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Question Number 152874 by mnjuly1970 last updated on 02/Sep/21

solve: I := ∫_0 ^( ∞) ((( tanh (x) )/x) )^( 2) dx = ? m.n.

$$ \\ $$$$\:\:\:{solve}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{I}\::=\:\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\:{tanh}\:\left({x}\right)\:}{{x}}\:\right)^{\:\mathrm{2}} {dx}\:=\:? \\ $$$$\:\:{m}.{n}. \\ $$

Answered by Ar Brandon last updated on 02/Sep/21

I=∫_0 ^∞ (((tanhx)/x))^2 dx=∫_0 ^∞ ((tanh^2 x)/x^2 )dx =[−((tanh^2 x)/x)+2∫((tanhx(1−tanh^2 x))/x)dx]_0 ^∞ =2∫_0 ^∞ (((tanhx)/x)−((tanh^3 x)/x))dx=2∫_0 ^∞ (1/x)(((1−e^(−2x) )/(1+e^(−2x) ))−(((1−e^(−2x) )/(1+e^(−2x) )))^3 )dx =2∫_0 ^∞ (1/x)[Σ_(n≥0) (−1)^n e^(−2nx) (1−e^(−2x) ) [−(1/2)Σ_(n≥0) (−1)^n n(n−1)e^(−2(n−2)x) (1−e^(−2x) )^3 ]dx =2Σ_(n≥0) (−1)^n ∫_0 ^∞ x^(−1) (e^(−2nx) −e^(−2(n+1)x) )dx −Σ_(n≥0) (−1)^n (n^2 −n)∫_0 ^∞ x^(−1) (e^(−2(n−2)x) −3e^(−2(n−1)x) +3e^(−2nx) −e^(−2(n+1)x) )dx (1/((1−ax)))=Σ_(n≥0) a^n x^n (x/((1−ax)^2 ))=Σna^(n−1) x^n ((2x^2 )/((1−ax)^3 ))=Σn(n−1)a^(n−2) x^n

$${I}=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{tanh}{x}}{{x}}\right)^{\mathrm{2}} {dx}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tanh}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:=\left[−\frac{\mathrm{tanh}^{\mathrm{2}} {x}}{{x}}+\mathrm{2}\int\frac{\mathrm{tanh}{x}\left(\mathrm{1}−\mathrm{tanh}^{\mathrm{2}} {x}\right)}{{x}}{dx}\right]_{\mathrm{0}} ^{\infty} \\ $$$$\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{tanh}{x}}{{x}}−\frac{\mathrm{tanh}^{\mathrm{3}} {x}}{{x}}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{1}−{e}^{−\mathrm{2}{x}} }{\mathrm{1}+{e}^{−\mathrm{2}{x}} }−\left(\frac{\mathrm{1}−{e}^{−\mathrm{2}{x}} }{\mathrm{1}+{e}^{−\mathrm{2}{x}} }\right)^{\mathrm{3}} \right){dx} \\ $$$$\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}}\left[\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} {e}^{−\mathrm{2}{nx}} \left(\mathrm{1}−{e}^{−\mathrm{2}{x}} \right)\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left[−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} {n}\left({n}−\mathrm{1}\right){e}^{−\mathrm{2}\left({n}−\mathrm{2}\right){x}} \left(\mathrm{1}−{e}^{−\mathrm{2}{x}} \right)^{\mathrm{3}} \right]{dx} \\ $$$$\:\:=\mathrm{2}\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\infty} {x}^{−\mathrm{1}} \left({e}^{−\mathrm{2}{nx}} −{e}^{−\mathrm{2}\left({n}+\mathrm{1}\right){x}} \right){dx} \\ $$$$\:\:−\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \left({n}^{\mathrm{2}} −{n}\right)\int_{\mathrm{0}} ^{\infty} {x}^{−\mathrm{1}} \left({e}^{−\mathrm{2}\left({n}−\mathrm{2}\right){x}} −\mathrm{3}{e}^{−\mathrm{2}\left({n}−\mathrm{1}\right){x}} +\mathrm{3}{e}^{−\mathrm{2}{nx}} −{e}^{−\mathrm{2}\left({n}+\mathrm{1}\right){x}} \right){dx} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{ax}\right)}=\underset{{n}\geqslant\mathrm{0}} {\sum}{a}^{{n}} {x}^{{n}} \\ $$$$\frac{{x}}{\left(\mathrm{1}−{ax}\right)^{\mathrm{2}} }=\Sigma{na}^{{n}−\mathrm{1}} {x}^{{n}} \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left(\mathrm{1}−{ax}\right)^{\mathrm{3}} }=\Sigma{n}\left({n}−\mathrm{1}\right){a}^{{n}−\mathrm{2}} {x}^{{n}} \\ $$

