I have the integral $$\int_{t=1}^{u}\min(1,t^{\,n-1}(u-1)^{n-1})\,dt .$$Can somebody kindly help with splitting the interval of integration $(1,u)$ so as to get rid of the $min$ function? Since $1 \leq t \leq u$, $$ 1\leq t^{n-1}(u-1)^{n-1} \implies 1\leq u^{n-1}(u-1)^{n-1 }. $$I don't know what to do further.Thank you for any hints/suggestions
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It depends on $u$, Can you please add more context to where and how you are coming across this integral? – Math Lover Jan 10 '22 at 18:48
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u lies between 1 and 2 – AgnostMystic Jan 10 '22 at 18:56
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You can see from inequality that for $1 \lt t \lt \frac{1}{u-1}$, integrand is $1$ and for $t \gt \frac{1}{u-1}$, is the other expression but it depends whether $u \gt \frac{1}{u-1}$ or not . Equating $u = \frac{1}{u-1}. $ we get $u = \frac{1 + \sqrt5}{2}$. So if $u$ is less than this, $t$ is always less than $\frac{1}{u-1}$ so integrand is $1$ and you do not need to split the integral. But if $u$ takes value between $\frac{1 + \sqrt5}{2}$ and $2$, you split the integral for $1 \lt t \lt \frac{1}{u-1}$ with integrand $1$ and then to $u$ with the other expression as integrand – Math Lover Jan 10 '22 at 19:07