Thread: Improper Integral xe^-(x^2)..

1. Improper Integral xe^-(x^2)..

I cant seem to get this in any way??

inttegral of xe^(-x^2)dx a=-infinite,b=infinite

The Answer in my textbook is 0, i dont know how to approach the problem? I know this i have to break it up into to several integrals, but i dont know how to integrate the function. Thank You for Any Help

2. Hello Kurosaki.

You just have to use this little fact: If $f(x)$ is an odd function, then $\int_{-a}^a f(x) \ dx = 0$

3. Originally Posted by zangestu888
I cant seem to get this in any way??

inttegral of xe^(-x^2)dx a=-infinite,b=infinite

The Answer in my textbook is 0, i dont know how to approach the problem? I know this i have to break it up into to several integrals, but i dont know how to integrate the function. Thank You for Any Help
What is the derivative of $e^{-x^2}$ ?

Then use the fundamental theorem of calculus.

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4. I dont quite understand? So thier is no actual way of doing it only just by the fact that its an odd function? am confused lol

5. Originally Posted by zangestu888
I cant seem to get this in any way??

inttegral of xe^(-x^2)dx a=-infinite,b=infinite

The Answer in my textbook is 0, i dont know how to approach the problem? I know this i have to break it up into to several integrals, but i dont know how to integrate the function. Thank You for Any Help
Because of the symmetry in the origin you should get zero.. To show this split the integral up into two pieces

$\int_{-\infty}^0 x e^{-x^2}\,dx + \int_0^{\infty} x e^{-x^2}\,dx$

and do each separately.

6. Originally Posted by o_O
Hello Kurosaki.

You just have to use this little fact: If $f(x)$ is an odd function, then $\int_{-a}^a f(x) \ dx = 0$
That does not prove that the double limit exists, it only tells you what its value is if it does.

.

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7. Wel i know how to split it up i got that part, my problem is integrating the functioon after the split?

8. i tried subs. and parts i didnt get anything after the split

9. Originally Posted by zangestu888
Wel i know how to split it up i got that part, my problem is integrating the functioon after the split?
Let $u = -x^2,\;\;\;du = -2x \,dx$ and switch the limits.

10. okay idid that but then i would have to integrate e^-(u)^2how can i do that

11. i have this integral of -1/2(e^-(u^2))

12. Originally Posted by zangestu888
okay idid that but then i would have to integrate e^-(u)^2how can i do that
Actually no. Under $u = -x^2, du = -2x\,dx$

$\int x\, e^{-x^2}\, dx = - \frac{1}{2} \int e^u \,du$

13. omg lol am so stupied i didnt see that thanks!!

14. Originally Posted by zangestu888
okay idid that but then i would have to integrate e^-(u)^2how can i do that

$\frac{d}{dx}e^{-x^2}=-2x e^{-2x}$

so:

$\int x e^{-2x} \ dx= -\frac{1}{2} e^{-x^2}$

and so:

$
\int_{-\infty}^{\infty}x e^{-2x} \ dx=\lim_{a\to -\infty,\ b\to \infty}\int_{a}^{b}x e^{-2x} \ dx=\lim_{a\to -\infty,\ b\to \infty}\left(e^{-a^2}-e^{-b^2}\right)=0
$

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integral of x.e^(-x/2) from 0 to infinity

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