Here are a few more thoughts on this intriguing problem. First,

$\displaystyle \int_0^1\frac{\log(1+tz)}tdt = \int_0^1\int_0^z\frac1{1+tx}dxdt = \int_0^z\Bigl[\frac{\log(1+tx)}x\Bigr]_0^1dx = \int_0^z\frac{\log(1+x)}xdx$

(reversing the order of integration in the double integral). Therefore

$\displaystyle f(z) = \exp\left\{-\int_0^1\frac{\log(1+tz)}tdt\right\} = \exp\left\{-\int_0^z\frac{\log(1+x)}xdx\right\}

$.

Next, let A denote the set of all infinitely-differentiable function on the positive reals whose derivatives alternate in sign: $\displaystyle A := \{f\in C^\infty(\mathbb{R}_{>0})

-1)^kf^{(k)}(x)>0\text{ for all }x>0\text{ and all }k\geqslant0\}$.

**Proposition.** If $\displaystyle u\in A$ then $\displaystyle f(x) = \exp\bigl(-{\textstyle\int}u(x)\,dx\bigr) \in A$.

*Outline proof:* The n'th derivative of f is of the form $\displaystyle f^{(n)}(x) = u_n(x)\exp\bigl(-{\textstyle\int}u(x)\,dx\bigr)$, where $\displaystyle u_1(x) = -u(x)$. Also, you can see by differentiating $\displaystyle f^{(n)}(x)$ that $\displaystyle u_{n+1}(x) = u_1(x)u_n(x) + u_n'(x)$. It is then straightforward to prove by induction (using the condition $\displaystyle u\in A$) that the functions $\displaystyle u_n(x)$ alternate in sign, with $\displaystyle (-1)^{n-1}u_n(x)>0$ for all x>0 and all n.

It follows from that Proposition and the previous remarks that if the function $\displaystyle u(z) = z^{-1}\log(1+z)$ is in A then so is the function $\displaystyle f(z) = \exp\left\{-\int_0^1\frac{\log(1+tz)}tdt\right\}$.

I suspect that the function u(z) is indeed in A, but I haven't been able to prove that. However, this seems like a step in the right direction, because u(z) is a considerably simpler-looking function than f(z).

One final thought: functions whose derivatives alternate in sign seem to occur in

**Boltzmann's approach** to the theory of entropy. I wonder if that is where this problem originated?