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Math Help - Logarithmic Integral

  1. #1
    Member RedBarchetta's Avatar
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    Logarithmic Integral

    Here's another logarithmic integral. I am in question about the answer given in the book.

    <br />
\int\limits_2^3 {\frac{{2\log (x - 1)}}<br />
{{x - 1}}dx}  = (\ln 2)^2 <br />

    Methodology:

    <br />
\begin{gathered}<br />
  \frac{2}<br />
{{\ln 10}}\int\limits_2^3 {\frac{{\ln (x - 1)}}<br />
{{x - 1}}dx}  \hfill \\<br />
  u = \ln (x - 1) \hfill \\<br />
  du = \tfrac{{dx}}<br />
{{x - 1}} \hfill \\ <br />
\end{gathered} <br />

    <br />
\frac{2}<br />
{{\ln 10}}\int\limits_0^{\ln 2} {udu}  = \tfrac{2}<br />
{{\ln 10}}\left[ {\tfrac{{u^2 }}<br />
{2}} \right]_0^{\ln 2}  = \tfrac{2}<br />
{{\ln 10}}\left[ {\tfrac{1}<br />
{2}(\ln 2)^2 } \right] = \frac{{(\ln 2)^2 }}<br />
{2}<br />

    Does this look right?
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  2. #2
    Moo
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    Hello,

    \tfrac{2}{\ln 10}\left[\tfrac 12 (\ln 2)^2 \right]=\frac{(\ln 2)^2}{\ln(10)} \neq \frac{(\ln 2)^2}{2}

    Can you check if they really used the log in the integral and the ln in the answer ? That's pretty weird.
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  3. #3
    Member RedBarchetta's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    \tfrac{2}{\ln 10}\left[\tfrac 12 (\ln 2)^2 \right]=\frac{(\ln 2)^2}{\ln(10)} \neq \frac{(\ln 2)^2}{2}

    Can you check if they really used the log in the integral and the ln in the answer ? That's pretty weird.
    Oops. That last flub was an error on my part. You're right. This is my work, in case you thought I copied this from an answer book.

    All I have down for an answer is (ln2)^2. I don't have a solutions manual. Either way, that answer minus my last error, is not equal. Is my work correct aside from that?

    Another solution manual error?
    Last edited by RedBarchetta; September 8th 2008 at 10:44 PM.
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  4. #4
    Moo
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    Quote Originally Posted by RedBarchetta View Post
    Oops. That last flub was an error on my part. You're right. This is my work, in case you thought I copied this from an answer book.

    All I have down for an answer is (ln2)^2. I don't have a solutions manual. Either way, that answer minus the last error, is not equal. Is my work correct aside from that?

    Another solution manual error?
    Oh I thought the result (\ln 2)^2 was from a book. (I must say, it's really good that you show your working )

    The problem in that is I don't know what log stands for, because it depends on the books !

    So, assuming that this "log" is base 10, the answer is \frac{(\ln 2)^2}{\ln(10)}

    If this "log" is the napierian (natural) logarithm, then the answer is simply (\log 2)^2


    Your work is correct aside that. There's just the confusion log / ln that just affects the result.
    Last edited by Moo; September 8th 2008 at 11:13 PM.
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  5. #5
    Member RedBarchetta's Avatar
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    Quote Originally Posted by Moo View Post
    Oh I thought the result (\ln 2)^2 was from a book. (I must say, it's really good that you show your working )

    The problem in that is I don't know what log stands for, because it depends on the books !

    So, assuming that this "log" is base 10, the answer is \frac{(\ln 2)^2}{\ln(10)}+c

    If this "log" is the napierian (natural) logarithm, then the answer is simply (\log 2)^2+c


    Your work is correct aside that. There's just the confusion log / ln that just affects the result.
    Well this is from a study sheet from my professor. I'm also assuming that log is base 10 as well. I'll have to inquire about this one. Thanks for the help Moo, as always.
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  6. #6
    Moo
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    Quote Originally Posted by RedBarchetta View Post
    Well this is from a study sheet from my professor. I'm also assuming that log is base 10 as well. I'll have to inquire about this one.
    Send me a pm or reply here when you have the answer please
    Thanks for the help Moo, as always.
    It's my pleasure
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  7. #7
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    Quote Originally Posted by RedBarchetta View Post
    Here's another logarithmic integral. I am in question about the answer given in the book.

    <br />
\int\limits_2^3 {\frac{{2\log (x - 1)}}<br />
{{x - 1}}dx} = (\ln 2)^2 <br />

    Methodology:

    <br />
\begin{gathered}<br />
\frac{2}<br />
{{\ln 10}}\int\limits_2^3 {\frac{{\ln (x - 1)}}<br />
{{x - 1}}dx} \hfill \\<br />
u = \ln (x - 1) \hfill \\<br />
du = \tfrac{{dx}}<br />
{{x - 1}} \hfill \\ <br />
\end{gathered} <br />

    <br />
\frac{2}<br />
{{\ln 10}}\int\limits_0^{\ln 2} {udu} = \tfrac{2}<br />
{{\ln 10}}\left[ {\tfrac{{u^2 }}<br />
{2}} \right]_0^{\ln 2} = \tfrac{2}<br />
{{\ln 10}}\left[ {\tfrac{1}<br />
{2}(\ln 2)^2 } \right] = \frac{{(\ln 2)^2 }}<br />
{2}<br />

    Does this look right?
    Assuming that log means \log_{10} then \frac{(\ln 2)^2}{\ln 10} is correct. This can be written as (\ln 2) (\log 2).

    Quote Originally Posted by Moo View Post
    [snip]

    If this "log" is the napierian (natural) logarithm, then the answer is simply (\log 2)^2{\color{red}+c}

    [snip]
    *Ahem* Did you type this bit with your tongue ..... lol!!
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  8. #8
    Moo
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    Quote Originally Posted by mr fantastic View Post
    *Ahem* Did you type this bit with your tongue ..... lol!!
    Yes... my tongue was very tired of doings *things* so it put this accidentally
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  9. #9
    Member RedBarchetta's Avatar
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    Yes, indeed my teacher did have a typo. The common log was supposed to be the natural log.
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