For c=−1 , it can be evaluated using the Taylor Series for lnx centered at 1 to get ζ(2). For c=1 you enforce the substitution x=1−u then use the Taylor series centered at 0.
Can we generalize for c∈R?
I've completed calc 1 and 2 but currently taking no math classes besides Ap statistics. I try to use this website to expand my math toolbox and learn tricks here and there. So can someone incorporate a technique that's not to far away for me to understand.
Thanks.
Idea for c<0:
Taylor series for lnx about x=−c is:
ln(x)=ln(−c)+∞∑n=1(n−1)!(−1)n+1(−c)n(x+c)nn!
I'm good from there... divide by (x+c) and integrate.. and I'm left with a pretty weird sum, which I see as acceptable. Here's what I've got from this method:
ln(−c)∫10dxx+c−∞∑n=1(1+c)n−cnn2cn
Which can be simplified if c∉(−1,0)
Now all that is left is c≥0
Answer
The Dilogarithm function can be defined as −∫10log(1−xt)xdx=Li2(t) so in your case we have, taking t=−1c,c∉(−1,0] and integrating by parts, Li2(−1c)=−∫10log(1+x/c)xdx =−[log(x)log(1+x/c)]10+∫10log(x)x+cdx=∫10log(x)x+cdx.
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