Integrate by parts to prove that this integral provides an analytic continuation , 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.
 A: (a). Since $f(z) = \sum_0^\infty a_nz^n$ converges for $|z| \le 1$, $\varlimsup_{n\to\infty}{\sqrt[n]{|a_n|}}<1$. So $\exists \:0<r<1,\:M>0$ such that $|a_n|<Mr^n,\:\forall n$. Also 
$$
\lim_{n\to\infty}{\sqrt[n]{\frac{|a_n|}{n!}}}=\lim_{n\to\infty}{\frac{\sqrt[n]{|a_n|}}{n{\dfrac{\sqrt[n]{n!}}{n}}}}<\lim_{n\to\infty}\frac{e}{{n}}=0
$$
So $\phi(z) = \sum\limits_{n=0}^\infty \dfrac{a_n}{n!}z^n$ is entire, and
$$
|\phi(z)|\leqslant \sum\limits_{n=0}^\infty \dfrac{|a_n|}{n!}|z|^n<M\sum\limits_{n=0}^\infty \dfrac{r^n|z|^n}{n!}<Me^{|z|}
$$
(b).
\begin{align}
\int_0^\infty e^{-s}\phi(sz)ds&=-\phi(sz)e^{-s}\bigg|_0^{\infty} + z\int_0^\infty e^{-s}\phi'(sz)ds
\\
&=-\sum\limits_{n=0}^\infty \dfrac{a_n}{n!}s^nz^ne^{-s}\bigg|_0^{\infty} + z\int_0^\infty e^{-s}\phi'(sz)ds
\\
&=a_0+z\int_0^\infty e^{-s}\sum\limits_{n=1}^{\infty}\dfrac{a_n}{(n-1)!}(sz)^{n-1}ds
\\
&=a_0+z\int_0^\infty e^{-s}\sum\limits_{n=0}^{\infty}\dfrac{a_{n+1}}{n!}(sz)^{n}ds
\\
&=a_0+\sum\limits_{n=0}^{\infty}\dfrac{a_{n+1}}{n!}z^{n+1}\int_0^\infty e^{-s}s^{n}ds
\\
&=a_0+\sum\limits_{n=0}^{\infty}\dfrac{a_{n+1}}{n!}z^{n+1}\Gamma(n+1)
\\
&=f(z)
\end{align}
Now consider $f(z) = \sum_{n=0}^\infty z^{2n}$. For $|z|<1$
$$
f(z) = \sum_{n=0}^\infty z^{2n}=\frac1{1-z^2}
$$
For $|z|\geqslant1$, since $\phi(z)=\sum_{n=0}^\infty \dfrac{z^{2n}}{2n!}$ is entire
$$
\int_0^\infty e^{-s}\phi(sz)ds=\sum_{n=0}^\infty \frac{z^{2n}}{2n!}=\cosh{z}
$$
Finally we prove that $\cosh{z}$ is the analytic continuation of $f(z)=\dfrac1{1-z^2}$ on $|z|<1$.
\begin{align}
\int_0^\infty e^{-s}\cosh(sz)ds&=\dfrac1{2}\int_0^\infty (e^{s(z-1)}+e^{-s(z+1)})ds
\\
&=\dfrac1{2}\left(\dfrac{e^{s(z-1)}}{z-1}-\dfrac{e^{-s(z+1)}}{z+1}\right)\bigg |_0^{\infty}
\\
&=\dfrac1{2}\left(\dfrac{e^{s(Re(z)-1)}e^{sIm(z)i)}}{z-1}-\dfrac{e^{-s(Re(z)+1)}e^{-sIm(z)i)}}{z+1}\right)\bigg |_0^{\infty}
\\
&=\dfrac1{2}\left(\dfrac1{1-z}-\dfrac1{z+1}\right)
\\
&=\dfrac1{1-z^2}
\end{align}
Note: $Re(z)-1<0$ on $|z|<1$ and so $e^{s(Re(z)-1)}\to0$ as $s\to\infty$. Also $Re(z)+1>0$ on $|z|<1$ and so $e^{-s(Re(z)+1)}\to0$ as $s\to\infty.$ 
