Let $r>3/4,$ and $p>1/2.$
My Question: Can we expect $\int_0^{\infty} t^{2r-2p-1} (e^{2t}-e^{-2t})^{-1/2} dt < \infty$?
I am trying to analyze the above integral. Any suggestions/comments are well-come .
My Vague Ideas: (I) My be some how I have to relate with the fact that: $\int_0^1 t^{-\alpha} dt < \infty$, for $0<\alpha <1$ (so near origin we may handle this way), and $\int_{a}^{\infty} t^{-\beta} dt < \infty$ for $\beta >1$ ($a>0$.) (II) I think: some where I need to use that : $g(t)=(e^{2t}-e^{-2t})^{-1}$ is zero at origin and at $1/2$ it is infinity, so it goes to infinity very fast.
The answer is dependent on $r$ and $p$. The issue is not at the upper limit since the exponential decay with increasing $t$ overwhelms any possible increasing power of $t$.
There is, however, a singularity at $t=0$ that requires attention. Note that the term $\left(e^{2t}-e^{-2t}\right)^{-1/2}$ can be expanded at $t=0$ as
$$\left(e^{2t}-e^{-2t}\right)^{-1/2}= \frac12 t^{-1/2}+O(t^{3/2})$$
Therefore, the integrand behaves asymptotically near the origin as $t^{2r-2p-3/2}$.
We conclude that the integral converges if $2r-2p-3/2>-1$ and diverges otherwise. Simplifying, the integral converges for $r-p>1/4$ and diverges otherwise.