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Question Number 174696 by mnjuly1970 last updated on 08/Aug/22

$$$$$$\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{t}\mathit{h}\mathit{a}\mathit{t}:$$$$$$$$\mathbf{\Omega }={\int }_0^{\mathrm{\infty }}{\left(\frac{\mathit{x}}{\mathit{s}\mathit{i}\mathit{n}\mathit{h}\left(\mathit{x}\right)}\right)}^3\mathit{d}\mathit{x}=\frac{{\mathit{\pi }}^2}{16}(12-{\mathit{\pi }}^2)$$$$\mathit{w}\mathit{r}\mathit{i}\mathit{t}\mathit{t}\mathit{e}\mathit{n}\mathit{a}\mathit{n}\mathit{d}\mathit{p}\mathit{r}\mathit{e}\mathit{p}\mathit{a}\mathit{r}\mathit{e}\mathit{d}\mathit{b}\mathit{y}:\mathit{m}.\mathit{n}$$$$$$

Answered by aleks041103 last updated on 08/Aug/22

$$\frac{1}{{(1-x)}^3}=\frac{1}{2}\frac{d^2}{{dx}^2}\left(\frac{1}{1-x}\right)=\frac{1}{2}\frac{d^2}{{dx}^2}\left(\underset{i=0}{\sum ^{\mathrm{\infty }}}{x}^i\right)=$$$$=\frac{1}{2}\left(\underset{i=2}{\sum ^{\mathrm{\infty }}}\frac{i!}{(i-2)!}{x}^{i-2}\right)$$$$\frac{1}{{sinh}^{3}x}={\left(\frac{2e^{-x}}{1-e^{-2x}}\right)}^3=8e^{-3x}{\left(\frac{1}{1-r}\right)}^3=$$$$=4e^{-3x}\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{(i+2)!}{i!}{r}^i=$$$$=4e^{-3x}\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{(i+2)!}{i!}{e}^{-2ix}=4\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{(i+2)!}{i!}{e}^{-(2i+3)x}$$$${\int }_0^{\mathrm{\infty }}\frac{x^{3}dx}{{sinh}^{3}x}=4\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{(i+2)!}{i!}{\int }_0^{\mathrm{\infty }}{x}^3{e}^{-(2i+3)x}dx$$$${\int }_0^{\mathrm{\infty }}{x}^3{e}^{-(2i+3)x}dx=\frac{1}{{(2i+3)}^4}{\int }_0^{\mathrm{\infty }}{z}^3{e}^{-z}dz=\frac{6}{{(2i+3)}^4}$$$$\Rightarrow Ans.=24\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{(i+1)(i+2)}{{(2i+3)}^4}=$$$$=24\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{i(i+1)}{{(2i+1)}^4}$$$$\frac{i(i+1)}{{(2i+1)}^4}=\frac{i^2+i}{{(2i+1)}^4}=\frac{i^2+2.\frac{1}{2}.i+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}{{(2i+1)}^4}=$$$$=\frac{{(i+\frac{1}{2})}^2-\frac{1}{4}}{{(2i+1)}^4}=\frac{1}{4}\frac{{(2i+1)}^2}{{(2i+1)}^4}-\frac{1}{4}\frac{1}{{(2i+1)}^4}=$$$$=\frac{1}{4}(\frac{1}{{(2i+1)}^2}-\frac{1}{{(2i+1)}^4})$$$$\Rightarrow Ans.=6[\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2i+1)}^2}-\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2i+1)}^4}]$$$$\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2i+1)}^n}=\underset{i=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{{(2i+1)}^n}+\frac{1}{{\left(2i\right)}^n})-\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\left(2i\right)}^n}=$$$$=\underset{i=2}{\sum ^{\mathrm{\infty }}}\frac{1}{{\left(i\right)}^n}-2^{-n}\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{i^n}=$$$$=\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{i^n}-1-2^{-n}\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{i^n}=$$$$=(1-2^{-n})\zeta \left(n\right)-1$$$$\Rightarrow Ans.=6((1-2^{-2})\zeta \left(2\right)-(1-2^{-4})\zeta \left(4\right))=$$$$=6(\frac{3}{4}\frac{{\pi }^2}{6}-\frac{15}{16}\frac{{\pi }^4}{90})=$$$$=\frac{3}{4}{\pi }^2-\frac{1}{16}{\pi }^4=$$$$=\frac{{\pi }^2}{16}(12-{\pi }^2)$$

