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Question Number 156991 by MathSh last updated on 18/Oct/21

Answered by mindispower last updated on 18/Oct/21

$$x^n=u$$$${\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{n}-1}}{n(1+u)}du,\beta (x,y)={\int }_0^{\mathrm{\infty }}\frac{t^{y-1}}{n{(1+t)}^{x+y}}$$$$weget=\frac{1}{n}\beta (1-\frac{1}{n},\frac{1}{n})=\frac{1}{n}.\frac{\pi }{sin\left(\frac{\pi }{n}\right)},n⩾2$$$$\frac{1}{sin\left(\frac{\pi }{n}\right)}⩾1\Rightarrow ⩾\frac{\pi }{n}$$

Commented by MathSh last updated on 18/Oct/21

$$\mathrm{Thank}\mathrm{you}\mathrm{dear}\mathbf{S}\mathrm{er},\mathrm{very}\mathrm{nice}$$

Commented by mindispower last updated on 19/Oct/21

$$witheplesur$$

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