integration

**romtik** · March 19th 2008, 10:58 PM

What is the integal of e^(x^2)x^4 from negative infinity to infinity?

Thank you!

**Moo** · March 19th 2008, 11:54 PM

Hello,

The function is an even function ie $f(x) = e^{x^2} x^4 = e^{(-x)^2} (-x)^4 = f(-x)$

---------> $\int_{- \infty}^{+\infty} f(x) dx = 2 \int_{-\infty}^0 f(x) dx$

Now, let's integrate.

At first sight, you have to use the integration by parts, this must be almost a reflex when you see an exponential. The exponential will very often take the place of the v' function in :

$\int_a^b u(x)v'(x) dx = [u(x)v(x)]_a^b - \int_a^b u'(x)v(x) dx$

So you must make appear the derivate of $x^2$ , 2x, so as to be able to integrate :

the integrate of $u'(x)e^{u(x)}$ will simply be $e^{u(x)}$

$2 \int_{-\infty}^0 e^{x^2} x^4 dx = \int_{-\infty}^0 (2x e^{x^2}) x^3 dx$

------ integration by parts ------

the derivate of $x^3$ is $3x^2$

$\int_{- \infty}^{+\infty} e^{x^2} x^4 dx = [e^{x^2} x^3]_{-\infty}^0 - \int_{-\infty}^0 3x^2 e^{x^2} dx$ $= \lim_{x \to 0} e^{x^2} x^3 - \lim_{x \to - \infty} e^{x^2} x^3 - \int_{-\infty}^0 3x^2 e^{x^2} dx$ $= - \int_{-\infty}^0 3x^2 e^{x^2} dx$

(the two limits tend to 0)

In the same way, make appear the derivate of $x^2$ so as to integrate once again by parts.

$- \int_{-\infty}^0 3x^2 e^{x^2} dx = - \frac{3}{2} \int_{-\infty}^0 (2x)e^{x^2} x dx = - \frac{3}{2} [e^{x^2} x]_{-\infty}^0 + \frac{3}{2} \int_{-\infty}^0 e^{x^2} dx$

And there, i can't help anymore because of the last integral : i don't remember how to do

**Moo** · March 20th 2008, 12:17 AM

Ay ! I found ^^

This integrate is not defined, because though the limit in $- \infty$ exists (0), the limit in $+ \infty$ doesn't.

$\lim_{x \to +\infty} e^{x^2} x^4 = + \infty$ , so the integrate diverges.

Are you sure it's the correct text ?

**romtik** · March 20th 2008, 07:42 AM

Hi,

Yea that is the question..ahhhhh I needed the answer to be zero! Ah well, thanks anyways!

**ThePerfectHacker** · March 20th 2008, 08:27 AM

Note, $x^2 \geq 0 \implies e^{x^2} \geq 1 \implies x^4 e^{x^2} \geq x^4$ .

Thus, $\int_{-\infty}^{\infty} x^4 e^{x^2} dx \geq \int_{-\infty}^{\infty}x^4 dx$ which clearly diverges since RHS diverges.

Maybe the poster wanted, $x^4 e^{-x^2}$ ? (Note the negative).

In that case,
$\int_{-\infty}^{\infty} x^4 e^{-x^2} dx = \int_{-\infty}^{\infty} \left( - \frac{1}{2} x^3 \right) \cdot \left( e^{-x^2} \right)' ~ dx = -\frac{1}{2} x^3 e^{-x^2}\bigg|_{-\infty}^{\infty} +\int_{-\infty}^{\infty} \left(-\frac{3x}{4}\right)\cdot \left( e^{-x^2} \right) ' dx$
$=-\frac{3}{4}xe^{-x^2} \bigg|_{-\infty}^{\infty} + \frac{3}{4}\int_{-\infty}^{\infty} e^{-x^2} dx = \frac{3\sqrt{\pi}}{8}$

**wingless** · March 20th 2008, 10:17 AM

Originally Posted by Moo

Ay ! I found ^^

This integrate is not defined, because though the limit in $- \infty$ exists (0), the limit in $+ \infty$ doesn't.

$\lim_{x \to +\infty} e^{x^2} x^4 = + \infty$ , so the integrate diverges.

No, $\lim_{x \to - \infty}e^{x^2} x^4 = +\infty$ too.

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