I have no idea how to do this :confused:

I have to write this integral:

http://img62.imageshack.us/img62/8049/msplq5.gif

as a power series. Please help!

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- April 17th 2007, 01:15 PMphackIndefinite Integral as Power Series
I have no idea how to do this :confused:

I have to write this integral:

http://img62.imageshack.us/img62/8049/msplq5.gif

as a power series. Please help! - April 17th 2007, 01:48 PMCaptainBlack
x Log[1+n^2]/n^2 + c

unless you really want integral log(1+x^2) / x^2 dx

Then we can write:

log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...

for |x|<1.

Then:

integral log(1+x^2) / x^2 dx = sum_{n=1 to infnty} integral (-1)^{n+1}x^{2n-1}/n dx

.................= sum_{n=1 to infnty} (-1)^{n+1}x^{2n}/(n (2n-1)

RonL - April 17th 2007, 03:10 PMphack
I'm not sure I'm following your steps. Why did you place an x infront in your first step?

And how did you get this?...

log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...

Please clarify. - April 17th 2007, 03:53 PMThePerfectHacker
- April 17th 2007, 09:33 PMCaptainBlack
Because in your question you have an integral over x, but no x in the

integrand. Therefore the integral is x times whatever you had under

the integral sign (as this was independent of x). Therefore the first step

is the answer to the question you asked. After that I proceeded to answer

the question I assumed that you realy wanted to ask.

Quote:

And how did you get this?...

log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...

Please clarify.

**Im**PerfectHacker says, I took a standard series representation of

log(1+u) and evaluated it at u=x^2. You can find this series either by

just knowing it, or taking the Taylor series of log(1+x) about 0, or by

integrating term by term the series expansion of 1/(1+x).

RonL