1. ## diverging or converging?

determine whether the integral diverges or converges
if it converges then evaluate the integral

integral from 1 to infinity: (ln t) / (t^2) dt

i got the answer up to this form

lim T->infinity (-1/T)*(ln T) - (1/T) + 1

how do i solve this limit
i have negative infinity times infinity if im not mistaken?

2. Originally Posted by razorfever
determine whether the integral diverges or converges
if it converges then evaluate the integral

integral from 1 to infinity: (ln t) / (t^2) dt

i got the answer up to this form

lim T->infinity (-1/T)*(ln T) - (1/T) + 1

how do i solve this limit
i have negative infinity times infinity if im not mistaken?
As T tends towards infinite ln(T) also tends towards infinity... hence that term of your integral is not finite. The integral diverges.

3. are you sure?
because doesn't
lim x-> infinity -ln(T) / T = 0

therefore the series converges at 1
since the ln function approaches infinity more slowly than any power of T

4. Originally Posted by razorfever
determine whether the integral diverges or converges
if it converges then evaluate the integral

integral from 1 to infinity: (ln t) / (t^2) dt

i got the answer up to this form

lim T->infinity (-1/T)*(ln T) - (1/T) + 1

how do i solve this limit
i have negative infinity times infinity if im not mistaken?
Originally Posted by Mush
As T tends towards infinite ln(T) also tends towards infinity... hence that term of your integral is not finite. The integral diverges.
Mush has made an uncommon careless mistake.

$\lim_{a \rightarrow +\infty} \int_1^a \frac{\ln t}{t^2} \, dt = - \lim_{a \rightarrow +\infty} \left( \frac{\ln (t) + 1}{t} \right) + 1$.

The value of the limit is easily shown to be equal to zero using l'Hopital's Rule.

5. Originally Posted by razorfever

determine whether the integral diverges or converges
if it converges then evaluate the integral

integral from 1 to infinity: (ln t) / (t^2) dt
You just need this little fact: $\ln t<\sqrt t$ which holds for all $t\ge1,$ so $\int_{1}^{\infty }{\frac{\ln t}{{{t}^{2}}}\,dt}< \int_{1}^{\infty }{\frac{{{t}^{1/2}}}{{{t}^{2}}}\,dt}=\int_{1}^{\infty }{\frac{dt}{{{t}^{3/2}}}}<\infty$ which means that your integral does converge.

Now, let's find its value, so, just integrate by parts:

$\int_{1}^{\infty }{\frac{\ln t}{{{t}^{2}}}\,dt}=\int_{1}^{\infty }{\left( -\frac{1}{t} \right)'\ln t\,dt}=\left. -\frac{\ln t}{t} \right|_{1}^{\infty }+\int_{1}^{\infty }{\frac{dt}{{{t}^{2}}}}=1.$

The fact $\underset{t\to \infty }{\mathop{\lim }}\,\frac{\ln t}{t}=0$ is well known.