Solutions to $\sum_{k=0}^{\infty}\int_0^1f^k(x)dx=1$

Question

I am looking for $f:\mathbb{R^+} \to \mathbb{R}$ differentiable on $(0,1)$ satisfying $$\sum_{k=0}^{\infty}\int_0^1f^k(x)dx=1$$ We have solutions $f(x)=1$ and $f(x)=\frac23x$ and etcetera, but I am interested in solutions which are infinitely non-zero differentiable (functions such that for no $k>0$ is $f^k(x)$ constant). I cannot produce any such solutions by the same method I used to get $f(x)=1$ or $\frac23x$, and am not quite sure how to proceed.

Answer 1

Let $f(x)=\dfrac{1}{2}e^{x/2}$, then $f^{(k)}(x)=\dfrac{1}{2^{k+1}}e^{x/2}$ for $k=0,1,...$ and \begin{align*} \sum_{k\geq 0}\int_{0}^{1}f^{(k)}(x)dx=\int_{0}^{1}\sum_{k\geq 0}\dfrac{1}{2^{k+1}}e^{x/2}dx=C\cdot\sum_{k\geq 0}\dfrac{1}{2^{k+1}}=C, \end{align*} normalize it and you get it done.