I must integrate the function $f(z)=\frac{e^{z^2}-1}{z\sin(z)}$ long the circle $\gamma$ of radius $r=1/4 $ and center in the origin. I thing that $\int_{\gamma}\frac{e^{z^2}-1}{z\sin(z)}dz=0$, because $lim_{z\to 0}f(z)=1$.
Is this solution correct?
Other solutions are very appreciated .
The only zero of the denominator of $f(z)=\frac{e^{z^2}-1}{z\sin(z)}$ for $|z|\le 1/4$ is at the origin. We can expand the numerator as
$$\begin{align} e^{z^2}-1&=\sum_{n=1}^\infty\frac{z^{2n}}{n!}\\\\ &=z^2\left(1+\frac12z^2+O(z^4)\right) \tag 1 \end{align}$$
while we can expand the denominator as
$$\begin{align} z\sin(z)&=\sum_{n=0}^\infty\frac{(-1)^n\,z^{2n+2}}{(2n+1)!}\\\\ &=z^2\left(1-\frac16z^2+O(z^4)\right) \tag 2 \end{align}$$
From $(1)$ and $(2)$ we see that for $z\ne 0$
$$\frac{e^{z^2}-1}{z\sin(z)}=1+\frac23z^2+O(z^4)$$
So, $f(z)$ has a removable discontinuity and is analytic for $|z|\le 1/4$. Therefore, from Cauchy's Integral Theorem,
$$\oint_{|z|=1/4}\frac{e^{z^2}-1}{z\sin(z)}\,dz=0$$