Commented by aleks041103 last updated on 08/Aug/22

$$Ididitfinally...$$🤣🤣

Commented by mnjuly1970 last updated on 09/Aug/22

$$thanksalot$$$$siraleks...iappreciate$$

Commented by Tawa11 last updated on 09/Aug/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by princeDera last updated on 09/Aug/22

$$$$$$$$$$\mathrm{\Omega }{\int }_0^{\mathrm{\infty }}{\left(\frac{x}{\sinh \left(x\right)}\right)}^3=8{\int }_0^{\mathrm{\infty }}{\left(\frac{{xe}^{-x}}{1-e^{-2x}}\right)}^3$$$$\mathrm{\Omega }=8{\int }_0^{\mathrm{\infty }}\frac{x^3{e}^{-3x}}{{(1-e^{-2x})}^3}=4{\int }_0^{\mathrm{\infty }}{x}^3{e}^{-3x}\sum _{k⩾0}(2+k)(1+k)e^{-2kx}dx$$$$\mathrm{\Omega }=4\sum _{k⩾0}(2+3k+k^2){\int }_0^{\mathrm{\infty }}{x}^3{e}^{-(3+2k)x}dx$$$$\mathrm{\Omega }=8\sum _{k⩾0}\frac{6}{{(3+2k)}^4}+24\sum _{k⩾0}\frac{{(k+\frac{3}{2})}^2-\frac{9}{4}}{{(2k+3)}^4}$$$$$$$$\mathrm{\Omega }=48\sum _{k⩾1}\frac{1}{{(2k+1)}^4}+6\sum _{k⩾0}\frac{{(2k+3)}^2}{{(2k+3)}^4}-54\sum _{k⩾0}\frac{1}{{(2k+3)}^4}$$$$\mathrm{\Omega }=48(\lambda \left(4\right)-1)+6\lambda \left(2\right)-6-54\lambda \left(4\right)+54$$$$\mathrm{\Omega }=48\{\frac{{\pi }^4}{96}-1\}+\frac{6{\pi }^2}{8}-6-54.\frac{{\pi }^4}{96}+54$$$$\mathrm{\Omega }=\frac{{\pi }^4}{96}(48-54)+\frac{3{\pi }^2}{4}=\frac{3{\pi }^2}{4}-\frac{{\pi }^4}{16}$$$$\mathrm{\Omega }=\frac{{\pi }^2}{16}(12-{\pi }^2)$$$$\lambda \left(x\right)isdirichletlambdafunction$$$$$$

Commented by mnjuly1970 last updated on 09/Aug/22

$$gratefulsir$$$$$$

Commented by Tawa11 last updated on 09/Aug/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by Ar Brandon last updated on 09/Aug/22

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{\left(\frac{x}{\sinh x}\right)}^{3}dx={\int }_0^{\mathrm{\infty }}\frac{x^3}{{\sinh }^{3}x}dx=8{\int }_0^{\mathrm{\infty }}\frac{x^3}{{(e^x-e^{-x})}^3}dx$$$$=8{\int }_0^{\mathrm{\infty }}\frac{x^3{e}^{-3x}}{{(1-e^{-2x})}^3}dx=4{\int }_0^{\mathrm{\infty }}{x}^3{e}^{-3x}\underset{n=0}{\sum ^{\mathrm{\infty }}}(n+1)(n+2)e^{-2nx}dx$$$$=======================================$$$$\frac{1}{1-t}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{t}^n\Rightarrow \frac{1}{{(1-t)}^2}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{nt}^{n-1}=\underset{n=0}{\sum ^{\mathrm{\infty }}}(n+1)t^n$$$$\Rightarrow \frac{2}{{(1-t)}^3}=\underset{n=0}{\sum ^{\mathrm{\infty }}}n(n+1)t^{n-1}=\underset{n=0}{\sum ^{\mathrm{\infty }}}(n+1)(n+2)t^n$$$$=======================================$$$$=4\underset{n=0}{\sum ^{\mathrm{\infty }}}(n+1)(n+2){\int }_0^{\mathrm{\infty }}{x}^3{e}^{-(2n+3)x}dx=4\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{(n+1)(n+2)}{{(2n+3)}^4}{\int }_0^{\mathrm{\infty }}{x}^3{e}^{-x}dx$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{(2n+3-1)(2n+3+1)}{{(2n+3)}^4}{\int }_0^{\mathrm{\infty }}{x}^3{e}^{-x}dx=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(2n+3)}^2-1}{{(2n+3)}^4}\mathrm{\Gamma }\left(4\right)$$$$=6\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{{(2n+3)}^2}-\frac{1}{{(2n+3)}^4})=6(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2n+1)}^2}-1-(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2n+1)}^4}-1))$$$$=6(\frac{3}{4}\times \zeta \left(2\right)-\frac{15}{16}\times \zeta \left(4\right))=6(\frac{3}{4}\times \frac{{\pi }^2}{6}-\frac{15}{16}\times \frac{{\pi }^4}{90})=\frac{3{\pi }^2}{4}-\frac{{\pi }^4}{16}=\frac{{\pi }^2}{16}(12-{\pi }^2)★$$

Commented by mnjuly1970 last updated on 09/Aug/22

$$thankyousir$$

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