INTEGRAL-prove-that-0-x-3-ln-1-e-x-x-dx-45-8-5-

Question Number 125684 by mnjuly1970 last updated on 13/Dec/20

.... INTEGRAL... prove that : ∫_0 ^( ∞) x^3 {ln(1+e^x ) −x}dx=((45)/8) ζ( 5 )

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\:\mathrm{INTEGRAL}… \\ $$$$\:\:\:\:\mathrm{prove}\:\:\mathrm{that}\:: \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} {x}^{\mathrm{3}} \left\{{ln}\left(\mathrm{1}+{e}^{{x}} \right)\:−{x}\right\}{dx}=\frac{\mathrm{45}}{\mathrm{8}}\:\zeta\left(\:\mathrm{5}\:\right) \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 13/Dec/20

∫_0 ^∞ x^3 log(((1+e^x )/e^x ))dx =∫_0 ^∞ x^3 log(1+e^(−x) )dx =Σ_(n=1) ^∞ (−1)^(n+1) (1/n)∫_0 ^∞ x^3 e^(−nx) dx =Σ_(n=1) ^∞ (−1)^(n+1) (1/n^5 )∫_0 ^∞ Λ^3 e^(−Λ) dx nx=Λ =Σ_(n=1) ^∞ (((−1)^(n+1) )/n^5 )Γ(4)=6Σ_(n=1) ^∞ (((−1)^(n+1) )/n^5 )=((90)/(16))ζ(5)=((45)/8)ζ(5) Note : 1+(1/2^5 )+(1/3^5 )+(1/4^5 )+...=ζ(5) 1−(1/2^5 )+(1/3^5 )−(1/4^5 )+..=S S+ζ(5)=2(1+(1/3^5 )+(1/5^5 )+...) (1+(1/3^5 )+(1/5^5 )+...)=ζ(5)(1−(1/2^5 )) S=2ζ(5)−((ζ(5))/2^4 )−ζ(5)=ζ(5)((15)/(16))

$$\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{3}} {log}\left(\frac{\mathrm{1}+{e}^{{x}} }{{e}^{{x}} }\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{3}} {log}\left(\mathrm{1}+{e}^{−{x}} \right){dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{3}} {e}^{−{nx}} {dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{n}^{\mathrm{5}} }\int_{\mathrm{0}} ^{\infty} \Lambda^{\mathrm{3}} {e}^{−\Lambda} {dx}\:\:\:\:\:\:\:{nx}=\Lambda \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}^{\mathrm{5}} }\Gamma\left(\mathrm{4}\right)=\mathrm{6}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}^{\mathrm{5}} }=\frac{\mathrm{90}}{\mathrm{16}}\zeta\left(\mathrm{5}\right)=\frac{\mathrm{45}}{\mathrm{8}}\zeta\left(\mathrm{5}\right) \\ $$$$\boldsymbol{{Note}}\:: \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{5}} }+…=\zeta\left(\mathrm{5}\right) \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{5}} }+..={S}\:\:\: \\ $$$${S}+\zeta\left(\mathrm{5}\right)=\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{5}} }+…\right)\:\:\:\:\:\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{5}} }+…\right)=\zeta\left(\mathrm{5}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{5}} }\right) \\ $$$${S}=\mathrm{2}\zeta\left(\mathrm{5}\right)−\frac{\zeta\left(\mathrm{5}\right)}{\mathrm{2}^{\mathrm{4}} }−\zeta\left(\mathrm{5}\right)=\zeta\left(\mathrm{5}\right)\frac{\mathrm{15}}{\mathrm{16}} \\ $$

Commented by Dwaipayan Shikari last updated on 13/Dec/20

$${But}\:{i}\:{haven}'{t}\:{find}\:{the}\:{same}\:{sir}! \\ $$

Commented by mnjuly1970 last updated on 13/Dec/20

$${thank}\:{you}\:{mr}\:{payan} \\ $$$${you}\:{solution}\:{is}\:{very} \\ $$$${nice} \\ $$$${to}\:{me}\:\:{final}\:{answer}\:{isn}'{t}\:{very} \\ $$$${important}… \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 13/Dec/20