Answered by Kamel last updated on 04/Sep/21

Ω =∫_0 ^(+∞) (((tanh(x))/x))^2 dx =^(t=tanh(x)) 4∫_0 ^1 (t^2 /(Ln^2 (((1−t)/(1+t)))(1−t^2 )))dt =^(IBP) −4∫_0 ^1 (t/(Ln(((1−t)/(1+t)))))dt=^(u=((1−t)/(1+t))) −8∫_0 ^1 ((1−u)/((1+u)^3 Ln(u)))du =8∫_0 ^1 ∫_0 ^1 (u^a /((1+u)^3 ))duda=−1+1−4∫_0 ^1 a(a−1)∫_0 ^1 (u^(a−1) /((1+u)))duda =2∫_0 ^1 a(1−a)(Ψ(((a+1)/2))−Ψ((a/2)))da =2∫_0 ^1 a(1−a)(Ψ(1−(a/2))−Ψ(((1−a)/2)))da ∴ Ω=π∫_0 ^1 a(1−a)(cot(((πa)/2))+tg(((πa)/2)))da =2π∫_0 ^1 ((a(1−a))/(sin(πa)))da=^(IBP) 2∫_0 ^1 (1−2a)Ln(cot(((πa)/2)))da =−4Σ_(n=0) ^(+∞) (1/((2n+1)))∫_0 ^1 (1−2a)cos((2n+1)πa)da =(8/π)Σ_(n=0) ^(+∞) (1/((2n+1)^2 ))∫_0 ^1 sin((2n+1)πa)da =((16)/π^2 )Σ_(n=0) ^(+∞) (1/((2n+1)^3 ))=((14ζ(3))/π^2 ) ∴ ∫_0 ^(+∞) (((tanh(x))/x))^2 dx=((14𝛇(3))/𝛑^2 ) BENAICHA KAMEL

$$ \\ $$$$\Omega\:\:\:=\int_{\mathrm{0}} ^{+\infty} \left(\frac{{tanh}\left({x}\right)}{{x}}\right)^{\mathrm{2}} {dx} \\ $$$$\overset{{t}={tanh}\left({x}\right)} {=}\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} }{{Ln}^{\mathrm{2}} \left(\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{dt} \\ $$$$\:\:\:\:\overset{{IBP}} {=}−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}}{{Ln}\left(\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\right)}{dt}\overset{{u}=\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}} {=}−\mathrm{8}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{u}}{\left(\mathrm{1}+{u}\right)^{\mathrm{3}} {Ln}\left({u}\right)}{du} \\ $$$$\:\:\:\:\:=\mathrm{8}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{{a}} }{\left(\mathrm{1}+{u}\right)^{\mathrm{3}} }{duda}=−\mathrm{1}+\mathrm{1}−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} {a}\left({a}−\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{{a}−\mathrm{1}} }{\left(\mathrm{1}+{u}\right)}{duda} \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {a}\left(\mathrm{1}−{a}\right)\left(\Psi\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\Psi\left(\frac{{a}}{\mathrm{2}}\right)\right){da} \\ $$$$\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {a}\left(\mathrm{1}−{a}\right)\left(\Psi\left(\mathrm{1}−\frac{{a}}{\mathrm{2}}\right)−\Psi\left(\frac{\mathrm{1}−{a}}{\mathrm{2}}\right)\right){da} \\ $$$$\therefore\:\Omega=\pi\int_{\mathrm{0}} ^{\mathrm{1}} {a}\left(\mathrm{1}−{a}\right)\left({cot}\left(\frac{\pi{a}}{\mathrm{2}}\right)+{tg}\left(\frac{\pi{a}}{\mathrm{2}}\right)\right){da} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{2}\pi\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{a}\left(\mathrm{1}−{a}\right)}{{sin}\left(\pi{a}\right)}{da}\overset{{IBP}} {=}\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{2}{a}\right){Ln}\left({cot}\left(\frac{\pi{a}}{\mathrm{2}}\right)\right){da} \\ $$$$\:\:\:\:\:\:\:\:=−\mathrm{4}\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{2}{a}\right){cos}\left(\left(\mathrm{2}{n}+\mathrm{1}\right)\pi{a}\right){da} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{8}}{\pi}\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left(\left(\mathrm{2}{n}+\mathrm{1}\right)\pi{a}\right){da} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{16}}{\pi^{\mathrm{2}} }\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{14}\zeta\left(\mathrm{3}\right)}{\pi^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{+\infty} \left(\frac{\boldsymbol{{tanh}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{x}}}\right)^{\mathrm{2}} \boldsymbol{{dx}}=\frac{\mathrm{14}\boldsymbol{\zeta}\left(\mathrm{3}\right)}{\boldsymbol{\pi}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{BENAICHA}}\:\boldsymbol{{KAMEL}} \\ $$

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