1. ## Logarithmic Integral

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

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

Methodology:

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

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

Does this look right?

2. Hello,

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

Can you check if they really used the log in the integral and the ln in the answer ? That's pretty weird.

3. Originally Posted by Moo
Hello,

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

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?

4. 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 $\displaystyle (\ln 2)^2$ 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 $\displaystyle \frac{(\ln 2)^2}{\ln(10)}$

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

Your work is correct aside that. There's just the confusion log / ln that just affects the result.

5. Originally Posted by Moo
Oh I thought the result $\displaystyle (\ln 2)^2$ 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 $\displaystyle \frac{(\ln 2)^2}{\ln(10)}+c$

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

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.

6. 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.
Thanks for the help Moo, as always.
It's my pleasure

7. Originally Posted by RedBarchetta
Here's another logarithmic integral. I am in question about the answer given in the book.

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

Methodology:

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

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

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

Originally Posted by Moo
[snip]

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

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

8. 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

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