Finding a $f(x)=\sum_{k\geq 0}a_k x^{4k+2}$ such that $f(x) \to 0$ as $x \to \infty$

Question

To provide a little context for this question, I recently proved that for a suitable function $f(x)$ the following identity holds

$$\sum_{k=0}^{\infty}f'(k)=-2\sum_{n=1}^{\infty}\frac{f'(n)}{e^{2\pi n}-1}$$

The criteria for suitability of the function $f(x)$ essentially comes down to two major factors. Firstly, $f(x)$ must possess a power series expansion of the form

$$f(x)=\sum_{k=0}^{\infty}a_k x^{4k+2}$$

such the series

$$\sum_{k=0}^{\infty}\frac{\vert{a_k}\vert}{(2\pi)^{4k}}(4k+2)! < \infty$$

Secondly, we must have $f(x) \to 0$ as $x \to \infty$.

After proving this, I wanted to find a specific example so that I could compute both of the series to ensure the identity actually holds and I didn't just make some error in my proof. I didn't think this would be too difficult of a task, but frustratingly I have not been able to come up with a single example!

My first thought was to try something Bessel-like as $J_v(x) \to 0$ as $x \to \infty$. Along these lines I figured the most sensible function to start with was the Wright function $z^2\phi(4,3,-z^4)$ where

$$\phi(\alpha,\beta,z) = \sum_{n=0}^{\infty}\frac{z^n}{n!\Gamma(\alpha n + \beta)}$$

so that

$$z^2\phi(4,3,-z^4) = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!(4n+2)!}z^{4n+2}$$

Unfortunately, I quickly became aware of the fact that the asymptotics of more general hypergeometric functions can become quite complicated. More specifically to my needs, $z^2\phi(4,3,-z^4)$ does not go to $0$ as $z \to \infty$. This paper gives the asymptotics for the Wright function.

I am wondering, does anyone have any suggestions on constructing a function with my desired properties? Or showing that such a function does indeed exist?

Answer 1

It's straightforward to check that such function $f$ is entire, has order less or equal to 1 and is of exponential type, i.e. satisfy the estimate $|f(z)|\le C e^{c|z|}$ for all $z\in \mathbb{C}$.

Since $f(i z)=-f(z)$ function $f$ also tends to zero on the imaginary axis as $|z|\to \infty$.

Consider $f$ in the angle $A=\{0<\arg z<\pi/2\}$. The function is bounded on its sides and grows no faster than exponentially inside. Applying the Phragmén–Lindelöf principle one gets that $f$ is bounded in $A$. By the same argument $f$ is bounded on the entire plane and therefore is a constant. From the condition at the infinity it follows that $f(z)\equiv0$.