How to prove that $\int_{0}^{\infty} e^{1/2 - x/2} \sqrt{x} dx = \sqrt{2 e \pi}$ Hello related to this Inequality with power and logarithm we have got this :
$$\lim\limits_{n \to \infty}(\frac{ln(x)+2n}{2n-1+x})^n=e^{1/2 - x/2} \sqrt{x}=f(x) $$
So I was wondering : What's the value of the integral of $f(x)$? And Wolfram Alpha says :
$$\int_{0}^{\infty} e^{1/2 - x/2} \sqrt{x} dx =  \sqrt{2 e \pi}≈4.13273$$
But I have not the level to prove this . So what's your finest method to prove this ? 
Thanks a lot.
 A: $$\int_0^{+\infty}e^{1/2 - x/2}\sqrt{x}\ dx = e^{1/2}\int_0^{+\infty}x^{1/2} e^{-x/2}$$
We can call without any loss, $\alpha = \frac 12$ and write more generally:
$$e^{1/2}\int_0^{+\infty} x^{\alpha}e^{-\alpha x}\ dx$$
That integral, you will surely face it later with more calculus notions, is a special function called the Gamma Function. More generally, your integral result is:
$$a^{-\alpha} \Gamma (\alpha)$$
hence eventually
$$\sqrt{e}\alpha^{-\alpha} \Gamma (a)$$
and being $\alpha = 1/2$, you get
$$\sqrt{e}\ \Gamma[1/2] (1/2)^{-1/2}$$
Now, $\Gamma[1/2] = \sqrt{\pi}$ so that at the end:
$$\sqrt{e}\sqrt{\pi}\sqrt{2} \to \sqrt{2e\pi}$$
A: A probabilistic approach.
Use the substitution $u^2=x$ so that $2udu = dx$. We get:
\begin{align}
\int^\infty_0 e^{1/2-x/2} \sqrt[]{x}dx = \int^\infty_0 2\sqrt[]{e} u^2 e^{-u^2/2}du = \sqrt[]{2e\pi} \int^\infty_{-\infty}u^2 \frac{1}{\sqrt[]{2\pi}} e^{-u^2/2}du =  \sqrt[]{2e\pi} \ \mathbb{E}[Z^2] = \sqrt[]{2e\pi}
\end{align}
Where $Z\sim \mathcal{N}(0,1)$ (standard normal distributed).
A: We want to show that 
\begin{eqnarray*}
\sqrt{e} \int_{0}^{\infty} e^{- x/2} \sqrt{x} dx =  \sqrt{2 e \pi}.
\end{eqnarray*}
Substitute $x=t^2$ and integrate by parts
\begin{eqnarray*}
2 \sqrt{e} \int_{0}^{\infty} t^2 e^{- t^2/2}  dt =2 \sqrt{e} \left(  \underbrace{\left[ -t e^{-t^2/2} \right]_0^{\infty}}_{=0} + \int_{0}^{\infty}  e^{- t^2/2}  dt \right)
\end{eqnarray*}
Now let 
\begin{eqnarray*}
I=  \int_{0}^{\infty}  e^{- t^2/2}  dt = \frac{1}{2} \int_{-\infty}^{\infty}  e^{- t^2/2}  dt
\end{eqnarray*}
Square this and change to polar coordinates
\begin{eqnarray*}
I^2 &=&  \frac{1}{4} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{- (x^2+y^2)/2}  dx dy \\
&=& \frac{1}{4}  \int_{0}^{\infty} \int_{0}^{2\pi} e^{- r^2/2}  r dr d\theta \\
&=& \frac{\pi}{2} \\
\end{eqnarray*}
So $I= \sqrt{\frac{\pi}{2}}$ and the result follows.
