Suppose $f(z) = \sum_0^\infty a_nz^n$ converges for $|z| \le 1$.
a) Prove $\phi(z) = \sum_0^\infty \frac{a_n}{n!}z^n$ is entire and $|\phi(z)|\le Me^{|z|}$.
b) Prove $f(z) = \int_0^\infty e^{-s}\phi(sz)ds$. (Hint: Integrate by parts.)
Apply this result to $f(z) = \sum_0^\infty z^{2n}$, which converges in $|z| < 1$ and show that the integral provides an analytic continuation of $f(z)$.
I have proved part(a), but am stuck on part(b).
With integration by parts, I am currently at:
$$-\phi(sz)e^{-s}|_0^{\infty} + z\int_0^\infty e^{-s}\phi'(sz)ds$$
Thanks,
EDIT: if I ignore the convergence of $f(z)$ on the boundary, and just assume that the convergence is for $|z|<1$, then repeated integration by parts, and the fact that $\phi$ is infinitely differentiable, pushes out the terms
$a_0 + a_1z + a_2z^2 + ... = \sum_0^\infty a_nz^n = f(z)$, which is what we needed to prove for part(b). But, is it ok to ignore the convergence on the boundary? I.e., only consider $z$ such that $|z|<1$, but the I feel that we haven't proved that the integral is exactly $f(z)$. Or, I might just be interpreting the question incorrectly, too.