MAT334-2018F > Final Exam


<< < (2/2)

Victor Ivrii:
There is a single pole of $f(z)=\frac{1}{1+z^n}$ inside $\Gamma$, namely $z=e^{i\pi/n}$, which is a simple pole and the residue is $\frac{1}{(1+z^n)'}\bigr|_{z=e^{i\pi/n}}= \frac{1}{nz^{n-1}}\bigr|_{z=e^{i\pi/n}}=-\frac{1}{n}e^{i\pi/n}$.

Therefore due to the residue theorem $I_R+J_R+K_R= -\frac{2}{n}\pi i \times  e^{i\alpha/2}$, where $K_R$ is an integral over an arc and $J_R$ is an integral over the second straight segment.

J_R=\int _{R}^0 \frac{e^{i\alpha}\,dt }{1+e^{in\alpha}t^n}=-e^{i\alpha}I_R
with $I_R=\int_0^R\frac{dx}{1+x^n}$.

On the other hand,
|K_R|=|\int_0^\alpha \frac{iRe^{it}\,dt}{1+e^{itn}R^n }|\le \frac{R}{R^n-1}\int_0^\alpha\, dt=\frac{\alpha R}{R^n-1}\to 0
\qquad\text{as }\ \ R\to \infty.$$

Then $(1-e^{i\alpha}) I =-\frac{2}{n}\pi i \times  e^{i\alpha/2}$ and
I =& -\frac{2}{n}\pi i \times  \frac{e^{i\alpha/2}}{1-e^{i\pi \alpha}}=
&&-\frac{2}{n}\pi i \times  \frac{1}{e^{-i\alpha/2}-e^{i\pi \alpha/2}}=\\
&-\frac{2}{n}\pi i \times  \frac{1}{-2i\sin(\alpha/2)}=
&&\frac{\pi}{n\sin(\pi /n)}.


[0] Message Index

[*] Previous page

Go to full version