I'm looking to find $t$ that satisfies the following equation, and treat it as $g(p)$, a function of $p>1$
$$ \frac{\Gamma \left(2-\frac{1}{p}\right) \Gamma (t+1)}{\Gamma \left(t-\frac{1}{p}+2\right)}=0.01 $$
Empirically, it seems like the following expression is an upper bound on $\log_{10} g$ $$5(p-1)^\frac{3}{4}$$
- Is this correct/tight upper bound?
- What is the asymptotic behavior of $g(p)$ as $p\to 1$?
- What is asymptotic behavior of $g(p)$ as $p\to\infty$?
We can make use of a few asymptotic expressions for the gamma function as $p \to \infty$:
$$ \Gamma\left(2-\frac1p\right) \sim 1-\frac{1-\gamma}{p} $$
where $\gamma$ is the Euler-Mascheroni constant, and
$$ \frac{\Gamma\left(t+2-\frac1p\right)}{\Gamma(t+1)} \sim (t+1)^{1-\frac1p} $$
Then we have
$$ \frac{1-\frac{1-\gamma}{p}}{(t+1)^{1-\frac1p}} \approx \frac{1}{100} $$
$$ (t+1)^{1-\frac1p} \approx 100\left(1-\frac{1-\gamma}{p}\right) $$
\begin{align} t+1 & \approx 100^\frac{p}{p-1} \left(1-\frac{1-\gamma}{p-1}\right) \\ & \approx 100 \left(1+\frac{\ln 100}{p-1}\right) \left(1-\frac{1-\gamma}{p-1}\right) \\ & \approx 100 \left(1+\frac{\ln 100-1+\gamma}{p-1}\right) \\ & \approx 100 \left(1+\frac{\ln 100-1+\gamma}{p}\right) \end{align}
since $\frac{1}{p-1} \sim \frac{1}{p}$, and finally we get
$$ t \approx 100 \left(1+\frac{\ln 100-1+\gamma}{p}\right) - 1 $$
Similarly, as $p \to 1$, we let $N = \frac{p}{p-1}$, and then as $N \to \infty$, we have
$$ \Gamma\left(1+\frac1N\right) \sim 1-\frac{\gamma}{N} $$
and
$$ \frac{\Gamma\left(t+1+\frac1N\right)}{\Gamma(t+1)} \sim \sqrt[N]{t} $$
Then we can have
$$ \frac{1-\frac{\gamma}{N}}{\sqrt[N]{t}} \approx \frac{1}{100} $$
$$ \sqrt[N]{t} \approx 100\left(1-\frac{\gamma}{N}\right) $$
$$ t \approx 100^N e^{-\gamma} $$
It should be clear how to generalize this for ratios other than $\frac{1}{100}$.