Did I Solve the Integral Test Correctly for This Question?

[QuarkCharmer]

Homework Statement

Took a test today, one question I am not sure I got right.

$\int_{0}^{\infty} te^{-at}dt$, when a>0

Homework Equations

The Attempt at a Solution

I set let infinity be b, then took the limit as b went to infinity of the integral with new bounds from 0 to b. My solution claimed that it converges to 1/a^2. Is this correct or should I expect to miss that question?

 

[gb7nash]

Homework Statement

Took a test today, one question I am not sure I got right.

$\int_{0}^{\infty} te^{-at}dt$, when a>0

Homework Equations

The Attempt at a Solution

I set let infinity be b, then took the limit as b went to infinity of the integral with new bounds from 0 to b. My solution claimed that it converges to 1/a^2. Is this correct or should I expect to miss that question?

List your steps and we can tell you if you made a mistake.

 

[QuarkCharmer]

Ok, I'll take that as an "it's incorrect".

I'll redo the problem now and post it here in a little while.

 

[gb7nash]

Ok, I'll take that as an "it's incorrect".

I'll redo the problem now and post it here in a little while.

I'm not saying that. The onus is on you to show your work so we can see what you did and if your method/answer is correct.

 

[QuarkCharmer]

It's no problem. I just got home so I will redo it, I made sure to memorize the question after I turned it in because it was giving me a hard time with it's recursive limit.

I keep on L'hopitaling, and it keeps on repeating! Finally I somehow worked it out to tend towards zero, so the solution was that it converges to $\frac{1}{a^{2}}$. Unless I made another mistake that is, and provided, that limit actually does tend to zero. I think my logic was that numerator went to zero, and the denominator went to zero, so really it didn't matter if it was a type o/o indeterminate, I believe that it what I was thinking anyway.

Here is the limit that I don't know how to solve.
$$lim_{b \to \infty} \frac{be^{-ab}}{-a}$$

 

[Dick]

The only part you need to care about is limit b->inf b*exp(-ab). Take the log. So you've got log(b)-ab. Factor it into b*(log(b)/b-a). Use l'Hopital to show limit log(b)/b goes to zero. So you are left with b*(-a) since a>0. Now goes to -infinity, yes? If log goes to -infinity then b*exp(-ab) goes to zero. Still ok?

 

[QuarkCharmer]

The only part you need to care about is limit b->inf b*exp(-ab)...

But it's $lim_{b \to \infty}be^{-ab}$ ?

 

[Dick]

But it's $lim_{b \to \infty}be^{-ab}$ ?

I mean exp(-ab) to be e^(-ab). Not sure what the question is? Sorry about the failure to correctly TeX things up.

 

[QuarkCharmer]

I read that as $b^{-ab}$, I didn't know that pxp(m) means p^m, sorry.

I don't understand what you mean when you say to take the log? It looks like you just take the log of the numerator?

 

[Dick]

I read that as $b^{-ab}$, I didn't know that pxp(m) means p^m, sorry.

I don't understand what you mean when you say to take the log? It looks like you just take the log of the numerator?

I was just writing exp(x) instead of e^x. Don't worry if you haven't seen it before. Yes, you can just work with the numerator. The denominator is just a constant.