I am working on a question asking me to find which of the following three functions are entire.
(1) $$f(z):=\int_{0}^{1}\dfrac{e^{z^{2}t}}{\sqrt{t}}dt$$
(2) $$\sum_{n=0}^{\infty}\dfrac{z^{n^{2}}}{(n!)^{2}}$$
(3)$$\dfrac{\sin^{2}z}{z(e^{iz}-1)}.$$
I made some progress in (1), had no idea about (2), and proved (3).
For (1):
replace $u=t^{1/2}$ so that $$t=u^{2},\ dt=2t^{1/2}du,\ t=0\implies u=0,\ t=1\implies u=1,$$ and thus the original integral can be re-written as $$f(z):=\int_{0}^{1}\dfrac{e^{z^{2}t}}{\sqrt{t}}dt=2\int_{0}^{1}e^{u^{2}z^{2}}du.$$
Now we use power series to expand the integrand, so that we have \begin{align*} f(z)&=2\int_{0}^{1}\sum_{n=0}^{\infty}\dfrac{(u^{2}z^{2})^{n}}{n!}du\\ &=2\sum_{n=0}^{\infty}\Big(\dfrac{z^{2n}}{n!}\int_{0}^{1}u^{2n}du\Big)\\ &=2\sum_{n=0}^{\infty}\dfrac{z^{2n}}{(2n+1)n!}. \end{align*}
However, I don't know if I can directly conclude here that it is entire, or I have to "return" the power series back as a function, since this new series may not converge..
For (2):
No idea..which power series should I use? I tried something similar to $e^{z}$, but had a hard time to create $z^{n^{2}}$ and $(n!)^{2}$.
For (3):
Using power series, we can see that $$z(e^{iz}-1)=z\Big(\sum_{n=0}^{\infty}\frac{(iz)^{n}}{n!}-1\Big)=z^{2}\Big(i+\frac{(i)^{2}z}{2!}+\cdots\Big),$$ where $\Big(i+\frac{(i)^{2}z}{2!}+\cdots\Big)$ is a non-vanishing holomorphic function on a neighborhood of $z_{0}=0$.
Therefore, the fraction $\dfrac{1}{z(e^{iz}-1)}$ has a pole at $z_{0}=0$ of order $2$.
However, using power series, we also have $$\sin^{2}z=\Big(\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}z^{2n+1}\Big)^{2}=z^{2}(1-z^{2}+z^{4}-\cdots)^{2}, $$ where $(1-z^{2}+z^{4}-\cdots)^{2}$ is a non-vanishing holomorphic function on a neighborhood of $z_{0}=0$.
Therefore, $\sin^{2}z$ has a zero at $z_{0}=0$ of order $2$.
Thus, the original fraction can be re-written as $$f(z):=\dfrac{\sin^{2}z}{z(e^{iz}-1)}=\dfrac{(1-z^{2}+z^{4}-\cdots)^{2}}{(i+\frac{(i)^{2}z}{2!}+\cdots)},$$ whence entire.
Thank you in advance for any hints, ideas or corrections of my finished proof!
Edit 1:
As what was discussed between me and Jose Carlos Santos, for part (2), Santos said that the series converges everywhere. However, after doing some calculation, I found that the radius of convergence is $1$. Thus, I posted my computation in this edition so that Santos and other users can have a look at it and point out where I got wrong.
Set $a_{n}:=\dfrac{1}{(n!)^{2}}$. Then the ratio test says that the series converges if $$\lim_{n\rightarrow\infty}\Big|\dfrac{a_{n+1}(z-0)^{(n+1)^{2}}}{a_{n}(z-0)^{n^{2}}}\Big|=\lim_{n\rightarrow\infty}\Big|\dfrac{z^{2n+1}}{(n+1)^{2}}\Big|<1, $$ so that the radius of convergence is $$|z-0|:=r=\lim_{n\rightarrow\infty}(n+1)^{2/(2n+1)}.$$
Set $x_{n}=(n+1)^{2/(2n+1)}$. Then $$\lim_{n\rightarrow\infty}\log x_{n}=\lim_{n\rightarrow\infty}\dfrac{2\log(n+1)}{2n+1}=\lim_{n\rightarrow\infty}\dfrac{1}{n+1}=0.$$
This implies that $$\lim_{n\rightarrow\infty}x_{n}=1.$$
Thus, the radius of convergence is $r=1$, and thus the function is not entire.
Edit 2:
As confirmed by Jose Carlos Santos, the solution in Edit 1 is correct.
Thus totally we have $(1)$ and $(3)$ entire and $(2)$ not entire with radius of convergence $=1$.
I'd like to express my appreciation to everyone who is participated in this post, especially to Jose Carlos Santos who patiently answered all my dumb questions.
