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Question Number 209918 by hardmath last updated on 25/Jul/24

$$\mathrm{Find}:{\int }_0^{\mathrm{\infty }}\frac{{\left\{\mathrm{x}\right\}}^{\left[\mathbf{x}\right]}}{\left[\mathrm{x}\right]+1}\mathrm{dx}=?$$$$\left\{\mathrm{x}\right\}\to \mathrm{fractional}\mathrm{part}$$$$\left[\mathrm{x}\right]\to \mathrm{full}\mathrm{part}$$

Answered by MM42 last updated on 25/Jul/24

$$\left\{x\right\}=x-\left[x\right]$$$$I_n={\int }_0^{1}dx+{\int }_1^2\frac{(x-1)}{2}dx+{\int }_2^3\frac{{(x-2)}^2}{3}dx+...+{\int }_{n-1}^n\frac{{(x-n+1)}^{n-1}}{n}dx$$$${={x{]}_0^1{+\frac{{(x-1)}^2}{4}]}_1^2+\frac{{(x-2)}^3}{9}]}_2^3+...+\frac{{(x-n+1)}^n}{n^2}]}_{n-1}^n$$$$=1+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{n^2}$$$$\Rightarrow I=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\frac{{\pi }^2}{6}✓$$$$$$

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