Homework Help: Question regarding certain standard integrals.

1. Feb 2, 2012

Salt

1. The problem statement, all variables and given/known data

OK, this is something that stumped me.

2. Relevant equations

$\int_{}^{}{\frac{dZ}{ A^{2}-Z^{2}}} = - \int_{}^{}{\frac{dZ}{ Z^{2}-A^{2} }}$

Right?

$\int_{}^{}{\frac{dZ}{ A^{2}-Z^{2} }}=\; \frac{1}{2A}\ln \left\{ \frac{A+Z}{A-Z} \right\}+\mbox{C}$

$\int_{}^{}{\frac{dZ}{ Z^{2}-A^{2} }}=\; \frac{1}{2A}\ln \left\{ \frac{Z-A}{Z+A} \right\}+\mbox{C}$

3. The attempt at a solution

But I don't see how,

$\frac{1}{2A}\ln \left\{ \frac{A+Z}{A-Z} \right\}+\mbox{C} = - \frac{1}{2A}\ln \left\{ \frac{Z-A}{Z+A} \right\}+\mbox{C}$

It should be,

$\frac{1}{2A}\ln \left\{ \frac{A+Z}{A-Z} \right\}+\mbox{C} = - \frac{1}{2A}\ln \left\{ \frac{A-Z}{A+Z} \right\}+\mbox{C}$

No?

I derived both (OK one of them; the other I just read from the textbook), the integrations "check out", but I don't understand why!? Why the term inside "ln" is "multiplied by -1"?

2. Feb 2, 2012

Nytik

Well, the integral actually comes out as the logarithm of the absolute value of something. So regardless of whether it is Z-A or A-Z, the two values will give the same answer; i.e. it doesn't matter which way around they are. (Remember there is no ln(-1).)

3. Feb 2, 2012

HallsofIvy

Are you forgetting the absolute value?
$$\int \frac{1}{z}dz= ln|z|+ C$$
NOT ln z+ C unless you are sure z is positive.

4. Feb 2, 2012

Salt

Oh ... that would explain it.

My textbook didn't put the absolute value brackets**. -_-

Thanks.

**Ya, it's not the best book, quite a few obvious typos, but it's easy enough to read. Should really get a more rigorous and accurate "second opinion" ...