Divergent or Convergent? Evaluating an Integral with Exponential Function

[cathy]

Homework Statement

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it.

∫ from negative infinity to infinity of (x^8*e^-x^9)

The Attempt at a Solution

The answer is diverged to infinity. But I got that by guessing. Can someone explain to me why, please?

100% lost on this problem.
I changed the interval to 2∫ from 0 to infinity of (x^8*e^-x^9). I don't even know if I needed to do that or not.

 

[jackarms]

I'm not so sure that you can change the interval, actually -- you can only do that for even functions, and I think the $$e^{-x^{9}}$$ messes that up. Have you been able to do the anti-derivative of the function? If so, what did you get? If not, u-substitution looks like a good option.

 

[cathy]

u=-x^9
du=-9x^8

-1/9∫from -inf to inf of (x^8e^u)

 

[jackarms]

Close -- with the exponential function you copy down the original function, then do the derivative of the exponent. So you would keep the e^-x^9, and then multiply that by the derivative of $-x^{9}$

Edit: Sorry, saw (I think you edited) your other stuff. Yes, you would have u be -x^9, and then that would be the correct du.

 

[cathy]

I edited my above post. Could I also do it like that?

 

[cathy]

and then how do i actually take integral from -inf to inf?

 

[jackarms]

Yes, that looks right. Make sure you replace $x^{8}$ with du in the integral though.

 

[cathy]

okay, so would the integral = [1/(9x^9)e^-x^9)] if I did that correctly?

 

[cathy]

in bounds from -inf to inf. This was the major part I was confused on. How to I do this bounds part to let me know it is diverging to infinity?

 

[jackarms]

No, I think you have an error in substituting. If you have $u = -x^{9}$, then $du = -9x^{8}dx$, so substituting into your integral you should get:$$-\frac{1}{9}\int e^{u}du$$

 

[cathy]

but this is what I got that integral equaled. like after taking the antiderivative and subsituting back in u.
[1/(9x^9)e^-x^9)]

 

[jackarms]

Well, first you want to integrate and get everything back in terms of x. Then you essentially just plug in the infinities as you would with other bounds, and then if it converges you'll get a finite answer, and an infinite answer otherwise.

 

[jackarms]

No, that isn't the right antiderivative though. You're missing a negative, and the $x^{9}$ shouldn't be there in the front.

 

[cathy]

but that's my problem. when i plug in infinity, wouldn't it give me infinity, and then subtract from when i plug in -inf, and that gives me -inf and don't those cancel to equal 0?

 

[cathy]

ok okay, the right antiderviative is -1/9e^-x^9

 

[cathy]

i got it. thank you so much jackarms.

 

[jackarms]

Well, make sure you have the right antiderivative, to start. And no, not necessarily. $\infinity - \infinity$ is not zero -- it's indeterminate. Improper integrals like this one are really limits. Really you should evaluate it like this:
$$\stackrel{lim}{a \rightarrow -\infty}\int^{0}_{a}f(x)dx + \stackrel{lim}{b \rightarrow \infty}\int^{b}_{0}f(x)dx$$

 

[jackarms]

Ah, okay. Glad you could work it out.