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Question Number 80764 by M±th+et£s last updated on 06/Feb/20

show that ∫_0 ^∞ x arctanh(e^(−αx) )dx=((7ζ(3))/(8α^2 ))

$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}\:{arctanh}\left({e}^{−\alpha{x}} \right){dx}=\frac{\mathrm{7}\zeta\left(\mathrm{3}\right)}{\mathrm{8}\alpha^{\mathrm{2}} } \\ $$

Answered by ~blr237~ last updated on 06/Feb/20

let be f(α)=∫_0 ^∞ xargth(e^(−αx) )dx f(α) exist ⇔∀ x>0 e^(−αx) <1 ⇔ α>0 cause argthx=(1/2)ln(((1+x)/(1−x))) let state u=e^(−αx) ⇔ x=−((lnu)/α) f(α)=(1/(2α^2 )) ∫_0 ^1 −((lnu)/u) ln(((1+u)/(1−u)))du 2α^2 f(α)= ∫_0 ^1 lnu ((ln(1−u))/u) du −∫_0 ^1 lnu((ln(1+u))/u)du 2α^2 f(α)= −Σ_(n=1) ^∞ ∫_0 ^1 (u^(n−1) /n)lnudu−Σ_(n=1) ^∞ ∫_0 ^1 (((−u)^(n−1) )/n)lnudu 2α^2 f(α)=Σ_(n=1) ^∞ (1/n^3 ) +Σ_(n=1) ^∞ (((−1)^(n−1) )/n^3 ) 2α^2 f(α)=Σ_(n=1) ^∞ (1/((2n)^3 )) +Σ_(n=1) ^∞ (1/((2n+1)^3 )) +Σ_(n=1) ^∞ (((−1)^(2n−1) )/((2n)^3 ))+Σ_(n=1) ^∞ (((−1)^(2n+1−1) )/((2n+1)^3 )) 2α^2 f(α)= 2Σ_(n=1) ^∞ (1/((2n+1)^3 )) =2(Σ_(n=1) ^∞ (1/n^3 ) −Σ_(n=1) ^∞ (1/((2n)^3 )) )=2×(1−(1/8))Σ_(n=1) ^∞ (1/n^3 ) so f(α)=((7ζ(3))/(8α^2 ))

$${let}\:{be}\:\:{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} {xargth}\left({e}^{−\alpha{x}} \right){dx} \\ $$$${f}\left(\alpha\right)\:{exist}\:\Leftrightarrow\forall\:{x}>\mathrm{0}\:\:\:\:\:{e}^{−\alpha{x}} \:<\mathrm{1}\:\Leftrightarrow\:\alpha>\mathrm{0}\:\:\:{cause}\:\:{argthx}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$$${let}\:{state}\:\:\:{u}={e}^{−\alpha{x}} \:\Leftrightarrow\:{x}=−\frac{{lnu}}{\alpha}\: \\ $$$${f}\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}\alpha^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\mathrm{1}} \:−\frac{{lnu}}{{u}}\:{ln}\left(\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\right){du}\: \\ $$$$\mathrm{2}\alpha^{\mathrm{2}} {f}\left(\alpha\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{lnu}\:\frac{{ln}\left(\mathrm{1}−{u}\right)}{{u}}\:{du}\:−\int_{\mathrm{0}} ^{\mathrm{1}} {lnu}\frac{{ln}\left(\mathrm{1}+{u}\right)}{{u}}{du} \\ $$$$\mathrm{2}\alpha^{\mathrm{2}} {f}\left(\alpha\right)=\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{{n}−\mathrm{1}} }{{n}}{lnudu}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(−{u}\right)^{{n}−\mathrm{1}} }{{n}}{lnudu}\: \\ $$$$\mathrm{2}\alpha^{\mathrm{2}} {f}\left(\alpha\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{3}} }\: \\ $$$$\:\:\mathrm{2}\alpha^{\mathrm{2}} {f}\left(\alpha\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{3}} }\:+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\:+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{n}−\mathrm{1}} }{\left(\mathrm{2}{n}\right)^{\mathrm{3}} }+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}−\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\: \\ $$$$\mathrm{2}\alpha^{\mathrm{2}} {f}\left(\alpha\right)=\:\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\mathrm{2}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{3}} }\:\right)=\mathrm{2}×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\: \\ $$$${so}\:\:{f}\left(\alpha\right)=\frac{\mathrm{7}\zeta\left(\mathrm{3}\right)}{\mathrm{8}\alpha^{\mathrm{2}} }\: \\ $$$$ \\ $$

Commented by M±th+et£s last updated on 06/Feb/20

thank you sir. nice solution

$${thank}\:{you}\:{sir}.\:{nice}\:{solution} \\ $$

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