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Math Help - diverging or converging?

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by razorfever View Post
    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.
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  3. #3
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    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
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  4. #4
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    Quote Originally Posted by razorfever View Post
    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 View Post
    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.
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  5. #5
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    Quote Originally Posted by razorfever View Post

    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.
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