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How would one apply Green's Theorem to the following complex contour integral: $\oint_\gamma $ $\frac{u^{s-1}}{e^{-u}-1)}du$.

Where $\gamma$ is the Hankel Contour (counterclockwise) and R is the region inside.

Note: Essentially, we are coming in from $-\infty$and going around the unit circle and then back out again to $-\infty$. The two paths coming in and out will cancel each other, as the only pole is at $u=0$. Attempt:

Rewrite $f(u)$ in the form $a(x,y)+i(b,xy)$

Let $u=x+iy$ and $s=\rho+i\sigma$.

$$\frac{u^{s-1}}{e^{-u}-1}=\frac{(x+iy)^{(\rho-1)+i\sigma}}{e^{-x-iy}-1}=\frac{\exp\big[((\rho-1)+i\sigma)\log(x+iy)\big]}{e^{-x-iy}-1}$$

$$=\frac{\exp\big[((\rho-1)+i\sigma)\log\left(\sqrt{x^2+y^2}e^{i\arctan(y/x)}\right)\big]}{e^{-x}(\cos(-y)+i\sin(-y))-1} $$

$$=\frac{\exp\big[((\rho-1)+i\sigma)\left(\frac{1}{2}\log(x^2+y^2)+i\arctan\frac{y}{x}\right)\big]}{(e^{-x}\cos y-1)-ie^{-x}\sin y}$$

$$\frac{\exp\left[\left(\frac{\rho-1}{2}\log(x^2+y^2)-\sigma\arctan\frac{y}{x}\right)+i\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)\right]}{(e^{-x}\cos y-1)-ie^{-x}\sin y}$$

$$\scriptsize =\exp\left[\frac{\rho-1}{2}\log(x^2+y^2)-\sigma\arctan\frac{y}{x}\right]\frac{\cos\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)+i\sin\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)}{(e^{-x}\cos y-1)-ie^{-x}\sin y} $$

$$\tiny =\frac{\exp\left(\frac{\rho-1}{2}\log(x^2+y^2)-\sigma\arctan\frac{y}{x}\right)}{(e^{-x}\cos y-1)^2+(e^{-x}\sin y)^2}\left[\cos\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)+i\sin\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)\right]\left((e^{-x}\cos y-1)+ie^{-x}\sin y\right)$$

$$\tiny = \frac{\exp\left(\frac{\rho-1}{2}\log(x^2+y^2)-\sigma\arctan\frac{y}{x}\right)}{(e^{-x}\cos y-1)^2+(e^{-x}\sin y)^2}\left[\begin{matrix}\cos\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)(e^{-x}\cos y-1)-\sin\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)e^{-x}\sin y \\ +i\left(\cos\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)e^{-x}\sin y+\sin\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)(e^{-x}\cos y-1)\right)\end{matrix}\right].$$

Therefore,

$$\small a=\frac{\exp\left(\frac{\rho-1}{2}\log(x^2+y^2)-\sigma\arctan\frac{y}{x}\right)}{(e^{-x}\cos y-1)^2+(e^{-x}\sin y)^2}\left[\begin{matrix}\cos\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)(e^{-x}\cos y-1) \\ -\sin\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)e^{-x}\sin y\end{matrix} \right]$$

and

$$\small b=\frac{\exp\left(\frac{\rho-1}{2}\log(x^2+y^2)-\sigma\arctan\frac{y}{x}\right)}{(e^{-x}\cos y-1)^2+(e^{-x}\sin y)^2}\left[\begin{matrix}\cos\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)e^{-x}\sin y \\ +\sin\left(\frac{\sigma}{2}\log(x^2+y^2)+(\rho-1)\arctan\frac{y}{x}\right)(e^{-x}\cos y-1)\end{matrix}\right] $$

Apply Green's Theorem by rewriting contour integral:

$\int_c f(u)du=$ $\iint_R (-\frac {\partial b}{\partial x}-\frac {\partial a}{\partial y})dA+i\iint_R (\frac {\partial a}{\partial x}-\frac {\partial b}{\partial y})dA$

