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Math Help - Improper Integral

  1. #1
    DBA
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    Improper Integral

    I need to integrate
    <br />
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx <br />

    The integral is impoper because x cannot be 1. Since 1 is at the left hand, I rewrite it to...

    <br />
\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx <br />

    Then I integrate and get...
    Let u=lnx, du=1/x dx

    <br />
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx = \frac{3}{2} (lnx)^{2/3} +C<br />

    Then
    <br />
\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx <br />
    =
    <br />
\lim t \rightarrow 1^+ [\frac{3}{2}*[(ln4)^{2/3} - (lnt)^{2/3}]]<br />

    Now I do not know how to find out what the answer is. Lnt would be Ln1+, so it gets closer and closer to ln1. Ln1 = 0. Does the term (lnt)^2/3 gets zero then?
    And the limit goes to 3/2((ln4)^{2/3} - 0 ?

    Thanks for any help.
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  2. #2
    Ted
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    Quote Originally Posted by DBA View Post
    I need to integrate
    <br />
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx <br />

    The integral is impoper because x cannot be 1. Since 1 is at the left hand, I rewrite it to...

    <br />
\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx <br />

    Then I integrate and get...
    Let u=lnx, du=1/x dx

    <br />
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx = \frac{3}{2} (lnx)^{2/3} +C<br />

    Then
    <br />
\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx <br />
    =
    <br />
\lim t \rightarrow 1^+ [\frac{3}{2}*[(ln4)^{2/3} - (lnt)^{2/3}]]<br />

    Now I do not know how to find out what the answer is. Lnt would be Ln1+, so it gets closer and closer to ln1. Ln1 = 0. Does the term (lnt)^2/3 gets zero then?
    And the limit goes to 3/2((ln4)^{2/3} - 0 ?

    Thanks for any help.
    Its pretty easy. (lnt)^{\frac{2}{3}} \rightarrow 0 as t \rightarrow 1.
    0^{\frac{2}{3}} = 0 .

    Quote Originally Posted by DBA View Post
    And the limit goes to 3/2((ln4)^{2/3} - 0 ?
    Yes. So the improper integral converges to \frac{3}{2} ln^{\frac{2}{3}}(4).

    Quote Originally Posted by DBA View Post
    <br />
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx = \frac{3}{2} (lnx)^{2/3} +C<br />
    This is wrong!
    1- as you said its improper integral. You can not evaluate it directly.
    2- Its and definite integral. And you solve it as an indefinite integral!
    You should write it as a indefinite integral. Without the upper/lower limits.
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  3. #3
    Super Member Aryth's Avatar
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    Quote Originally Posted by DBA View Post
    I need to integrate
    <br />
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx <br />

    The integral is impoper because x cannot be 1. Since 1 is at the left hand, I rewrite it to...

    <br />
\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx <br />

    Then I integrate and get...
    Let u=lnx, du=1/x dx

    <br />
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx = \frac{3}{2} (lnx)^{2/3} +C<br />

    Then
    <br />
\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx <br />
    =
    <br />
\lim t \rightarrow 1^+ [\frac{3}{2}*[(ln4)^{2/3} - (lnt)^{2/3}]]<br />

    Now I do not know how to find out what the answer is. Lnt would be Ln1+, so it gets closer and closer to ln1. Ln1 = 0. Does the term (lnt)^2/3 gets zero then?
    And the limit goes to 3/2((ln4)^{2/3} - 0 ?

    Thanks for any help.

    First, Note that the integral via substitution should look like this:

    \lim_{t \to 0^+} \int_t^{ln(4)} \frac{1}{\sqrt[3]{u}} ~du

    You must change the limits, so you get:

    u(1) = ln(1) = 0
    u(4) = ln(4)

    So, your final solution looks like this:

    \lim_{t \to 0^+} \left[\frac{3u^{\frac{2}{3}}}{2}\right]^{\ln{(4)}}_t

    = \frac{3}{2}\lim_{t \to 0^+} (ln(4)^{\frac{2}{3}} - t^{\frac{2}{3}})

    From here it is very obvious that your answer is correct.

    Hope this helps.
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