help regarding an integral I have the following integral:
$$ I=\int_{V=0}^{1} \min (1,(u-1)^{n-1}(u-V)^{n-1}) dV, $$
where u is a constant which can take values between 1 and 2 and $V$ is variable whose range is (0,1).
putting $u-V=t,$ we get :
$$ I= \int_{t=u-1}^{u} \min (1,t^{n-1}(u-1)^{n-1}) dt$$
When $1 \le t \le \frac{1}{u-1},$ the integrand reduces to 1 .I don't know how to proceed further.I will be highly obliged for any help
 A: Here is my attempt based on the above comment .kindly correct me if I have made some mistake:\begin{aligned}
& I=\int_{t=u-1}^u \min \left(1, t^{n-1}(u-1)^{n-1}\right) d t \\
& \text { Now, } t^{n-1}(n-1)^{n-1}>1 \Rightarrow t>\frac{1}{n-1} \\
& \text { So if } u>\frac{1}{u-1}, \min \left(1, t^{n-1}(u-1)^{n-1}\right)=1 \\
& \text { and if } u<\frac{1}{u-1}, \min \left(1, t^{n-1}(u-1)^{n-1}\right)=t^{n-1}(u-1)^{n-1} \\
& \text { Case I: } \quad 1 \leqslant u \leqslant \phi, u \leqslant \frac{1}{u-1} \\
& I=\int_{t=u-1}^u t^{n-1}(u-1)^{n-1} d t \\
& =\frac{(u-1)^{n-1}}{n}\left[u^n-(u-1)^n\right] \\
& \text { Case II } \phi<u \leq 2, u>\frac{1}{u-1} \\
& \therefore I=\int_{u-1}^{\frac{1}{u-1}} t^{n-1}(u-1)^{n-1} d t+\int_{\frac{1}{u-1}}^u 1 \cdot d t \\
& =\frac{(u-1)^{n-1}}{n}\left[\left(\frac{1}{u-1}\right)^n-(u-1)^n\right] 
& +u-\frac{1}{u-1} \\
& =\frac{1}{n}\left[1-(u-1)^{2 n-1}\right]+\frac{u^2-u-1}{u-1} \\
& \therefore I=\left\{\begin{array}{l}
\frac{(u-1)^{n-1}}{n}\left[u^n-(u-1)^n\right], 1 \leqslant u<\phi \\
\frac{1}{n}\left[\frac{1}{u-1}-(u-1)^{2 n-1}\right]+\frac{u^2-u-1}{u-1}, \phi \leq u \leq 2 .
\end{array}\right. \\
&
\end{aligned}where,$ \phi= \frac{\sqrt{5}+1}{2}$ denotes the golden ratio.
