Some tricky path integrals without residue theory

Question

I am trying to work out the following path integrals without residue theory and have a really hard time doing so: \begin{align*} \int_{\partial K_2(0)}\frac{e^{iz^2}-1}{z^2}dz \end{align*} \begin{align*} \int_{\partial K_1(i)}\frac{\cos(\pi z)}{(z^2+1)^2}dz \end{align*} \begin{align*} \int_{\partial K_3(0)}\left(e^z+\frac{\cos(z)}{z}\right)^3dz \end{align*} Where $\partial K_i(z_0)$ is the contour of a circle wtih radius i around the point $z_0$.
I tried applying Cauchy's integral formula, but I always get stuck. Does anyone have an advice on how to solve these? Another idea was to show the functions inside of the integral are holomorphic, so all of the integrals would be $0$. E.g. the first function $\frac{e^{iz^2}-1}{z^2}$ is holomorphic on $\mathbb{C}\setminus\{0\}$ as a composition of holomorphic functions on $\mathbb{C}\setminus\{0\}$, so the integral disapears.
UPDATE
After getting the hint to use the Cauchy integral formula for derivatives, I tried the same for the second integral. Notice that: \begin{align} \int_{\partial K_1(i)}\frac{\cos(\pi z)}{(z^2+1)^2}dz =\int_{\partial K_1(i)}\frac{\cos(\pi z)}{((z-i)(z+i))^2}\end{align} So I tried using $f(z)=\frac{\cos(\pi z)}{(z+i)^2}$ as my function. For the formula (see answer below) and with that I get \begin{align}f'(i)=\frac{-\pi sin(\pi i)(2i)-cos(\pi i)\cdot 2}{-8i}=-\frac{1}{4}i cosh(\pi) + \frac{1}{4} π i sinh(π)\end{align} From here on I can't seem to work it out any further. Do you have any suggestions?

Answer 1

Maybe those integrals disapear since $0$ and $i$ are interior points of your contours, so it could be a priori wrong. You could use Cauchy's formula for derivatives. For the first integral we use, $$f^{(n)} (z) = \dfrac{n!}{2 \pi i } \int_C \dfrac{f(w)}{(w - z)^{n + 1}} \ dw, \ \ n = 0, 1, 2, ... \nonumber,$$ with $f(z)=e^{iz^2}-1$ and $n=1$. So $$ \int_{\partial K_2(0)}\frac{e^{iz^2}-1}{z^2}dz=2\pi i f'(0)=0,$$ that makes sense since the limit at $z=0$ exists. For the last one lets take $f(z):=(ze^z+\cos z)^3.$ By the Cauchy's formula $$\int_{\partial K_3(0)}\left(e^z+\frac{\cos(z)}{z}\right)^3dz=\pi if''(0)=9\pi i.$$