- #1

- 184

- 0

[itex]\int_0^{\infty}(t^{-2}e^t) dt[/itex]

What would be the first step here in determining convergence or divergence?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter VitaX
- Start date

- #1

- 184

- 0

[itex]\int_0^{\infty}(t^{-2}e^t) dt[/itex]

What would be the first step here in determining convergence or divergence?

- #2

- 81

- 1

[itex]\int_0^{\infty}(t^{-2}e^t) dt[/itex]

What would be the first step here in determining convergence or divergence?

You can apply several tests. If you take integral calculus..... you use the comparsion test.. testing the bounds or evaluating a similar function via comparsion, test. You should find whether or not these converges.

- #3

lurflurf

Homework Helper

- 2,444

- 139

[itex]\int_0^{\infty}(t^{-2}e^t) dt[/itex]

What would be the first step here in determining convergence or divergence?

Try determining convergence or divergence of

[itex]\int_0^{1}t^{-2} dt[/itex]

and

[itex]\int_1^{\infty}e^t dt[/itex]

then deduce the convergence by comparison or otherwise of

[itex]\int_0^{\infty}t^{-2}e^t dt[/itex]

- #4

- 33

- 0

- #5

lurflurf

Homework Helper

- 2,444

- 139

- #6

- 184

- 0

Try determining convergence or divergence of

[itex]\int_0^{1}t^{-2} dt[/itex]

and

[itex]\int_1^{\infty}e^t dt[/itex]

then deduce the convergence by comparison or otherwise of

[itex]\int_0^{\infty}t^{-2}e^t dt[/itex]

So if both have no limit and diverge, then by comparison test the whole integrand diverges?

Edit: Just noticed you have the limits a little different. How did you determine the limits to be those when evaluating them separately?

I get [itex]\int_0^1 t^{-2} dt = -1[/itex]

And [itex]\int_1^{\infty} e^t dt = \infty[/itex]

What would I determine from these exactly?

Last edited:

- #7

lurflurf

Homework Helper

- 2,444

- 139

Share: