$\int\limits_0^\infty \frac{e^{-x}}{\sqrt{x}} dx$ How to compute $$\int\limits_0^\infty \frac{e^{-x}}{\sqrt{x}}\ dx?$$
I tried integration by parts, but it gave me division by 0. Should I substitute variable? 
 A: First substitute $x=y^2$ so your integral is $2\int_0^\infty e^{-y^2}dy$. Then pick your favourite proof this is $\sqrt{\pi}$. The first is by far the most common; the sixth doesn't even require the above substitution.
A: Using the $u$-substitution $u=\sqrt{x}$, we find $$\int_0^\infty\frac{e^{-x}}{\sqrt{x}}\,dx=2\int_0^\infty e^{-u^2}\,du.$$ This latter integral is well known so that we have $$\int_0^\infty \frac{e^{-x}}{\sqrt x}\,dx=\sqrt{\pi}.$$
A: What works here is to get this into the form of a Gaussian integral by letting $x = u^2$.
Then $dx = 2u \,du$ and the integral becomes
$$
\int_{u=0}^\infty \frac1 u e^{-u^2} 2u\,du = \int_{u=0}^\infty  e^{-u^2} du = \sqrt{\pi}
$$
The way that Gaussian integral is found is by first saying that it is half the integral from $-\infty$ to $\infty$, then saying that integral is the square root of the double integral $dx\,dy$ of $e^{-x^2} e^{-y^2}$, then changing to polar coordinates to get 
$$\int_{r=0}^\infty e^{-r^2} r\,dr\,d\theta$$
and because the $r$ from the transformation of $dy\,dx$ has come into play, this integral is easy.
A: 
For $\Re[z] > 0, $
$$\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}dt$$
  For $z \in \mathbb N,$ $$\Gamma(z) = (z - 1)!$$
For $z \notin \mathbb Z,$ $$\Gamma(1-z)\Gamma(z) = \frac{\pi}{sin(\pi z)}$$

So, to solve $\int_{0}^{\infty}x^{-1/2}e^{-x}dx$, let $z = \frac{1}{2}$ in the equation above, thus we have 
$$\int_{0}^{\infty}x^{-1/2}e^{-x}dx = \Gamma(\frac{1}{2})$$
Now, since $z = \frac{1}{2}$, $$\Gamma(1 - \frac{1}{2})\Gamma(\frac{1}{2}) = \frac{\pi}{sin(\pi z)}$$ $$\therefore \space \Gamma(\frac{1}{2})^2 = \frac{\pi}{sin(\frac{\pi}{2})} = \pi$$ $$ \therefore \space \Gamma(\frac{1}{2}) = \sqrt{\pi}$$
Thus, we have that $$\int_{0}^{\infty}x^{-1/2}e^{-x}dx = \sqrt{\pi}. \quad \blacksquare$$