Evaluating with Mathematica (Where the radius of the circle $\epsilon<2 \pi in$$\forall n \in\mathbb z$:$\int _{-\epsilon }^{\epsilon }\int _{-\sqrt{\epsilon ^2-x^2}}^{\sqrt{\epsilon ^2-x^2}}\frac{e^{-3 x} \left(x^2+y^2\right)^{\frac{\rho -1}{2}} \left(e^{-\text{$\sigma $ArcTan}\left(\frac{y}{x}\right)} \left(\frac{e^{-\sigma \tan ^{-1}\left(\frac{y}{x}\right)} \left(e^x \left(e^{\text{$\sigma $ArcTan}\left(\frac{y}{x}\right)} (-\rho y+y+x \sigma ) \left(-2 e^x \cos (y)+e^{2 x}+1\right)-2 e^{x+\sigma \tan ^{-1}\left(\frac{y}{x}\right)} y (\rho -1) \cos (y)\right) \cos \left((\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)+2 e^{2 x+\sigma \tan ^{-1}\left(\frac{y}{x}\right)} y (\rho -1) \cosh (x) \cos \left((\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)-\left(e^{\sigma \tan ^{-1}\left(\frac{y}{x}\right)} y (\rho -1)+e^{\text{$\sigma $ArcTan}\left(\frac{y}{x}\right)} (-\rho y+y+x \sigma )\right) \left(-2 e^x \cos (y)+e^{2 x}+1\right) \cos \left(y+(\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)+\left(e^{\sigma \tan ^{-1}\left(\frac{y}{x}\right)}-e^{\text{$\sigma $ArcTan}\left(\frac{y}{x}\right)}\right) \left(-e^{3 x} (x (\rho -1)+y \sigma ) \sin \left((\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)+e^x \left((-\rho x+x-y \sigma ) \cos \left((\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right) \sin (2 y)-(2 (x (x+\rho -1)+y (y+\sigma ))+(x (\rho -1)+y \sigma ) \cos (2 y)) \sin \left((\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)\right)+(x (x+\rho -1)+y (y+\sigma )) \sin \left(y+(\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)+e^{2 x} \left((-x (x+\rho -1)-y (y+\sigma )) \sin \left(y-(\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)-\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)+2 (x (\rho -1)+y \sigma ) \sin \left(y+(\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)\right)\right)\right)}{x^2+y^2}-\frac{\left(-2 e^x \cos (y)+e^{2 x}+1\right) \left(e^x \cos \left((\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)-\cos \left(y+(\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)\right) \text{$\sigma $ArcTan}'\left(\frac{y}{x}\right)}{x}\right)+i \left(-2 e^{-\text{$\sigma $ArcTan}\left(\frac{y}{x}\right)} \left(e^x \cos (y)-1\right) \left(\cos \left(y+(\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)-e^x \cos \left((\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)\right)+\frac{2 e^{x-\sigma \tan ^{-1}\left(\frac{y}{x}\right)} \left((x (x+\rho -1)+y (y+\sigma )) \cos \left(y+(\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)-e^x (x (\rho -1)+y \sigma ) \cos \left((\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)\right) (\cos (y)-\cosh (x))}{x^2+y^2}-\frac{e^{-\text{$\sigma $ArcTan}\left(\frac{y}{x}\right)} \left(-2 e^x \cos (y)+e^{2 x}+1\right) \left(\left(x^2+y^2\right) \cos \left(y+(\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)+(y (\rho -1)-x \sigma ) \left(e^x \sin \left((\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)-\sin \left(y+(\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)\right)\right)}{x^2+y^2}-\frac{e^{-\sigma \tan ^{-1}\left(\frac{y}{x}\right)} (-\rho y+y+x \sigma ) \left(-2 e^x \cos (y)+e^{2 x}+1\right) \left(e^x \sin \left((\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)-\sin \left(y+(\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)\right)}{x^2+y^2}+2 e^{x-\sigma \tan ^{-1}\left(\frac{y}{x}\right)} \sin (y) \left(\sin \left(y+(\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)-e^x \sin \left((\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)\right)-\frac{e^{-\text{$\sigma $ArcTan}\left(\frac{y}{x}\right)} \left(-2 e^x \cos (y)+e^{2 x}+1\right) \left(e^x \cos \left((\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)-\cos \left(y+(\rho -1) \tan ^{-1}\left(\frac{y}{x}\right)+\frac{1}{2} \sigma \log \left(x^2+y^2\right)\right)\right) \left((\rho -1) x^3+y \left(x^2+y^2\right) \text{$\sigma $ArcTan}'\left(\frac{y}{x}\right)\right)}{x^2 \left(x^2+y^2\right)}\right)\right)}{\left(-2 e^{-x} \cos (y)+e^{-2 x}+1\right)^2}dydx$. However, I am not sure how to integrate this, or how changing $\epsilon$ would affect the integral. Essentially, does $\epsilon$ depend on anything (do you have to make it a certain number?) or do you just integrate with it?

Best Regards, Thank You!

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  • $\begingroup$ 0) Why Green's theorem??? 1) In general, the two path coming in and out won't cancel unless $z$ is an integer. 2) if $z$ is an integer, the integral can be evaluated by calculating the residue of the integrand at $u = 0$. 3) if $z$ is not an integer but $\Re z > 1$, you can shrink the "circle" part of the contour to infinitesimal small and the integral reduces to some standard integral over $[0,\infty)$ and the result is expressible in terms of Gamma and Zeta functions. 4) If $\Re z \le 1$ and $\ne 1, 0, -1,\ldots$, just analytic continue what you get in (3). $\endgroup$ – achille hui Sep 3 '14 at 3:16
  • $\begingroup$ Okay, thank you. In this case would $z=s$? ...Is there a way to directly apply Green's Theorem, instead of using the Residue Theorem or The Cauchy Integral Formula. I am doing this because it will make a step in my process easier.@achillehui $\endgroup$ – VectorCalculus Sep 3 '14 at 3:39
  • $\begingroup$ I don't think there is an easy way to directly apply Green's theorem. Green's theorem works on a more general class of function. The price to pay is it has much less property you can use to pin down the final value of the integral. $\endgroup$ – achille hui Sep 3 '14 at 3:46
  • $\begingroup$ Okay, thank you for the help, I will try the standard method.@achillehui $\endgroup$ – VectorCalculus Sep 3 '14 at 3:54
  • $\begingroup$ How would you express this integral in terms of the residues?@achillehui $\endgroup$ – VectorCalculus Sep 4 '14 at 3:21

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