Logarithmic Integral

**RedBarchetta** · September 8th 2008, 10:04 PM

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

$ \int\limits_2^3 {\frac{{2\log (x - 1)}} {{x - 1}}dx} = (\ln 2)^2 $

Methodology:

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

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

Does this look right?

**Moo** · September 8th 2008, 10:11 PM

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.

**RedBarchetta** · September 8th 2008, 10:27 PM

Originally Posted by Moo

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?

**Moo** · September 8th 2008, 10:45 PM

Originally Posted by RedBarchetta

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.

**RedBarchetta** · September 8th 2008, 10:48 PM

Originally Posted by Moo

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.

**Moo** · September 8th 2008, 10:52 PM

Originally Posted by RedBarchetta

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

**mr fantastic** · September 8th 2008, 11:09 PM

Originally Posted by RedBarchetta

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

$ \int\limits_2^3 {\frac{{2\log (x - 1)}} {{x - 1}}dx} = (\ln 2)^2 $

Methodology:

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

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

Does this look right?

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

Originally Posted by Moo

[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!!

**Moo** · September 8th 2008, 11:13 PM

Originally Posted by mr fantastic

*Ahem* Did you type this bit with your tongue ..... lol!!

Yes... my tongue was very tired of doings *things* so it put this accidentally

**RedBarchetta** · September 12th 2008, 10:15 PM

Yes, indeed my teacher did have a typo. The common log was supposed to be the natural log.

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