Where does this approximation for the expected value of an exponential come from? I have seen the following approximation in a number of papers recently, but I cannot figure out where it comes from. The approximation is as follows

Let $W_t$ be a standard Brownian motion and $f$ some smooth, positive function. Then we can make the following approximation $$\mathbb{E}\left[\exp\left(\int^t_0 \frac{1}{f(W_u)} du\right)\Big|W_t = z\right] = \exp\left(\frac{1}{2}\left(\frac{1}{f(0)} + \frac{1}{f(z)}\right)t\right) + \mathcal{O}(t^2)$$ where $\mathcal{O}(t^2)$ is a term that groups together all terms of order $t^2$ or higher.

I figured this must be some sort of Taylor series approximation, but then terms are always missing w.r.t. the expression on the right hand side above.
Another thought I had was that the approximation looks a lot like some moment generating function, more specifically that of a normal distribution, but this also led to nothing.
Any tips are appreciated!

This looks like the Trapezoid Rule approximation of the integral of a continuous function, with one interval: If $g$ is twice differnetiable on $[0,t]$ then
$$
\int_0^t g(u)\,du = {g(0)+g(t)\over 2}\cdot t + {t^3\over 12}g''(\xi),
$$
for some $\xi\in (0,t)$. This can't be applied directly to $I(t):=\int_0^t 1/f(W_u)\,du$, but it can be used after taking expectations. As a start, write $J(t)=t[1/f(0)+1/f(z)]/2$ and $\Delta (t) =I(t)-J(t)$. Then $J(t)\to 0$ as $t\to 0$, and so
$$
\exp(I(t))-\exp(J(t))=\left[\exp(\Delta(t))-1\right]\cdot\exp(J(t)).
$$
and, as $J(t)$ is nonrandom,  it is enough to examine $\Bbb E\left[\exp(\Delta(t))-1\mid W(t)=z\right]$. For this expand the exponential in its Maclaurin series. The first order term is $\mathscr O(t^3)$ by the Trapezoid Rule with error term noted above. The sum of the remaining terms is easily seen to be $\mathscr O(t^2)$ because $|\Delta(t)|\le Ct\|1/f\|_\infty$ for a suitable constant $C$.
