Question-158207

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

Answered by qaz last updated on 01/Nov/21

I=∫_0 ^∞ ((x^5 (e^(3x) −e^x ))/((e^x −1)^4 ))dx =−∫_0 ^1 ((ln^5 x)/((1−x)^3 ))(1+x)dx =−(1/2)Σ_(n=0) ^∞ (n+1)(n+2)∫_0 ^1 x^n (1+x)ln^5 xdx =((5!)/2)Σ_(n=0) ^∞ (n+1)(n+2)((1/((n+1)^6 ))+(1/((n+2)^6 ))) =((5!)/2)Σ_(n=1) ((n^2 +n)/n^6 )+(1/((n+1)^4 ))−(1/((n+1)^5 )) =120ζ(4) −−−−−−−−−−−−−− (1/((1−x)^3 ))=(Σ_(k=0) ^∞ x^k )^3 =(Σ_(k=0) ^∞ x^k )(Σ_(k=0) ^∞ (k+1)x^k ) =Σ_(n=0) ^∞ Σ_(k=0) ^n x^n (k+1)=(1/2)Σ_(n=0) ^∞ (n+1)(n+2)x^n

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

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