# Improper integral $\int_{0}^{\infty } e^{-\sqrt{x}}dx$

Help me please with this improper integral:

$\int_{0}^{\infty } e^{-\sqrt{x}}dx$

Thanks.

I solved it partially, and stuck after integration by parts.

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Care to post what you have done so far? – soandos Mar 22 '12 at 12:27
How far in integration by parts did you get? – yunone Mar 22 '12 at 12:28
$\underset{b \to \infty }{\lim }\int_{0}^{b}e^{-\sqrt{x}}dx=2\underset{b \to \infty }{\lim }\int_{0}^{b}te^{-t}dt=2\underset{b \to \infty }{\lim }\left ( -e^{-t}(t+1) \right )$ – Lilly Mar 22 '12 at 12:40
In general, $\displaystyle\int_0^\infty e^{-\large\sqrt[n]x}~dx=n!~$, for $n>0$. – Lucian Feb 2 '15 at 0:14

$$I= \displaystyle \lim_{a \to \infty} \int \limits_{0}^{a} e^{-\sqrt{x}}\, dx= \displaystyle \lim_{a \to \infty} \left(2-2 \cdot\frac{\sqrt{a}+1}{e^{\sqrt a}}\right)=2-2\cdot \displaystyle \lim_{a \to \infty} \frac{\sqrt{a}+1}{e^{\sqrt a}}=2$$

The last limit can be evaluated using substitution $t=\sqrt{a}~$ and L'Hopital rule .

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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 $n\geq 0$.

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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.

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You can use \tag{$\ast$} instead of \qquad(\ast) :) – user2468 Mar 22 '12 at 17:42
@J.D.: Thanks for the suggestion, I had slipped into hand-formatting, which I had not done since early TeX days. – André Nicolas Mar 22 '12 at 17:48

(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 $$\int\limits_0^\infty e^{-y}dy \text{ ?}$$

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Isn't the LP too much here, if the OP is struggling with integration by parts? (I think your answer is OK, but maybe you can simply calculate the integral with $x=y^2$) – Pedro Tamaroff Mar 22 '12 at 19:52
@PeterT.off you're absolutely right. It is not pedagogical. – user2468 Mar 22 '12 at 20:00
Mind if I edit it? – Pedro Tamaroff Mar 22 '12 at 20:02
@PeterT.off No problem. Go ahead please. – user2468 Mar 22 '12 at 20:07
I think the real reason I wanted to do this in Laplace transform is the other question we had yesterday! – user2468 Mar 22 '12 at 20:08