I would like to know if the contour integral have no poles how do I solve it? Please explain with workings. Thank you.
$\displaystyle\oint_C z^5 \sin\left(\dfrac{1}{z^2}\right) \space dz$
[Math] How to find contour integral with no poles
complex-analysisintegration
Related Solutions
You could also use the following contour.
$$\int_{-\infty}^0 \frac{2x^2-1}{x^4+1}\operatorname dx =\int_{0}^{+\infty} \frac{2x^2-1}{x^4+1}\operatorname dx $$
Has an analytic continuation as $$\int_{\Gamma} \frac{2z^2-1}{z^4+1}\operatorname dz $$ with 4 poles, but just on pole inside of the contour.
$$\operatorname*{res}_{z=e^{i\frac{\pi}{4}}} f(z) = \frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8} $$
I think you made a calculation error here (?)
Using $$\int_{\Gamma} \frac{2z^2-1}{z^4+1}\operatorname dz = \color{blue}{\int_{\Gamma_1}\frac{2z^2-1}{z^4+1}\operatorname dz} + \int_{\Gamma_2}\frac{2z^2-1}{z^4+1}\operatorname dz + {\color{red}{\int_{\Gamma_3}\frac{2z^2-1}{z^4+1}\operatorname dz}}$$
- Now $\color{blue}{\int_{\Gamma_1} \to 0}$ as $R\to +\infty$ which can be proven using the triangle inequality.
- Use $\Gamma_2 \leftrightarrow z(x) = x$ and $x:0\to R$
- Use $\color{red}{\Gamma_3 \leftrightarrow z(y) = iy}$ and $y: R \to 0$
Which finally results in (as $R \to +\infty$) $$2\pi i \left(\frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8}\right) = \color{blue}{0}+\int_{0}^{+\infty}\frac{2x^2-1}{x^4+1}\operatorname dx + \color{red}{ i \int_{+\infty}^0\frac{-2y^2-1}{y^4+1}\operatorname dy}$$
Here you can read the real parts which results in:
$$\int_{0}^{+\infty}\frac{2x^2-1}{x^4+1}\operatorname dx = \frac{\pi\sqrt{2}}{4}$$
Alternatively, you can use the straight-forward expression for the residue of a function at a pole of order $n$
$${\rm Res}\left(f\left(z\right),z_{0}\right)=\dfrac{1}{\left(n-1\right)!}\dfrac{{\rm d}^{n-1}}{{\rm d}z^{n-1}}\lim_{z\rightarrow z_{0}}\Bigg[\left(z-z_{0}\right)^{n}f\left(z\right)\Bigg]$$
In your case $f\left(z\right)=\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}}$, thus $z_{0}=-1$, $n=3$ and so
$${\rm Res}\left(\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}},-1\right)=\dfrac{1}{2!}\dfrac{{\rm d}^{2}}{{\rm d}z^{2}}\lim_{z\rightarrow -1}\Bigg[\left(z+1\right)^{3}\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}}\Bigg]=\frac{1}{2}\lambda e^{-\lambda}\left(2-\lambda\right)$$
You then get from the residue theorem that
$$\oint_{C\left(\frac{1}{2},\sqrt{2}\right)}\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}}{\rm d}z=2\pi i{\rm Res}\left(\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}},-1\right)=\pi i\lambda e^{-\lambda}\left(2-\lambda\right)$$
exactly as @coreyman317 got.
Best Answer
Let's assume that you are integrating a simple closed contour that includes the origin. By the residue theorem, the integral is $i 2 \pi$ times the sum of the residues at the poles or other isolated singularities inside the integration contour (times a winding number). The residue is the coefficient of $1/z$ in the Laurent expansion of the integrand. So:
$$\sin{\frac1{z^2}} = \frac1{z^2} - \frac1{6 z^6} + \cdots...$$
The coeffficient of $1/z$ in the integrand then is $-1/6$; thus, the integral is $-i \pi/3$.