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Question Number 158207 by tounghoungko last updated on 01/Nov/21

Answered by qaz last updated on 01/Nov/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^5({\mathrm{e}}^{3\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}})}{{({\mathrm{e}}^{\mathrm{x}}-1)}^4}\mathrm{dx}$$$$=-{\int }_0^1\frac{{\ln }^5\mathrm{x}}{{(1-\mathrm{x})}^3}(1+\mathrm{x})\mathrm{dx}$$$$=-\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}(\mathrm{n}+1)(\mathrm{n}+2){\int }_0^1{\mathrm{x}}^{\mathrm{n}}(1+\mathrm{x}){\ln }^5\mathrm{xdx}$$$$=\frac{5!}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}(\mathrm{n}+1)(\mathrm{n}+2)(\frac{1}{{(\mathrm{n}+1)}^6}+\frac{1}{{(\mathrm{n}+2)}^6})$$$$=\frac{5!}{2}\sum _{\mathrm{n}=1}\frac{{\mathrm{n}}^2+\mathrm{n}}{{\mathrm{n}}^6}+\frac{1}{{(\mathrm{n}+1)}^4}-\frac{1}{{(\mathrm{n}+1)}^5}$$$$=120\zeta \left(4\right)$$$$--------------$$$$\frac{1}{{(1-\mathrm{x})}^3}={\left(\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{x}}^{\mathrm{k}}\right)}^3=\left(\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{x}}^{\mathrm{k}}\right)\left(\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}(\mathrm{k}+1){\mathrm{x}}^{\mathrm{k}}\right)$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{x}}^{\mathrm{n}}(\mathrm{k}+1)=\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}(\mathrm{n}+1)(\mathrm{n}+2){\mathrm{x}}^{\mathrm{n}}$$

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