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Question Number 79222 by mind is power last updated on 23/Jan/20

$${\int }_0^1\frac{x^n}{\underset{k=0}{\sum ^{n-1}}{x}^k}dx=?$$

Commented by mathmax by abdo last updated on 24/Jan/20

$$atformofserielet{A}_n={\int }_0^1\frac{x^n}{\sum _{k=0}^{n-1}{x}^k}dx\Rightarrow$$$$A_n={\int }_0^1\frac{x^n}{\frac{x^{n-1}}{x-1}}dx={\int }_0^1\frac{x^n(x-1)}{x^n-1}dx={\int }_0^1\frac{x^{n+1}-x^n}{x^n-1}dx$$$$={\int }_0^1(x^{n+1}-x^n)\sum _{k=0}^{\mathrm{\infty }}{x}^{kn}=\sum _{k=0}^{\mathrm{\infty }}{\int }_0^1(x^{n+1+kn}-x^{n+kn})dx$$$$=\sum _{k=0}^{\mathrm{\infty }}{\int }_0^1(x^{(k+1)n+1}-x^{(k+1)n})dx$$$$=\sum _{k=0}^{\mathrm{\infty }}{[\frac{1}{(k+1)n+2}{x}^{(k+1)n+2}-\frac{1}{(k+1)n+1}{x}^{(k+1)n+1}]}_0^1$$$$=\sum _{k=0}^{\mathrm{\infty }}\{\frac{1}{(k+1)n+2}-\frac{1}{(k+1)n+1}\}=\sum _{k=1}^{\mathrm{\infty }}\{\frac{1}{kn+2}-\frac{1}{kn+1}\}$$$$=-\sum _{k=1}^{\mathrm{\infty }}\frac{1}{(kn+1)(kn+2)}becontinued...(restvalueofthisserie!)$$

Commented by mind is power last updated on 24/Jan/20

$$nicesirthankyou$$

Commented by abdomathmax last updated on 24/Jan/20

$$youarewelcome$$

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