Improper Integral with Infinity in Limits

[johnhuntsman]

∞
∫xe^[-x^2] dx
-∞

So basically I've solved for everything in this problem and it looks like it should be an indeterminate form and thus divergent. My book and Wolfram both say it's 0 and convergent though.

I get it down into:
lim [[e^(-t^2)] - e^0]/2 + lim [e^0 - [e^(-v^2)]]/2
t->-∞__________________v->∞

When I plug stuff in I get:

[e^∞ - e^∞ - e^0 + e^0]/2

I can see why it might be 0 from the stuff above, but e^∞ - e^∞ should be indeterminate rather than 0. Can someone please explain what I'm not getting?Wolfram:
http://www.wolframalpha.com/input/?...=DefiniteIntegralCalculator.rangeend_infinity

 

[jbunniii]

The integrand is an odd function, so if you can show that $\int_{0}^{\infty} x e^{-x^2} dx$ is finite, then the integral from $-\infty$ to $\infty$ will be zero. Seems like a simple substitution will do the trick.

 

[johnhuntsman]

The integrand is an odd function, so if you can show that $\int_{0}^{\infty} x e^{-x^2} dx$ is finite, then the integral from $-\infty$ to $\infty$ will be zero. Seems like a simple substitution will do the trick.

I'm afraid I don't understand what you're saying.

 

[jackmell]

Well you know that sin(x) is an odd function right? An odd function means that f(-x)=-f(x) and the function just flips over when it crosses the origin like sin(x) does. So that if f(-x)=-f(x) then surely:

$$\int_{-1}^{1} f(x)dx=0$$

if f(x) is odd. Same diff for any symmetric interval like -2 to 2, -100 to 100 even -infty to infty.

So what about the function $x e^{-x^2}$? Is that one f(-x)=-f(x)? Then it would be an odd function then so that:

$$\int_{-a}^{a} f(x)dx=0$$

And would be likewise zero if a were infinity and the integral is finite in the interval

$$\int_0^{\infty} f(x)dx$$

And I think you can compute:

$$\int_0^{\infty}x e^{-x^2} dx$$

right?

 

[HallsofIvy]

You are trying to evaluate "$e^{\infty}$" when you should be looking at "$e^{-\infty}$.

 

[johnhuntsman]

Well you know that sin(x) is an odd function right? An odd function means that f(-x)=-f(x) and the function just flips over when it crosses the origin like sin(x) does. So that if f(-x)=-f(x) then surely:

$$\int_{-1}^{1} f(x)dx=0$$

if f(x) is odd. Same diff for any symmetric interval like -2 to 2, -100 to 100 even -infty to infty.

So what about the function $x e^{-x^2}$? Is that one f(-x)=-f(x)? Then it would be an odd function then so that:

$$\int_{-a}^{a} f(x)dx=0$$

And would be likewise zero if a were infinity and the integral is finite in the interval

$$\int_0^{\infty} f(x)dx$$

And I think you can compute:

$$\int_0^{\infty}x e^{-x^2} dx$$

right?

I see what you're talking about. Thanks. I messed up on my math anyway : D