# diverging or converging?

• Jan 28th 2009, 06:05 PM
razorfever
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?
• Jan 28th 2009, 06:17 PM
Mush
Quote:

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.
• Jan 28th 2009, 06:37 PM
razorfever
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
• Jan 28th 2009, 10:17 PM
mr fantastic
Quote:

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?

Quote:

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.

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

The value of the limit is easily shown to be equal to zero using l'Hopital's Rule.
• Jan 29th 2009, 06:15 AM
Krizalid
Quote:

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: $\displaystyle \ln t<\sqrt t$ which holds for all $\displaystyle t\ge1,$ so $\displaystyle \int_{1}^{\infty }{\frac{\ln t}{{{t}^{2}}}\,dt}< \int_{1}^{\infty }{\frac{{{t}^{1/2}}}{{{t}^{2}}}\,dt}=\int_{1}^{\infty }{\frac{dt}{{{t}^{3/2}}}}<\infty$ which means that your integral does converge.

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

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

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