Let $f(z)=\sum_{n=0}^\infty a_nz^n$ be a power series of radius of convergence $R>0.$

Denote, now $F(z)=\sum_{n=0}^\infty \frac{a_n}{n!}z^n$. I proved that the radius of convergence for this power series is $+\infty.$

Now I would like to prove that $f(z)=\int_{0}^\infty F(tz)e^{-t}dt$ for $\vert z\vert<R$

First I tried to prove that the integral converges. I didn't succeed, my main problem is that I get always $a_n$ somewhere. Any idea ?


2 Answers 2


Let us show that the integral converges absolutely, if $|z|<R$:

$$\int_0^\infty |F(tz)e^{-t}| dt \le \int_0^\infty \sum_{n=0}^\infty \frac{|a_n z^n|}{n!} t^n e^{-t} dt = \sum_{n=0}^\infty \frac{|a_n z^n|}{n!} \int_0^\infty t^n e^{-t} dt.$$

Note that the interchange of summation and integral is justified by Fubini-Tonelli (since the integrand is non-negative). Also please take care to understand that in case of divergence (which happens if $|z|>R$), this calculation still makes perfect sense, by positivity. That is, in case of divergence, all the terms simply equal $\infty$.

Now we recognize the Gamma function, $$\int_0^\infty t^n e^{-t} dt =: \Gamma(n+1) = n!,$$ which can be verified using repeated integration by parts.

Thus the sum on the right in the previous display equals

$$\sum_{n=0}^\infty |a_n z^n|$$

which converges to a finite value by assumption since $|z|<R$.

Also you see, that to prove your claimed relation $f(z)=\int_0^\infty F(tz) e^{-t} dt$, you just need to follow this argument without the absolute values.

  • $\begingroup$ I didn't "switch" that's why I didn't suceed, thanks. $\endgroup$
    – Alex
    Feb 1, 2017 at 11:53

See that $$ n! =\int_{0}^\infty t^n e^{-t}dt $$ then

\begin{split} f(z)=\sum_{n=0}^\infty \frac{a_n}{n!}z^n n!&=& \sum_{n=0}^\infty \int_{0}^\infty \frac{a_n}{n!}(zt)^ne^{-t} dt\\ &= &\int_{0}^\infty\sum_{n=0}^\infty \frac{a_n}{n!}(zt)^ne^{-t} dt\\ &=&\int_{0}^\infty F(tz)e^{-t}dt \end{split} Since your Domain is $D(0,R)$ we can exchange the sum with integral sign.

  • $\begingroup$ The interchange of $\int \sum$ and $\sum \int$ is, I think, the part that needs justifying, as it is, essentially, the Q. $\endgroup$ Feb 1, 2017 at 12:15
  • $\begingroup$ Because we are manipulating under the compact set $\bar{D}(0,R)$ $\endgroup$
    – Guy Fsone
    Oct 30, 2017 at 20:59

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