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Question Number 186321 by Mingma last updated on 03/Feb/23

Answered by normans last updated on 03/Feb/23

$${\int }_1^{e^{(\mathit{n}+1)}}\mathit{l}\mathit{o}\mathit{g}\left(\mathit{x}\right)={\mathit{e}}^{\mathit{n}+1}(\mathit{l}\mathit{o}\mathit{g}\left({\mathit{e}}^{\mathit{n}+1}\right)-1+1$$$$\mathit{f}\mathit{o}\mathit{r}{\mathit{e}}^{\mathit{n}}⩾0\wedge ({\mathit{e}}^{\mathit{n}+1}\notin \mathbb{R}\vee \mathbf{R}\mathit{e}\left({\mathit{e}}^{\mathit{n}+1}\right)>0)$$$$$$

Commented by MJS_new last updated on 03/Feb/23

$$\mathrm{this}\mathrm{is}\mathrm{no}\mathrm{answer}\mathrm{to}\mathrm{the}\mathrm{question}$$

Answered by MJS_new last updated on 03/Feb/23

$$\mathrm{assuming}n\in \mathbb{N}$$$$\underset{1}{\stackrel{{\mathrm{e}}^{n+1}}{\int }}\lfloor \ln x\rfloor dx=$$$$[t=\ln x\to dx={\mathrm{e}}^{t}dt]$$$$=\underset{0}{\stackrel{n+1}{\int }}{\mathrm{e}}^t\lfloor t\rfloor dt=0{\left[{\mathrm{e}}^t\right]}_0^1+1{\left[{\mathrm{e}}^t\right]}_1^2+...n{\left[{\mathrm{e}}^t\right]}_n^{n+1}=$$$$=\underset{k=0}{\sum ^n}\left(k({\mathrm{e}}^{k+1}-{\mathrm{e}}^k)\right)=(\mathrm{e}-1)\underset{k=0}{\sum ^n}\left(k{\mathrm{e}}^k\right)=$$$$=(\mathrm{e}-1)\frac{(1+(n\mathrm{e}-n-1){\mathrm{e}}^n)\mathrm{e}}{{(\mathrm{e}-1)}^2}=$$$$=n{\mathrm{e}}^{n+1}-\frac{({\mathrm{e}}^n-1)\mathrm{e}}{\mathrm{e}-1}$$

Answered by Mathspace last updated on 04/Feb/23

$$changementlnx=tgivex=e^t\Rightarrow$$$$I={\int }_0^{n+1}\left[t\right]e^{t}dt$$$$=\sum _{k=0}^{n-1}{\int }_k^{k+1}k{e}^{t}dt$$$$=\sum _{k=0}^{n-1}k(e^{k+1}-e^k)$$$$=(e-1)\sum _{k=1}^{n-1}{ke}^k$$$$f\left(x\right)=\sum _{k=0}^{n-1}{x}^k=\frac{x^n-1}{x-1}forx\ne 1$$$$f^{{}^{\prime }}\left(x\right)=\sum _{k=1}^{n-1}{kx}^{k-1}\Rightarrow$$$${xf}^{{}^{\prime }}\left(x\right)=\sum _{k=1}^{n-1}k{x}^k$$$$=x.\frac{{nx}^{n-1}(x-1)-(x^n-1)}{{(x-1)}^2}$$$$=x.\frac{{nx}^n-{nx}^{n-1}-x^n+1}{{(x-1)}^2}$$$$=x.\frac{(n-1)x^n-{nx}^{n-1}+1}{{(x-1)}^2}$$$$=\frac{(n-1)x^{n+1}-{nx}^n+x}{{(x-1)}^2}$$$$x=e\Rightarrow \sum _{k=1}^{n-1}{ke}^k$$$$=\frac{(n-1)e^{n+1}-{ne}^n+e}{{(e-1)}^2}\Rightarrow$$$$\Rightarrow I=(e-1).\frac{(n-1)e^{n+1}-{ne}^n+e}{{(e-1)}^2}$$$$=\frac{(n-1)e^{n+1}-{ne}^n+e}{e-1}$$

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