Maximum value of $\int_{0}^{1}e^x\log f(x)dx$ when $\int_{0}^{1}f(x)dx=1$

There is no lower bound. Pick $f(x) = (n+1)x^n$. Then $f(x)$ satisfies the hypotheses except at the origin (so one can simplify modify the function in a small neighborhood of the origin later), and

$$ \int_0^1 e^x \log(f(x)) \, dx = (e - 1) \log(n+1) + n \int_0^1 e^x \log{x} \, dx$$

We can observe that

$$ \int_0^1 e^x \log{x} \, dx < 0$$

So the first term is growing logarithmically, while the second term is decreasing linearly, so for arbitrarily large $n$, the integral will be arbitrarily negative.

To find the maximum, first pick $f(x) = e^x / (e - 1)$. Then one finds that

$$ \int_0^1 e^x \log(f(x)) \, dx = 1 - (e - 1) \log(e - 1) \approx 0.07$$

I claim that this is the maximizer. To do this, define the functional

$$ I(f) = \int_0^1 e^x \log(f(x)) \, dx $$

If $f_0(x) = e^x/(e - 1)$, then any admissable $f(x)$ can be written as $f(x) = f_0(x) + \epsilon g(x)$, where $g(x)$ is some function such that $\int_0^1 g(x) \, dx = 0$. Then

$$ \frac{d}{d\epsilon} I(f_0 + \epsilon g) = \int_0^1 e^x \frac{g(x)}{f_0(x) + \epsilon g(x)} \, dx $$

Therefore, by setting $\epsilon = 0$, the directional derivative of $I$ at $f_0$ in the direction of $g$ is given by

\begin{align*} \frac{d}{d\epsilon} \left. I(f_0 + \epsilon g) \right|_{\epsilon = 0} & = \int_0^1 e^x \frac{g(x)}{f_0(x)} \, dx \\ & = \int_0^1 (e -1) g(x) \, dx \\ & = 0. \end{align*} The derivative vanishes independent of $g$, and hence $f_0$ is a local extrema of $I$. Calculating the second derivative,

\begin{align*} \frac{d^2}{d\epsilon^2} I(f_0 + \epsilon g) & = - \int_0^1 e^x \frac{g(x)^2}{(f_0(x) + \epsilon g(x))^2} \, dx \\ & < 0, \end{align*} and hence $f_0(x)$ is at least a local maximum. However, the above calculation of the second derivative is valid even when $f_0(x)$ is replaced by any other admissable function $f(x)$, and hence this shows that $I$ is in fact strictly concave. Therefore $f_0(x)$ is the global maximum.