Question Number 132091 by rs4089 last updated on 11/Feb/21 | ||
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$${evaluate}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{da}.{db}.{dc}.{dd}.{df}}{\mathrm{1}−{abcdf}} \\ $$ | ||
Answered by Dwaipayan Shikari last updated on 11/Feb/21 | ||
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$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{5}} }=\zeta\left(\mathrm{5}\right) \\ $$ | ||
Answered by Dwaipayan Shikari last updated on 11/Feb/21 | ||
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$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \left({abcdf}\right)^{{n}} {dadbdcdddf} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{5}} }=\zeta\left(\mathrm{5}\right) \\ $$ | ||