Question Number 209918 by hardmath last updated on 25/Jul/24
![Find: ∫_0 ^( ∞) (({x}^([x]) )/([x] + 1)) dx = ? {x} → fractional part [x] → full part](https://www.tinkutara.com/question/Q209918.png)
$$\mathrm{Find}:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\left\{\mathrm{x}\right\}^{\left[\boldsymbol{\mathrm{x}}\right]} }{\left[\mathrm{x}\right]\:+\:\mathrm{1}}\:\mathrm{dx}\:=\:? \\ $$$$\left\{\mathrm{x}\right\}\:\rightarrow\:\mathrm{fractional}\:\mathrm{part} \\ $$$$\left[\mathrm{x}\right]\:\:\:\rightarrow\:\mathrm{full}\:\mathrm{part} \\ $$
Answered by MM42 last updated on 25/Jul/24
![{x}=x−[x] I_n =∫_0 ^1 dx+∫_1 ^2 (((x−1))/2)dx+∫_2 ^3 (((x−2)^2 )/3)dx+...+∫_(n−1) ^n (((x−n+1)^(n−1) )/n) dx =x]_0 ^1 +(((x−1)^2 )/4)]_1 ^2 +(((x−2)^3 )/9)]_2 ^3 +...+(((x−n+1)^n )/n^2 )]_(n−1) ^n =1+(1/4)+(1/9)+...+(1/n^2 ) ⇒I=Σ_(n=1) ^∞ (1/n^2 )=(π^2 /6) ✓](https://www.tinkutara.com/question/Q209921.png)
$$\left\{{x}\right\}={x}−\left[{x}\right] \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {dx}+\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\left({x}−\mathrm{1}\right)}{\mathrm{2}}{dx}+\int_{\mathrm{2}} ^{\mathrm{3}} \frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{3}}{dx}+…+\int_{{n}−\mathrm{1}} ^{{n}} \frac{\left({x}−{n}+\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{dx} \\ $$$$\left.=\left.{x}\left.\right]_{\mathrm{0}} ^{\mathrm{1}} \left.+\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\right]_{\mathrm{1}} ^{\mathrm{2}} +\frac{\left({x}−\mathrm{2}\right)^{\mathrm{3}} }{\mathrm{9}}\right]_{\mathrm{2}} ^{\mathrm{3}} +…+\frac{\left({x}−{n}+\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\right]_{{n}−\mathrm{1}} ^{{n}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}+…+\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\Rightarrow{I}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\checkmark \\ $$$$\: \\ $$