Is this function bounded? $g(n)=t>1\text{ s.t. } \int_1^n (1-i^{-p})^t i^{-p} \, di =\varepsilon$ Let $p>1,t>1,\varepsilon>0$. Assuming the function below exists, when is it bounded?
$$g(n)=t>1\text{ s.t. } \int_1^n (1-i^{-p} )^t i^{-p} \,di =\varepsilon$$
From simulations I suspect it's bounded for all $p>1$ and $\varepsilon>0$, can this be shown analytically?
For instance, for $p=2$ and $\varepsilon=0.01$ the graph obtained numerically


This is not an answer but it is too long for a comment.
I repeated the calculations for the general case and found that, if
$$I_p=\int_1^n \left(1-i^{-p}\right)^t \,i^{-p}\, di$$
$$(p-1)\,I_p=\frac{\Gamma \left(2-\frac{1}{p}\right) \Gamma (t+1)}{\Gamma
   \left(t-\frac{1}{p}+2\right)}-n^{1-p} \,\,
   _2F_1\left(\frac{p-1}{p},-t;2-\frac{1}{p};n^{-p}\right)$$ So, for $p=2$
$$I_2=\frac{\sqrt{\pi }\,\, \Gamma (t+1)}{2 \Gamma \left(t+\frac{3}{2}\right)}-\frac{1}{n}\,\,
   _2F_1\left(\frac{1}{2},-t;\frac{3}{2};\frac{1}{n^2}\right)$$ while your file shows
$$I_2=\frac{\sqrt{\pi }\,\, \Gamma (2 t+1)}{2 \Gamma \left(2 t+\frac{3}{2}\right)}-\frac{1}{n}\,\,
   _2F_1\left(\frac{1}{2},-2 t;\frac{3}{2};\frac{1}{n^2}\right)$$ which does not seem to be the same.
So, at this time, I prefer to not continue until we clarify.
