Complex Integral With Cosine

Let $f(z) = \cos(z)$ . This is holomorphic inside the circle $|z|=1$ , and the ratio $\cos (z)/z^3$ is integrable over this circle, since the only place the ratio is undefined is at $z=0$ , which is not on the circle. Hence, by the Cauchy integral formula:

$\int_C \frac {\cos(z)}{z^3} dz = \int_C \frac {f(z)}{z^3} dz = \frac {2\pi i f''(0)}{2!} = \pi i (-\cos (0)) = \boxed{-\pi i}$

Consider the Taylor series $\cos(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}-...$ and keep to mind that $\int_{C}z^ndz=0$ for $n\neq-1$ . Thus the given integral is $-\frac{1}{2}\int_{C}\frac{dz}{z}=-\frac{1}{2}\times2\pi i=\boxed{-\pi i}$ .

• Cauchy Residue theorem, f"(a)={frac(n!/(2.pi.i)}. integral \frac{f(z)/(z-a)^(n+1)} cosz/z^3 dz= {2Pi.i f"(0)} / 2! = - Pi. i