Let $p > 1$ be a number. How can I tell for which values of $p$ the integral $$\int_0^T \int_0^T \frac{1}{|t-s|^p}\;dtds$$ is finite? Here $T$ is a positive and finite number.
The singularity is when $t=s$ where we have the problem. but I don't know how to handle this.
$$I=(-1)^p\int_0^T \int_0^s \frac{1}{(t-s)^p}\;dtds+\int_0^T \int_s^T \frac{1}{(t-s)^p}\;dtds$$ $$I=(-1)^p\int_0^T \int_{-s}^0 \frac{1}{t^p}\;dtds+\int_0^T \int_{0}^{T-s} \frac{1}{t^p}\;dtds$$ Now we consider $\large \int_{-s}^0 \frac{1}{t^p}$, since $p>1$ then $1-p<0$ $$\int_{-s}^0 \frac{1}{t^p}=\frac{1}{1-p}\underset{t\to 0 }{\mathop{\lim }}\,t^{1-p}-\frac{(-1)^{1-p}}{1-p}s^{1-p}=?$$
