Question:
Let F(z) be the antiderivative of f(z)=e^(z^2)with F(0). Find F(z) as a power series around z=0.
Attempt at solution:
We know that, $\displaystyle e^z = \sum_{n=0}^{\infty} \frac{1}{n!}z^n$
Hence, $\displaystyle e^{z^2} = \sum_{n=0}^{\infty} \frac{1}{n!}z^{2n}$
Thus,
$\displaystyle F(z)=\int e^{z^2} = C+\sum_{n=0}^{\infty} \frac{1}{(2n+1)n!}z^{2n+1}$
Now this is where i have problems, to find C,
let $\displaystyle e^{z^2}= (cos\theta +isin\theta)^2$
Integrating we find,
$\displaystyle \int(cos\theta +isin\theta)^2 = \frac{-1}{2}icos^2\theta+\frac{1}{2} sin2\theta=F(z)$
Therefore,
$\displaystyle F(z)= \frac{-1}{2}icos^2\theta+\frac{1}{2} sin2\theta=C+\sum_{n=0}^{\infty} \frac{1}{(2n+1)n!}z^{2n+1}$
$\displaystyle => F(0)=\frac{-i}{2}= C$
Therefore,
$\displaystyle F(z)= \frac{-i}{2} +\sum_{n=0}^{\infty} \frac{1}{(2n+1)n!}z^{2n+1}$
Is my methodology correct? there is no answer in the book for some reason, but upon research i think my power series is correct, but none of them have a value for C when doing an indefinate integral.