Question

improper integral. \[\int^{\infty}_0 x^2e^{-x^2}\]

Answer 1

is there a quick method to show that this converges?

Answer 2

limits. they both go "0:"

Answer 3

huh?

Answer 4

\[\lim_{x\to0}f(x)=0\\{\rm and}\\
\lim_{x\to\infty}f(x)=0
\]

Answer 5

and finite in between.
so, the integral exists

Answer 6

do you see it?

Answer 7

no. I don't.

Answer 8

let \(u=x^2\), \(du=2xdx\)\[\int^{\infty}_0\frac{ue^{-u}}{2\sqrt{u}}du\]\[\frac{1}{2}\int^{\infty}_0u^{1/2}e^{-u}du\]\[\frac{1}{2}\Gamma(3/2)\]

Answer 9

\[\frac{1}{2}\Gamma(1+1/2)=\frac{1}{2}\Gamma(1/2)=\frac{\sqrt{\pi}}{4}\]

Answer 10

I am so fed up with the section on improper integrals. the only way I can ever figure them out is by evaluating them. surely there must be a quick way to find a larger function that converges?

Answer 11

\[\LARGE \int\limits^{\infty}_0 \frac{x^2} {e^{x^2}}dx\] hmm maybe the comparison test http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx

Answer 12

every resource I can find on the comparison test tells me the same thing. what I suck at is finding the function to compare!

Answer 13

ok here I think that \[\frac{1}{x^2}\geq \frac{1}{e^{x^2}} \quad \forall \; x\in [0,\infty )\]

Answer 14

and \[\int^{\infty}_0 \frac{1}{x^2}\text{d}x\]converges. So by comparison \[ \int\limits^{\infty}_0 \frac{x^2} {e^{x^2}}\text{d}x\]converges?

Answer 15

is this correct? is there a systematic way to find the comparison function for when taking an exam?

Answer 16

Que 1) Does the function converge?
\[\lim_{x\to0}x^2e^{-x^2}=0\\
\lim_{x\to\infty}x^2e^{-x^2}=0\times\lim_{x\to\infty}x^2=0\]
So, the integral is finite.

Answer 17

Que. 2) What is its value?
\[
I={1\over2}\Gamma(1/2)=\Large \boxed{\pi\over4}
\]

Answer 18

thanks. I really get worried because this seems to be straight out of calc 2! I have no idea what they are trying to teach me...

Answer 19

i meant
\[I=\Large{\sqrt{\pi}\over4}\]

Answer 20

what do you mean? what calc class are you taking now?

Answer 21

uh it's called "Advanced Calculus and Introductory Analysis"

Answer 22

textbook is folland

Answer 23

it's basically revisiting the first three calculus courses from a different perspective and then doing some extra vector calculus stuff.

Answer 24

I see. Then you should expect to see more of calc 1 & 2 combined. and calc 3 is nothing but the same and a little simpler.
In this problem, what we have learnt is that the function might look non-converging but you cannot be sure unless you test the limits and for intermediate asymptotes. This looks like a differential of Gaussian distribution function

Answer 25

"In this problem, what we have learnt is that the function might look non-converging " It looks like a converging function to me, because the e^-x^2 exponent term approaches zero much faster than x^2, so the integral should converge to me \[\LARGE \int\limits\limits^{\infty}_0 \frac{x^2} {e^{x^2}}dx \]

Answer 26

x^2 approaches infinity much slower than e^x^2 on the bottom, so it should approach zero quickly and thus converge.

Answer 27

compared to something like 1/x or 1/x^2:

https://www.google.com/search?q=x%5E2%2F(e%5E(x%5E2))%2C+1%2Fx%2C+1%2Fx%5E2&aq=f&oq=x%5E2%2F(e%5E(x%5E2))%2C+1%2Fx%2C+1%2Fx%5E2&aqs=chrome.0.57j0j62l3.561j0&sourceid=chrome&ie=UTF-8