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Question Number 126969 by mnjuly1970 last updated on 25/Dec/20

$$calculus$$$$pleaseevaluate::$$$$\varphi ={\int }_0^1\lfloor ln\left(x\right)\rfloor dx=?$$$$$$

Answered by mindispower last updated on 25/Dec/20

$$ln\left(x\right)=t$$$$={\int }_{-\mathrm{\infty }}^0\left[t\right]e^t$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}{\int }_{-k-1}^{-k}\left[t\right]e^{t}dt$$$$=-\underset{k=0}{\sum ^{\mathrm{\infty }}}{\int }_{-k-1}^{-k}(k+1)e^t=$$$$=-\underset{k=0}{\sum ^{\mathrm{\infty }}}(k+1)(e^{-k}-e^{-k-1})dt$$$$=-\mathrm{\Sigma }({ke}^{-k}-(k+1)e^{-(k+1)})-\underset{k=0}{\sum ^{\mathrm{\infty }}}{e}^{-k}$$$$=-(\underset{N\to \mathrm{\infty }}{\lim }(-(N+1)e^{-(N+1)})+\frac{1-e^{-(N+1)}}{1-e^{-1}})$$$$=-\frac{1}{1-e^{-1}}=-\frac{e}{e-1}$$

Commented by mnjuly1970 last updated on 25/Dec/20

$$merceysirmidspowdr$$$$grateful...$$

Answered by mathmax by abdo last updated on 27/Dec/20

$$\mathrm{\Phi }={\int }_0^1\left[\mathrm{lnx}\right]\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{lnx}=-\mathrm{t}\Rightarrow \mathrm{x}={\mathrm{e}}^{-\mathrm{t}}\Rightarrow$$$$\mathrm{\Phi }=-{\int }_0^{\mathrm{\infty }}[-\mathrm{t}](-{\mathrm{e}}^{-\mathrm{t}})\mathrm{dt}={\int }_0^{\mathrm{\infty }}[-\mathrm{t}]{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_{\mathrm{n}}^{\mathrm{n}+1}[-\mathrm{t}]{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\mathrm{we}\mathrm{have}\mathrm{n}⩽\mathrm{t}⩽\mathrm{n}+1\Rightarrow -\mathrm{n}-1⩽-\mathrm{t}⩽-\mathrm{n}\Rightarrow$$$$[-\mathrm{t}]=-(\mathrm{n}+1)\Rightarrow \mathrm{\Phi }=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_{\mathrm{n}}^{\mathrm{n}+1}-(\mathrm{n}+1){\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+1){[-{\mathrm{e}}^{-\mathrm{t}}]}_{\mathrm{n}}^{\mathrm{n}+1}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+1)({\mathrm{e}}^{-\mathrm{n}}-{\mathrm{e}}^{-(\mathrm{n}+1)})$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+1){\mathrm{e}}^{-\mathrm{n}}-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+1){\mathrm{e}}^{-(\mathrm{n}+1)}$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{\mathrm{e}}^{-(\mathrm{n}-1)}-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{\mathrm{e}}^{-\mathrm{n}}$$$$=(\mathrm{e}-1)\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{\left(\frac{1}{\mathrm{e}}\right)}^{\mathrm{n}}\mathrm{we}\mathrm{have}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}=\frac{1}{1-\mathrm{x}}\mathrm{for}\mid \mathrm{x}\mid <1\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}=\frac{1}{{(1-\mathrm{x})}^2}\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}=\frac{\mathrm{x}}{{(1-\mathrm{x})}^2}\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{\left(\frac{1}{\mathrm{e}}\right)}^{\mathrm{n}}=\frac{1}{\mathrm{e}{(\frac{1}{\mathrm{e}}-1)}^2}=\frac{\mathrm{e}}{{(1-\mathrm{e})}^2}\Rightarrow \mathrm{\Phi }=(\mathrm{e}-1)\times \frac{\mathrm{e}}{{(1-\mathrm{e})}^2}=\frac{\mathrm{e}}{1-\mathrm{e}}$$$$$$

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