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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

AgnostMystic
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  • $$t^{n-1}(u-1)^{n-1}>1,\Rightarrow, t>\frac{1}{u-1}$$ It means the interval should be splitted: $[u-1;\frac{1}{u-1}]$ (we integrate $t^{n-1}(u-1)^{n-1}$), and $[\frac{1}{u-1};u]$ (we integrate $1$) - if $u>\frac{1}{u-1}$. If $u<\frac{1}{u-1}$, we integrate $t^{n-1}(u-1)^{n-1}$ on the whole interval $[u-1;u]$ – Svyatoslav Mar 01 '23 at 15:02
  • What will be the respective limits of integration in the two cases.will be the integral be the sum of those two integrals or two diff functions depending on where u lies – AgnostMystic Mar 01 '23 at 16:11
  • As you split the interval - it'll be the sum of these two integrals. – Svyatoslav Mar 01 '23 at 16:25

1 Answers1

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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.

AgnostMystic
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