2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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

and

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

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.