For $a,b>0$ and $b>a+1$, consider the contour integral
$$\oint_{|z|=a+\epsilon}\frac{\ln((z-a)(z-b))}{(z-a)(z-b)}dz.$$
This essentially corresponds to taking the residue at $z=a$. What does this result in?
Does $\frac{\ln((z-a)(z-b))}{(z-a)(z-b)}$ have a residue at $z=a$?
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complex-analysis
contour-integration
residue-calculus
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0Under integral $|z|=a+\epsilon$.? – 2017-02-16
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0Yes, thanks! I fixed it. – 2017-02-16
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0Looks like it to me. It is not removable since the numerator goes to $-\infty$ while the denominator goes to $0$. – 2017-02-16
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0But you need to find the Laurent series to do this, sadly. – 2017-02-16
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0Actually, since there could be some contour deformation trick that would make it trivial, I just adjusted the integral to exclude that possibility. – 2017-02-16
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0@TheCount Is there a general procedure I can use to find the Laurent series? I'm not scared if it is a lot of work. Looking for any way to get the result. – 2017-02-16
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0It's the expansion of $1/(z-a)$ times the expansion of $\ln(z-a)$. Do-able, but gross. Your residue is the coefficient on the $z^{-1}$ term. – 2017-02-16
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0What series representation for the $\ln$ should I use though? – 2017-02-16
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0@TheCount, makes sense asking for the residue/Laurent series around a *branch* point? – 2017-02-19
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0@Martín-BlasPérezPinilla I'm just spitballin' here. – 2017-02-19