Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Help me please with this improper integral:

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


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

share|cite|improve this question
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
up vote 4 down vote accepted

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

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

share|cite|improve this answer

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 $n\geq 0$.

share|cite|improve this answer

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.

share|cite|improve this answer
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 $$\int\limits_0^\infty e^{-y}dy \text{ ?}$$

share|cite|improve this answer
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 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 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.