Let $[0,z]$ be the line in $\mathbb{C}$ starting at 0 and ending at $z$. Let $f(z)=\int_{[0,z]} sin(w^2)dw$. Find the power series expansion of $f$ at $0$ and its radius of convergence.
I tried to solve it as follow, and that's what I did and wonder whether it is correct or not. Thank you!
Let $r(t)=tz,\ t\in [0,1]$. Then
$f(z)=\int_{0}^1 sin(t^2z^2)zdt=\int_{0}^1 \sum_{n=0}^\infty \frac{(-1)^n (t^2z^2)^{2n+1}}{(2n+1)!}zdt$.
If we let $a_n=\sum_{n=0}^\infty \frac{(-1)^n (t^2z^2)^{2n+1}}{(2n+1)!}z $, then $lim_{n\rightarrow \infty} |\frac{a_{n+1}}{a_{n}}|=0$, which implies that the radius of convergence of $\sum a_n$ is infinite.
Hence, $\sum a_n$ uniformly converges on $[0,1].$ Therefore, we have
$f(z)=\sum_{n=0}^\infty \int_{0}^1 a_n dt = \sum_{n=0}^\infty \frac{(-1)^n z^{4n+3}}{(2n+1)!(4n+3)}$.
If we let $b_n=\frac{(-1)^n z^{4n+3}}{(2n+1)!(4n+3)}$, it is clear that $lim_{n\rightarrow \infty} |b_{n+1}/b_n|=0$. We can conclude that the radius of convergence of f is $\infty$.