# cauchy int formula, function not holomorphic

Use Cauchy's integral formula to evaluate the following integral, $$\int \limits_{\Gamma} \frac{\sin(\pi z^2)+\cos(\pi z^2)}{(z-1)(z-2)}dz$$where the contour $\Gamma$ is parameterised by $\gamma : [-\pi,\pi] \rightarrow \mathbb{C}$ given by $\gamma (\theta)=3e^{i\theta}+1$.

If you make $$f(z)=\frac{\sin(\pi z^2)+\cos(\pi z^2)}{(z-2)}$$ it wont be holomorphic so how can you do it? When z=2, it would be undefined. and 2 is in the region of gamma.

Also correct me if I am wrong but the region is just a circle centred at 1 and radius 3.

• What does the Cauchy integral formula tell us and how does that help here? – dustin Mar 8 '15 at 18:15
• Either split the contour into two circles or use partial fractions. – anon Mar 8 '15 at 18:17
• $$f(w)=\frac1{2\pi i} \int \limits_{\Gamma} \frac{f(z)}{z-w}dz$$ where f(z) is holomorphic and w is an interior point. If you make f(z) which I said and w=1, it wont be holomorphic. – snowman Mar 8 '15 at 18:17
• Note that the integrand is holomorphic except for simple poles at $1$ and $2$, and these poles both lie in the interior of the region bounded by $\Gamma$. Does this ring a bell? – MPW Mar 8 '15 at 18:18

Note

$$\frac{1}{(z - 1)(z - 2)} = \frac{1}{z - 2} - \frac{1}{z - 1},$$

so if $f(z) = \sin(\pi z^2) + \cos(\pi z^2)$, you can write

$$\int_{\Gamma} \frac{\sin(\pi z^2) + \cos(\pi z^2)}{(z - 1)(z - 2)}\, dz = \int_{\Gamma} \frac{f(z)}{z - 2} \, dz - \int_{\Gamma} \frac{f(z)}{z - 1}\, dz.$$

Since $f$ is entire and both $1$ and $2$ lie inside $\Gamma$, the Cauchy integral formula gives

$$\int_{\Gamma} \frac{f(z)}{z - 1} = 2\pi i f(1) = -2\pi i$$

and

$$\int_{\Gamma} \frac{f(z)}{z - 2} = 2\pi i f(2) = 2\pi i.$$

Hence

$$\int_{\Gamma} \frac{\sin(\pi z^2) + \cos(\pi z^2)}{(z - 1)(z - 2)}\, dz = (2\pi i) - (-2\pi i) = 4\pi i.$$

• what is your f(z)? – snowman Mar 8 '15 at 18:43
• @snowman as I wrote, $f(z) = \sin(\pi z^2) + \cos(\pi z^2)$. – kobe Mar 8 '15 at 18:44
• where does the half come from? when you don't have the 1/2, it equates to the LHS... (first line) – snowman Mar 8 '15 at 18:50
• @snowman that was an error on my part. Thanks, I made the corrections. – kobe Mar 8 '15 at 18:52