definition of a log contract:
A log contract on a stock $S_t$ with strike $K$ is a European derivative with expiry $T$ and payoff function $V(S_T, \tau = 0) = \log(S_T/K), K > 0$ where $\tau = (T-t)$ is the time to expiration of the contract.
If I want to hedge (i.e., replicate) this contract I can crate a portfolio $\Pi$ as continuum of call and put options with strikes $y$.
At expiry, the payoff of this replicating portfolio is $$\Pi(S_T, (T-t) = 0)= \int_0^K \underbrace{(y - S_T)^{+}}_{puts} \frac{dy}{y^2}+ \int_K^\infty\underbrace{(S_T - y)^{+}}_{calls} \frac{dy}{y^2}= I_1 + I_2$$
with the integrals given as \begin{align*} I_1 &= \Bigg[ - \log(S_T/K) + \frac{S_T - K}{K} \Bigg] 1_{\{S_T < K\}} \\ I_2 &= \Bigg[ - \log(S_T/K) + \frac{S_T - K}{K} \Bigg] 1_{\{S_T > K\}} \end{align*}
My question is now, how are those integrals calculated? I am having trouble in particular with the indicator functions, when rewriting $(y-S_T)^{+} = (y - S_T) 1_{\{y > S_T\}}$.
You just do the casework. Obviously, if $S_T > K$, then the integrand for $I_1$ is zero for all $0 < y \le K$, since $(y - S_T)^+ = 0$ if $y \le S_T$. So $I_1 = 0$ if $S_T > K$. Similarly, $I_2 = 0$ if $S_T < K$.
Consequently, we consider the case where $0 \le S_T < K$ for $I_1$, which simplifies the integral as follows:
$$\begin{align}I_1 &= \int_{y=0}^K (y - S_T)^+ \frac{dy}{y^2} \\ &= \int_{y=S_T}^K \frac{y-S_T}{y^2} \, dy \\ &= \int_{y=S_T}^K \frac{1}{y} \, dy - S_T \int_{y=S_T}^K \frac{1}{y^2} \, dy \\ &= \Bigl[\log y \Bigr]_{y=S_T}^K - S_T \left[-\frac{1}{y}\right]_{y=S_T}^K \\ &= \log \frac{K}{S_T} + S_T \left(\frac{1}{K} - \frac{1}{S_T}\right) \\ &= - \log \frac{S_T}{K} + \frac{S_T - K}{K}. \end{align}$$ The only thing left to do is reintroduce the indicator function, which applies because the above calculation is valid only when $S_T < K$.
The calculation of $I_2$ follows similarly and is left as an exercise.