Can anyone help me prove using Contour Integration that the value of integral log(1+z)/z from limit -1 to +1 is pi^2/4?
You may follow some related blogs & tutorials, e.g., https://math.stackexchange.com/questions/2680434/keyhole-contour-integration-of-int-0-infty-fracz1-2-logz1z2dz
https://www.youtube.com/watch?v=zyzlvT7BLCo
https://www.youtube.com/watch?v=22xZkShcpEw
Muhammad Ali, I am not able to use the Keyhole counter at -1, since it is a non-isolated singularity. Can you please give me a detailed solution to the question?
Peter Breuer , the problem I am facing is that the -1 is a non-isolated singularity. I am not able to form a keyhole around -1 to form the contour.
Peter Breuer, Muhammad Ali : Does it help using the definition of Polylogarithms? Please check definition 25.12.2 at (nist): https://dlmf.nist.gov/25.12
I think this question is ill posed. log(1+z) has a branch point at z=-1. The function log(1+z)/z has a removable singularity at z=0 and can be extended to an analytical function at B(0,1), but has again a branch point at z=-1. The function g(z)=int _0^z log(1+z)/z dz is know and is called -Polylog(2,-z) (see Mathematica), or ( Maple, -dilog(1+z) ).
integral log(1+z)/z from limit -1 to +1 can be intended as an improper real integral, because lim x->-1 log(1+x)/x is infinity. This integral is convergent as improper integral.
Because -1 is a logarithmic branch point and upper limit of the integral is 1, I think contour integration is difficult to use
Polylog seems to be relevant here. This is what I wrote before: see definition 25.12.2 at (nist): https://dlmf.nist.gov/25.12. We just set -Li_2(-z)) and get the integral.
This might be helpful as well: https://arxiv.org/pdf/1810.05865.pdf.
Hi all. Here is a more or less elementary solution. Since the function extends to a holomorphic one on the complement of the half line ]-\infty,-1] (of course, we are using the standard determination of log), we can use a contour which is slightly on the right of -1. See the attached file.
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