prove-to-ealier-problem-of-0-1-0-1-tanh-1-x-1-4-tanh-1-y-1-4-x-y-dxdy-pi-2-solution-let-I-0-1-tanh-1-x-1-4-x-dx-0-1-tanh-1-y-1-4-y-d

Question Number 110395 by mathdave last updated on 28/Aug/20

prove to ealier problem of ∫_0 ^1 ∫_0 ^1 ((tanh^(−1) ((x)^(1/4) )tanh^(−1) ((y)^(1/4) ))/(x(√y)))dxdy=π^2 solution let I=∫_0 ^1 ((tanh^(−1) ((x)^(1/4) ))/x)dx∫_0 ^1 ((tanh^(−1) ((y)^(1/4) ))/( (√y)))dy=A.B let m=x^(1/4) and dx=4m^3 A=∫_0 ^1 ((tanh^(−1) (m))/m^4 )×4m^3 dm=4∫_0 ^1 ((tanh^(−1) (m))/m)dm but series of tanh^(−1) (m)=Σ_(k=0) ^∞ (m^(2k+1) /(2k+1)) A=4Σ_(k=0) ^∞ (1/(2k+1))∫_0 ^1 (m^(2k+1) /m)dm=4Σ_(k=o) ^∞ (1/(2k+1))∫_0 ^1 m^(2k) dm A=4Σ_(k=0) ^∞ (1/((2k+1)^2 ))=4×(1/2)Σ_(k=−∞) ^∞ (1/((2k+1)^2 )) recall that Σ_(n=−∞) ^∞ (1/((an+1)^m ))=−(π/((a)^m ))lim_(z→−(1/a) ) (1/((m−1)!))(d^((m−1)) /dz^((m−1)) )[cot(πz)] ∵A=2[−(π/((2)^2 ))lim_(z→−(1/2)) (1/((2−1)!))(d/dz)[cot(πz)] A=−(π/2)lim_(z→−(1/2)) [−πcosec^2 (−(π/2))]=(π^2 /2).....(1) then B=∫_0 ^1 ((tanh^(−1) ((y)^(1/4) ))/( (√y)))dy let n=y^(1/4) and dy=4n^3 B=∫_0 ^1 ((tanh^(−1) (n))/n^2 )×4n^3 dn=4∫_0 ^1 ntanh^(−1) (n)dn B=4Σ_(k=0) ^∞ (1/(2k+1))∫_0 ^1 n^(2k+2) dn=4Σ_(k=0) ^∞ (1/((2k+1)(2k+3))) B=4×(1/2)Σ_(k=0) ^∞ ((1/(2k+1))−(1/(2k+3))) B=2lim_(k→∞) (1−(1/3)+(1/3)−(1/5)+(1/5).......+(1/(2k+1)))=2.....(2) the series is telescoping but I=A×B=(π^2 /2)×2=π^2 ..............(3) ∫_0 ^1 ∫_0 ^1 ((tanh^(−1) ((x)^(1/4) )tanh^(−1) ((y)^(1/4) ))/(x(√y)))dxdy=π^2 Q.E.D by mathdave(28/08/2020)

$${prove}\:{to}\:{ealier}\:{problem}\:{of}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{x}}\right)\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{y}}\right)}{{x}\sqrt{{y}}}{dxdy}=\pi^{\mathrm{2}} \\ $$$${solution}\: \\ $$$${let} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{x}}\right)}{{x}}{dx}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{y}}\right)}{\:\sqrt{{y}}}{dy}={A}.{B} \\ $$$${let}\:{m}={x}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:{and}\:\:{dx}=\mathrm{4}{m}^{\mathrm{3}} \:\: \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left({m}\right)}{{m}^{\mathrm{4}} }×\mathrm{4}{m}^{\mathrm{3}} {dm}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left({m}\right)}{{m}}{dm} \\ $$$${but}\:{series}\:{of}\:\mathrm{tanh}^{−\mathrm{1}} \left({m}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{m}^{\mathrm{2}{k}+\mathrm{1}} }{\mathrm{2}{k}+\mathrm{1}} \\ $$$${A}=\mathrm{4}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{m}^{\mathrm{2}{k}+\mathrm{1}} }{{m}}{dm}=\mathrm{4}\underset{{k}={o}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {m}^{\mathrm{2}{k}} {dm} \\ $$$${A}=\mathrm{4}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{4}×\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${recall}\:{that} \\ $$$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({an}+\mathrm{1}\right)^{{m}} }=−\frac{\pi}{\left({a}\right)^{{m}} }\underset{{z}\rightarrow−\frac{\mathrm{1}}{{a}}\:} {\mathrm{lim}} \\ $$$$\frac{\mathrm{1}}{\left({m}−\mathrm{1}\right)!}\frac{{d}^{\left({m}−\mathrm{1}\right)} }{{dz}^{\left({m}−\mathrm{1}\right)} }\left[\mathrm{cot}\left(\pi{z}\right)\right] \\ $$$$\because{A}=\mathrm{2}\left[−\frac{\pi}{\left(\mathrm{2}\right)^{\mathrm{2}} }\underset{{z}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\frac{{d}}{{dz}}\left[\mathrm{cot}\left(\pi{z}\right)\right]\right. \\ $$$${A}=−\frac{\pi}{\mathrm{2}}\underset{{z}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} {\mathrm{lim}}\:\left[−\pi\mathrm{cosec}^{\mathrm{2}} \left(−\frac{\pi}{\mathrm{2}}\right)\right]=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}…..\left(\mathrm{1}\right) \\ $$$${then}\:\:{B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{y}}\right)}{\:\sqrt{{y}}}{dy} \\ $$$${let}\:\:{n}={y}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:\:{and}\:\:\:{dy}=\mathrm{4}{n}^{\mathrm{3}} \\ $$$${B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left({n}\right)}{{n}^{\mathrm{2}} }×\mathrm{4}{n}^{\mathrm{3}} {dn}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} {n}\mathrm{tanh}^{−\mathrm{1}} \left({n}\right){dn} \\ $$$${B}=\mathrm{4}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {n}^{\mathrm{2}{k}+\mathrm{2}} {dn}=\mathrm{4}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)} \\ $$$${B}=\mathrm{4}×\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{3}}\right) \\ $$$${B}=\mathrm{2}\underset{{k}\rightarrow\infty} {\mathrm{li}{m}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}…….+\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right)=\mathrm{2}…..\left(\mathrm{2}\right) \\ $$$${the}\:{series}\:{is}\:{telescoping} \\ $$$${but}\:{I}={A}×{B}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}×\mathrm{2}=\pi^{\mathrm{2}} …………..\left(\mathrm{3}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{x}}\right)\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{y}}\right)}{{x}\sqrt{{y}}}{dxdy}=\pi^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:{Q}.{E}.{D}\:\:\:\: \\ $$$$\:{by}\:{mathdave}\left(\mathrm{28}/\mathrm{08}/\mathrm{2020}\right) \\ $$

Commented by Aziztisffola last updated on 28/Aug/20