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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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!
x Log[1+n^2]/n^2 + cQuote:
Originally Posted by phack
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
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.
Because,Quote:
Originally Posted by phack
ln(1+x)= x - x^2/2+x^3/3 -... for -1<x<=1
But you have,
ln(1+x^2)
Thus, instead of x you have x^2.
Because in your question you have an integral over x, but no x in theQuote:
Originally Posted by phack
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.
As ImPerfectHacker says, I took a standard series representation ofQuote:
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.
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