I encountered a certain family of infinite series in some work, which is given by $$F_r(x)=\frac1{2^r}\sum_{k=0}^r\binom{r}k\frac1{1+x(2k-r)^2}.$$ I've convincing date to believe the following is true, but it needs a proof.
Question. Does this hold true? For each $r\in\mathbb{N}$, the Taylor series for $F_r(x)$ has integer coefficients.
Yes. The coefficient of $x^n$ in $F_r$ is the $2n$-th derivative at $t = 0$ of the function $$ t \mapsto (\cos t)^{r} = \frac{1}{2^r} \sum_{k=0}^r \binom{r}{k} e^{it(2k-r)}. $$ But the successive derivatives of $(t \mapsto (\cos t)^r)$ are easily seen by induction to be polynomials with integer coefficients in $\cos(t),\sin(t)$.
As a follow-up to js21's answer, the following explicit formula for $F_r(x^2)$ holds: $$F_r(x^2) = \int_0^{\infty} \cos(xt)^r e^{-t} dt.$$
