Question

another e integration: xe^(-x^2)

x times e to the negative x squared

Answer 1

\[\int\limits_{-\infty}^{\infty}xe ^{-x ^{2}}\]

Answer 2

Try a change of variables: u = x^2. You'll find this comes out quite nicely.

Answer 3

good cause u=x was making me unhappy

Answer 4

u = - x^2 -> du = -2xdx => xdx = du/-2

Answer 5

yay and u=-x^2 made me even happier:\[ -(1/2) e ^{-x ^{2}}\]

Answer 6

( -1/2) integral ( e^u du)

Answer 7

I have to keep the du and the other thing separate or I get confused dx = -1/2 du/1

Answer 8

If you use the substitution u = -x^2, then the indefinite integral becomes
\[ \int \frac{-1}{2}e^u \ du = \frac{-1}{2}e^u + C = \frac{-1}{2}e^{-x^2} + C \]

Now what does that mean for your definite integral?

Answer 9

reverse the limits

Answer 10

It is gonna be even rougher than that because he is worse than definite he is improper....

Answer 11

Yes, but what's the limit
\[ \lim_{x \rightarrow \infty} e^{-x^2} \]
?

Answer 12

zero ish?

Answer 13

Zero exactly.

Similarly, what is
\[ \lim_{x \rightarrow -\infty} e^{-x^2} \]

Answer 14

negative negative exponent looks like infinity squared

Answer 15

No, think through it

Answer 16

everyone here knows thinking isn't my strongest skill......direct substitution no?

Answer 17

\[ \lim_{x \rightarrow -\infty} x^2 = +\infty \]
hence
\[ \lim_{x \rightarrow -\infty} -x^2 = -\infty \]
and therefore
\[ \lim_{x \rightarrow -\infty} e^{-x^2} = ... what? \]

Answer 18

aha negative numbers squared go back to positive....reminds me during class I decided 3*1 was 9 for some reason.......must remember to think of that as -(x^2) I have goofed that before

it is still infinity just the negative kind

Answer 19

zero minus negative infinity would seem divergent...I don't want to commit just yet

Answer 20

no no no that converges at zero because the lim at -infinity is also zero

Answer 21

whew thanks

Answer 22

Yes, at both ends the limit is zero.

You can also see that the integral is zero because the function \( f(x) = xe^{-x^2} \) is an odd function, i.e., \( f(-x) = -f(x) \). Hence for all \( R \),
\[ \int_{-R}^R f(x) \ dx = 0 .\]

Answer 23

Are you saying that all odd functions in an improper integral will be convergent at zero?

Answer 24

I am saying exactly what I wrote above. And that implies that if f is an odd function ___and if this integral exists___ then
\[ \int_{-\infty}^\infty f(x) \ dx = 0 \]

We need the second condition because \( f(x) = x \) is an odd function, but in this case, that improper integral does not exist.

Answer 25

ok I'll have to think on that...thanks for all the patience and help -- If I could give out more medals I'd do it