Improper integral $ \int_{0}^{\infty } e^{-\sqrt{x}}\text dx$ Help me please with this improper integral:
$ \int_{0}^{\infty } e^{-\sqrt{x}}\text dx$
Thanks.
I solved it partially, and stuck after integration by parts. 
 A: Let $u=\sqrt{x}$.  Then $du=\dfrac{1}{2\sqrt{x}}\,dx$.  But, using $\sqrt{x}=u$, we have $du=\dfrac{1}{2u}\, dx$.  So, $2u\,du=dx$.  You are also going to need that 
$$
\lim_{x\to\infty} x^ne^{-x}=0
$$
for any $n\geq 0$.
A: The integration by parts that you mention in a comment went bad for a truly minor reason. But I have seen versions of this slip before, so it is maybe worth commenting on.
We want $\int_0^b 2te^{-t}\,dt$. Let $u=2t$, and $dv=e^{-t}\,dt$. You decided to find an antiderivative of $2te^{-t}$. 
We get
$$\int 2te^{-t}\,dt=-2te^{-t}+\int 2e^{-t}\,dt=-2te^{-t}-2e^{-t}+C=-2(t+1)e^{-t}+C. \qquad(\ast)$$
This seems to be precisely what you did, apart from the $+C$ that I added because of excessive fussiness.
We now want to "plug in $b$, take away the result of plugging in $0$." But because we are so accustomed to the result of plugging in $0$ being $0$, it is all too easy not to see the $0$. However, in this case, and often with integration of exponentials, the important action is at $0$. We find that 
$$ \int_0^b 2te^{-t}\,dt=2-2(b+1)e^{-b}.$$
The rest is routine limit taking.
Remark: As a parenthetical remark, I would prefer to work with the definite integral, as in 
$$\int_0^b 2te^{-t}\,dt=\left.(-2te^{-t})\right|_0^b+\int_0^b 2e^{-t}\,dt.$$
Less algebra, and the first part dies at both ends. 
A: $$I= \displaystyle \lim_{a \to \infty} \int \limits_{0}^{a} e^{-\sqrt{x}}\, \text dx= \displaystyle \lim_{a \to \infty} \left(2-2 \cdot\frac{\sqrt{a}+1}{e^{\sqrt a}}\right)=2-2\cdot  \displaystyle \lim_{a \to \infty} \frac{\sqrt{a}+1}{e^{\sqrt a}}=2$$
The last limit can be evaluated using substitution $t=\sqrt{a}~$ and  L'Hopital rule .
A: (homework) so hints. 
First, substitute $x = y^2,$ we get
$$
\newcommand\L[1]{\mathcal{L}\left[#1\right]}
\int\limits
_{0}^{\infty} e^{-\sqrt{x}}\, dx = \int\limits_{0}^{\infty} 2y e^{-y}\, dy $$
Integration by parts gives ($y=u$, $dv =e^{-y}dy$)
$$\int\limits_{0}^{\infty} 2y e^{-y} dy =\left. -2e^{-y}y \right| _0^\infty +2\int\limits_0^\infty e^{-y}dy $$
So, to what does $-2e^{-y}y$ evaluate for $y \to \infty$ and $y \to 0$?
What is $$\int\limits_0^\infty e^{-y}dy \text{ ?}$$
