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Question Number 184523 by cortano1 last updated on 08/Jan/23

Answered by SEKRET last updated on 08/Jan/23

$$\mathbf{t}=\frac{1}{\mathbf{x}}\mathbf{dt}=-\frac{1}{{\mathbf{x}}^2}\mathbf{dx}\mathbf{dx}=-\frac{1}{{\mathbf{t}}^2}\mathbf{dt}$$$${\int }_{\mathrm{\infty }}^1\frac{\frac{1}{{\mathbf{t}}^5}}{\lfloor \mathbf{t}\rfloor }\cdot \left(\frac{-1}{{\mathbf{t}}^2}\right)\mathbf{dt}={\int }_1^{\mathrm{\infty }}\frac{1}{{\mathbf{t}}^7\lfloor \mathbf{t}\rfloor }\mathbf{dt}=$$$$\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}{\underset{\mathbf{k}}{\int }}^{\mathbf{k}+1}\frac{1}{\mathbf{k}\cdot {\mathbf{t}}^7}\mathbf{dt}=\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}{(\frac{1}{\mathbf{k}}\cdot \frac{-1}{6{\mathbf{t}}^6})}_{\mathbf{k}}^{\mathbf{k}+1}=$$$$\frac{1}{6}\cdot (\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathbf{k}}^7}-\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathbf{k}{(\mathbf{k}+1)}^6})=$$$$\approx \frac{1}{6}\cdot (1.0083-0.016415)\approx 0.165314$$

Answered by mr W last updated on 08/Jan/23

$$I={\int }_0^1\frac{x^5}{\lfloor \frac{1}{x}\rfloor }dx=?$$$$letx=\frac{1}{t}$$$$dx=-\frac{dt}{t^2}$$$$I={\int }_1^{\mathrm{\infty }}\frac{dt}{t^7\lfloor t\rfloor }$$$$=\underset{k=1}{\sum ^{\mathrm{\infty }}}{\int }_k^{k+1}\frac{dt}{t^7\lfloor t\rfloor }$$$$=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}{\int }_k^{k+1}\frac{dt}{t^7}$$$$=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}{[-\frac{1}{6t^6}]}_k^{k+1}$$$$=\frac{1}{6}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}[\frac{1}{k^6}-\frac{1}{{(k+1)}^6}]$$$$=\frac{1}{6}\underset{k=1}{\sum ^{\mathrm{\infty }}}[\frac{1}{k^7}-\frac{1}{k{(k+1)}^6}]$$$$=\frac{1}{6}(A-B)$$$$A=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^7}=\zeta \left(7\right)\leftarrow RiemannZetafunction\zeta \left(s\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^s}$$$$B=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k{(k+1)}^6}$$$$=\underset{k=2}{\sum ^{\mathrm{\infty }}}\frac{1}{(k-1)k^6}$$$$=\underset{k=2}{\sum ^{\mathrm{\infty }}}[\frac{1}{(k-1)}-\frac{1}{k}-\frac{1}{k^2}-\frac{1}{k^3}-\frac{1}{k^4}-\frac{1}{k^5}-\frac{1}{k^6}]$$$$=\underset{k=2}{\sum ^{\mathrm{\infty }}}\frac{1}{(k-1)}-\underset{k=2}{\sum ^{\mathrm{\infty }}}\frac{1}{k}-\underset{k=2}{\sum ^{\mathrm{\infty }}}(\frac{1}{k^2}+\frac{1}{k^3}+\frac{1}{k^4}+\frac{1}{k^5}+\frac{1}{k^6})$$$$=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}+1-\underset{k=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{k^2}+\frac{1}{k^3}+\frac{1}{k^4}+\frac{1}{k^5}+\frac{1}{k^6})+5$$$$=-[\zeta \left(2\right)+\zeta \left(3\right)+\zeta \left(4\right)+\zeta \left(5\right)+\zeta \left(6\right)]+6$$$$I=\frac{1}{6}[\zeta \left(7\right)+\zeta \left(2\right)+\zeta \left(3\right)+\zeta \left(4\right)+\zeta \left(5\right)+\zeta \left(6\right)-6]$$$$\Rightarrow I=\frac{\zeta \left(2\right)+\zeta \left(3\right)+\zeta \left(4\right)+\zeta \left(5\right)+\zeta \left(6\right)+\zeta \left(7\right)}{6}-1$$$$\approx 0.165322382919$$

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