consider a holomorphic function $f(z)$ and the paths $\gamma_1:(0,\pi)\rightarrow \mathbb{C}, t\mapsto r\cdot i\cdot e^{i t}$, $\gamma_2:(0,\pi)\rightarrow \mathbb{C}, t\mapsto r\cdot i\cdot e^{i (-t)}$ with $0<r<1$. Is it possible to tell something about the behavior of the following limit?

$$ \lim\limits_{r\downarrow 0} \left(\int\limits_{\gamma_1}\frac{f(z)}{\sin( z)}dz + \int\limits_{\gamma_2}\frac{f(z)}{\sin( z)}dz\right).$$

I have computed some examples and got zero as limit. Is it always true that the limit is zero?

Best wishes


As stated in the OP, on the curve $\gamma_1$, $z=ire^{it}$, beginning at $t=0$ and ending at $t=\pi$, while on $\gamma_2$, $z=ire^{-it}$ beginning at $t=0$ and ending at $t=\pi$. Then, we have

$$\begin{align} \lim_{r\to 0^+}\int_{\gamma_1}\frac{f(z)}{\sin(z)}\,dz&=\lim_{r\to 0^+}\int_0^\pi \frac{f(ire^{it})}{\sin(ire^{it})}\,(-r) e^{it}\,dt\\\\ &=i\pi f(0) \tag 1 \end{align}$$


$$\begin{align} \lim_{r\to 0^+}\int_{\gamma_2}\frac{f(z)}{\sin(z)}\,dz&=\lim_{r\to 0^+}\int_0^\pi \frac{f(ire^{-it})}{\sin(ire^{-it})}\,(r) e^{-it}\,dt\\\\ &=-i\pi f(0) \tag 2 \end{align}$$

Upon adding $(1)$ and $(2)$, the coveted limit is zero.

  • $\begingroup$ Thank you very much! Your arguments looks very reasonable. But I am confused because of the answer of User001 below, since it claims the opposite…Why did he had in mind and why is he wrong then?! $\endgroup$ – Hasti Musti Apr 18 '16 at 20:42
  • $\begingroup$ Hasti, You're welcome! My pleasure. I'm not really sure how others interpreted the question. I wrote my assumptions, under which I believe that this is correct. -Mark $\endgroup$ – Mark Viola Apr 18 '16 at 20:49

The limit should be non-trivial, integrating with perhaps semi-circular indents around the poles that will lie on the contour.

I.e., use the parametrization, and go for a principal value integral.

  • $\begingroup$ There are no poles on the contours since $f(z)$ is analytic and $\sin(z)$ has zeroes only at $\ell \pi$ for integer $\ell$. $\endgroup$ – Mark Viola Apr 18 '16 at 20:00
  • $\begingroup$ Hi @Dr.MV, using linearity of the integral, we combine the two integrals, which is now integration along a full circle. Do you agree? $\endgroup$ – User001 Apr 18 '16 at 20:02
  • $\begingroup$ The paths are traversed in opposite directions. $\endgroup$ – Mark Viola Apr 18 '16 at 20:06
  • $\begingroup$ Yes, I am aware of that. I figured that would just be reconciled by factoring out $-1$. @Dr.MV $\endgroup$ – User001 Apr 18 '16 at 20:08
  • $\begingroup$ And then going for P.V. integrals @Dr.MV. $\endgroup$ – User001 Apr 18 '16 at 20:09

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