

I am assuming you know the value if $\Gamma(\frac{1}{2})$ which is equal to $\sqrt{\pi}$ .

After that the problem becomes very trivial as we know that,

$\Gamma (x) = \int_{0}^{\infty} e^{-t}t^{-1+x} dt$

Applying Integration by parts,

$= [-t^{1-x}e^{-t}]_0^{\infty} + \int_{0}^{\infty} (x-1)e^{-t}t^{-2+x} dt$

$= (x-1)\int_{0}^{\infty} e^{-t}t^{-2+x} dt$

$= (x-1)\Gamma (x-1)$

Now, if we put $x= \frac{5}{2}$ ,

$\Gamma(\frac{5}{2}) = \frac{3}{2} \times \Gamma(\frac{3}{2})$

$= \frac{3}{2} \times \frac{1}{2} \times \Gamma(\frac{1}{2})$

$= \boxed{\frac{3\sqrt{\pi}}{4}}$

Now if you don't know that $\Gamma (\frac{1}{2}) = \sqrt{\pi}$ , we can prove it by the famous Euler's reflection formula which states that,

$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z}$ , where $z \in \mathbb{C}.$

Thus, plugging in $z= \frac{1}{2}$ , we get-

$\Gamma\left(\frac{1}{2}\right)^{2} = \left[\Gamma\left(\frac{1}{2}\right)\Gamma\left(1-\frac{1}{2}\right)\right] = \left(\frac{\pi}{\sin \frac{\pi}{2}}\right) = \pi$

And hence, $\Gamma(\frac{1}{2})= \sqrt{\pi}$

P.S. For more information regarding Gamma Function and its forms, click here

Cheers!