# Improper integral

• Aug 6th 2010, 01:08 PM
Ulysses
Improper integral
Hi. Well, I've got this problem where the statement exhorts me to determinate the character of some improper integrals, and it begins with this one $\displaystyle \displaystyle\int_{-\infty}^{\infty}xe^{-x^2}dx$

I really don't know how to work this out. I think its not possible to integrate it, cause its not an elemental function. So I think I should use something, I know it converges to zero, cause I used mathematica to calculate it, but I don't know how to reason it, but maybe using something like the sandwich theorem for function, I mean squeezing it, knowing that integral is not bigger to some other one I could determinate its character. The other thing I thought watching at its graph was demonstrating that its symmetrical to the origin, so thats why it converges to zero. But I'm not sure.

Bye there. And thanks.
• Aug 6th 2010, 01:29 PM
Also sprach Zarathustra
Hint:

Prove: $\displaystyle \int x e^{-x^2} dx = {-e^{\frac{-x^2}{2}}+C$
• Aug 6th 2010, 01:33 PM
11rdc11
$\displaystyle \int^{\infty}_{-\infty} xe^{-x^2}dx$

$\displaystyle u=-x^2$ and $\displaystyle du = -2xdx$

sub into integral

$\displaystyle \int \frac{e^u}{-2}du$

which is

$\displaystyle \frac{e^u}{-2}$

$\displaystyle \frac{e^{-x^2}}{-2}\bigg|^{\infty}_{-\infty}$

$\displaystyle \lim_{x \to \infty} (\frac{1}{-2e^{-x^2}}+\frac{1}{2e^{-x^2}}) -\lim_{x \to -\infty} (\frac{1}{-2e^{-x^2}}+\frac{1}{2e^{-x^2}})$

$\displaystyle 0+0=0$
• Aug 6th 2010, 02:04 PM
Quote:

Originally Posted by Ulysses
Hi. Well, I've got this problem where the statement exhorts me to determinate the character of some improper integrals, and it begins with this one $\displaystyle \displaystyle\int_{-\infty}^{\infty}xe^{-x^2}dx$

I really don't know how to work this out. I think its not possible to integrate it, cause its not an elemental function. So I think I should use something, I know it converges to zero, cause I used mathematica to calculate it, but I don't know how to reason it, but maybe using something like the sandwich theorem for function, I mean squeezing it, knowing that integral is not bigger to some other one I could determinate its character. The other thing I thought watching at its graph was demonstrating that its symmetrical to the origin, so thats why it converges to zero. But I'm not sure.

Bye there. And thanks.

$\displaystyle f(0)=0$

$\displaystyle f(-x)=-f(x)$

It's an "odd" function.
However, it crosses the x-axis only at x=0
and converges on zero as x approaches plus and minus infinity.
Hence, it has a completely different character to functions such as sinx that oscillate.
The integral can be evaluated on both sides of the y-axis from x=0 since those integrals converge.

$\displaystyle f(x)\ge\ 0$ for $\displaystyle x\ge\ 0$

$\displaystyle f(x)\le\ 0$ for $\displaystyle x\le\ 0$

Therefore the integration of the function from 0 to $\displaystyle \infty$ is positive,
and the integration of the function from 0 to $\displaystyle -\infty$ is the negative of that.

The entire area between the curve and x-axis is

$\displaystyle 2\int_0^{\infty}xe^{-x^2}dx$
• Aug 6th 2010, 02:31 PM
mr fantastic
A general remark is in order here:

The posted integral is an improper integral. It has to be dealt with using limits. You CANNOT substitute infinity into an expression. The fact that the integrand is odd is NOT a proof that the integral is equal to zero (as a case in point, members would be well advised to review the Cauchy distribution - in particular the fact that the mean does not exist). Without wanting to sound disparaging, none of the replies so far given provide what would be considered a correct solution.

The OP is advised to go back and read his/her class notes and/or textbook in the first instance to get some background on improper integrals (I do not have the time to conduct a tutorial on the subject). Google will also find useful links.
• Aug 6th 2010, 03:28 PM
Plato
This is a post script to the immediate preceding post.
In order for $\displaystyle \int_{ - \infty }^\infty f \;$ to exist then for any $\displaystyle a\in \mathbb{R}$ then $\displaystyle \int_a^\infty f \;\& \,\int_{ - \infty }^a f$ must exist.
• Aug 6th 2010, 03:42 PM
Also sprach Zarathustra
It's trivial(to me...)
• Aug 6th 2010, 03:47 PM
Plato
• Aug 6th 2010, 03:54 PM
Also sprach Zarathustra
Quote:

Originally Posted by Plato
This is a post script to the immediate preceding post.
In order for $\displaystyle \int_{ - \infty }^\infty f \;$ to exist then for any $\displaystyle a\in \mathbb{R}$ then(*) $\displaystyle \int_a^\infty f \;\& \,\int_{ - \infty }^a f$ must exist.

Quote:

Of curse the existence of two(*) guaranty you the existence of $\displaystyle \int_{ - \infty }^\infty f \;$ .
• Aug 6th 2010, 04:10 PM
Also sprach Zarathustra
By the way...

I can prove more general result:

Let $\displaystyle p(x)$ be polynomial and $\displaystyle s>0$ then the improper integral $\displaystyle \displaystyle\int_{0}^{\infty}|p(x)|e^{-x^s}dx$ converges.
• Aug 6th 2010, 06:23 PM
11rdc11
Yes I'm sorry. Thanks for the correction Mr. Fantastic. To evaluate the limit, realize the denominator is growing faster than the numerator since it is exp so therefore it goes to 0. On another note, how do I make the notation appear below the limit?
• Aug 6th 2010, 07:07 PM
mr fantastic
Quote:

Originally Posted by 11rdc11
[snip]
On another note, how do I make the notation appear below the limit?

Use the command \displaystyle inside your tags.
• Aug 7th 2010, 04:31 AM
HallsofIvy
As a final note, $\displaystyle \int_{-\infty}^\infty f(x) dx$ is defined as $\displaystyle \lim_{A\to-\infty}\lim_{B\to\infty}\int_A^B f(x)dx$.

$\displaystyle \lim_{A\to\infty}\int_{-A}^A f(x)dx$ is called the "Cauchy principal value" which may exist even when the improper integral does not. If the improper integral exists, then so does the Cauchy principal values and they are equal.

Noting that $\displaystyle \int_{-A}^A xe^{-x^2}dx= 0$ for all A, because the integrand is odd, shows that the Cauchy principal value is 0 but you still must show that the improper integral exists.
• Aug 7th 2010, 11:27 AM
Ulysses
I began with a big mistake thinking it wasn't an elemental function. All the answers been helpful. Thanks you all.