Determine whether the integral converges or diverges:
$\displaystyle \int_0^{\infty}(t^{-2}e^t) dt$
What would be the first step here in determining convergence or divergence?
Well the function will go to infinity wouldn't it? But, I'm a little confused about what you did here. First off, integrating this thing is pointless right? Because it's impossible to integrate it. That's how I checked for convergence and divergence up until now, integrated and substituted the limits in and found the limit. So what would be the exact method here? I know it diverges but how would I show it?
What do you know already? Do you know the Taylor expansion for $\displaystyle e^t$? If you do, then multiply every term of that expansion by $\displaystyle t^{-2}$ and try integrating term-by-term.
If you don't know that, then consider a series:
$\displaystyle \sum_{n=1}^\infty{\frac{e^n}{n^2}}$. What do you know about this series in comparison to the integral? (Hint: is $\displaystyle t^{-2}e^t$ increasing or decreasing? What does that tell you about the series? What about the integral? If the individual terms don't converge, what do you know?)
To find the limit of that would take the derivative of that twice? Using L'Hôpital's rule, I end end up with $\displaystyle \frac{e^t}{2}$. Then taking the limit of that as $\displaystyle t \rightarrow 0$ I get $\displaystyle \frac{e^0}{2} = \frac{1}{2}$
Or did you want the derivative of it as a whole like how TKHunny did it above?
No. $\displaystyle \frac{e^{t}}{t^{2}}$ is NOT an indeterminate form for t approaching zero (0).
But really, why do we care about zero, the other direction is clearly unbounded and the integral cannot converge. How clearly is it unbounded? Well, it IS an indeterminate form for t increasing without bound. NOW you may use L'Hôpital's rule.
Can't you just look at it like this:
$\displaystyle \int_0^1 t^{-2} dt = -1$
$\displaystyle \int_1^{\infty} e^t dt = \infty$
And conclude since one part of the integrand diverges that the integrand as a whole diverges?
Or is what you are saying to use L'Hopital's Rule on the integral 2 times then integrate and then put the limits in as t goes towards infinity. I guess that is probably the better approach then what I did above.
I do not understand what you are trying to do. Are we evaluating the integral or demonstrating convergence or divergence? There is no integration necessary to demonstrate divergence on the interval $\displaystyle (0,\infty)$. The Domain of $\displaystyle t^{-2}e^{t}$ is the entire interval and $\displaystyle \lim\limits_{t \to \infty} t^{-2}e^{t}} = \infty$. You are then done.
If you insist on using an integral, it is no trouble at all to demonstrate: $\displaystyle \int_{2}^{R}t^{-2}e^{t}\;dt > M$ for suitable R and arbitrary M. You are then done.
I thought you have to evaluate the integral first because that's how the other homework questions were. That's why I'm confused by your posts when I was looking back at previous examples done in my notes.
If no integration is needed, what is the purpose of the integral in the first place?
The real question is, why are you trying to evaluate the integral without knowing whether or not it converges. If it diverges, never spend another second trying to evaluate it. Finding the antiderivative and seeing that it fails to converge is an excellent way to proceed. This seems to be the basis of your notes. Showing that it can't possibly converge is even better.
I see, if only I could remember limits from Calculus II that dealt with convergence and divergence. Never thought I'd see them in Differential Equations.
Would not being able to integrate be equivalent to saying that it cannot possibly converge? As this problem is impossible to integrate. At least with all the methods I've learned thus far.