Question Number 222453 by Nicholas666 last updated on 27/Jun/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{tanh}^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} }\:{dx} \\ $$$$ \\ $$
Answered by MrGaster last updated on 28/Jun/25
![tanh x=−2Σ_(k=1) ^∞ (−1)^(k−1) e^(−2kx) ,x>0 tanh^2 x=(1−2Σ_(k=1) ^∞ (−1)^(k−1) e^(−2kx) )^2 =1+4Σ_(m=1) ^∞ m(−1)^m e^(−2mx) ∫_0 ^∞ ((tanh^2 x)/x^2 )dx=∫_0 ^∞ (1/x^2 )(1+4Σ_(m=1) ^∞ m(−1)^m e^(−2mx) )dx lim_(ε→0^+ ) ∫_ε ^∞ ((tanh^2 x)/x^2 )dx=lim_(ε→0^+ ) [∫_ε ^∞ (1/x^2 )dx+4Σ_(m=1) ^∞ m(−1)^m ∫_ε ^∞ (e^(−2mx) /x^2 )dx] ∫_ε ^∞ (e^(−2mx) /x^2 )dx=(e^(−2mε) /ε)−2m E_1 (2mc) E_1 (z)∼γ−ln z,z→0^+ Σ_(m=1) ^∞ (−1)^m =η(−2)=(1−2^(1−(2)) )ζ(−2)=(1−2^3 )ζ(−2)=(−7)∙0=0 Σ_(m=0) ^∞ (−1)^m m^2 ln m=η′(−2) η(s)=(1−2^(1−s) )ζ(s) η′(s)=(1−2^(1−s) )′ζ(s)+(1−2^(1−s) )ζ′(s)=(2^(1−s) ln 2)ζ(s)+(1−2^(1−s) )ζ′(s) s=−2: ζ(−2)=0,1−2^(1−(−2)) =1−2^3 =−7 η(−2)=−7ζ′(−2) ζ(s)=2^8 π^(s−1) sin(((πs)/2))Γ(1−s)ζ(1−s) g(s)−2^s π^(s−1) Γ(1−s)ζ(1−s) g(−2)=2^(−2) π^(−3) Γ(3)ζ(3)=(1/4)π^(−3) ∙2∙ζ(3)=(1/2)π^(−3) ζ(3) g′(s)sin(((πs)/2))+g(s)cos(((πs)/2))((π/2)) s=−2: sin(((π(−2))/2))=sin(−π)=0,cos(((π(−2))/2))=cos(−π)=−1 ζ′(−2)=g′(−2)sin(0)+g(−2)cos(−π)(π/2)=0+((1/2)π^(−3) ζ(3))∙(−1)∙(π/2)=−(1/4)π^(−2) ζ(3) ∫_ε ^∞ ((tanh^2 x)/x^2 )dx=((tanh^2 x)/ε)−8Σ_(m=1) ^∞ m^2 (−1)^m E_1 (2mε) ((tanh^2 ε)/ε)∼ε→0,ε→0^+ −8Σ_(m=1) ^∞ m^2 (−1)^m E_1 (2mε)∼8Σ_(m=1) ^∞ m^2 (−1)^m (−γ−ln(2mε))=8η′(−2)+const.∙0 η^′ (−2)=−7ζ′(−2)=−7(−(1/4)π^(−2) ζ(3))=(7/4)π^(−2) ζ(3) 8η′(−2)=8∙(7/4)π^(−2) ζ(3)=14π^(−2) ζ(3) ((14ζ(3))/π^2 )](Q222474.png)
$$\mathrm{tanh}\:{x}=−\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {e}^{−\mathrm{2}{kx}} ,{x}>\mathrm{0} \\ $$$$\mathrm{tanh}^{\mathrm{2}} {x}=\left(\mathrm{1}−\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {e}^{−\mathrm{2}{kx}} \right)^{\mathrm{2}} =\mathrm{1}+\mathrm{4}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}{m}\left(−\mathrm{1}\right)^{{m}} {e}^{−\mathrm{2}{mx}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tanh}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{4}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}{m}\left(−\mathrm{1}\right)^{{m}} {e}^{−\mathrm{2}{mx}} \right){dx} \\ $$$$\underset{\epsilon\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\int_{\epsilon} ^{\infty} \frac{\mathrm{tanh}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}=\underset{\epsilon\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left[\int_{\epsilon} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx}+\mathrm{4}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}{m}\left(−\mathrm{1}\right)^{{m}} \int_{\epsilon} ^{\infty} \frac{{e}^{−\mathrm{2}{mx}} }{{x}^{\mathrm{2}} }{dx}\right] \\ $$$$\int_{\epsilon} ^{\infty} \frac{{e}^{−\mathrm{2}{mx}} }{{x}^{\mathrm{2}} }{dx}=\frac{{e}^{−\mathrm{2}{m}\epsilon} }{\epsilon}−\mathrm{2}{m}\:{E}_{\mathrm{1}} \left(\mathrm{2}{mc}\right) \\ $$$${E}_{\mathrm{1}} \left({z}\right)\sim\gamma−\mathrm{ln}\:{z},{z}\rightarrow\mathrm{0}^{+} \\ $$$$\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{m}} =\eta\left(−\mathrm{2}\right)=\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−\left(\mathrm{2}\right)} \right)\zeta\left(−\mathrm{2}\right)=\left(\mathrm{1}−\mathrm{2}^{\mathrm{3}} \right)\zeta\left(−\mathrm{2}\right)=\left(−\mathrm{7}\right)\centerdot\mathrm{0}=\mathrm{0} \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{m}} {m}^{\mathrm{2}} \mathrm{ln}\:{m}=\eta'\left(−\mathrm{2}\right) \\ $$$$\eta\left({s}\right)=\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−{s}} \right)\zeta\left({s}\right) \\ $$$$\eta'\left({s}\right)=\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−{s}} \right)'\zeta\left({s}\right)+\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−{s}} \right)\zeta'\left({s}\right)=\left(\mathrm{2}^{\mathrm{1}−{s}} \mathrm{ln}\:\mathrm{2}\right)\zeta\left({s}\right)+\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−{s}} \right)\zeta'\left({s}\right) \\ $$$${s}=−\mathrm{2}: \\ $$$$\zeta\left(−\mathrm{2}\right)=\mathrm{0},\mathrm{1}−\mathrm{2}^{\mathrm{1}−\left(−\mathrm{2}\right)} =\mathrm{1}−\mathrm{2}^{\mathrm{3}} =−\mathrm{7} \\ $$$$\eta\left(−\mathrm{2}\right)=−\mathrm{7}\zeta'\left(−\mathrm{2}\right) \\ $$$$\zeta\left({s}\right)=\mathrm{2}^{\mathrm{8}} \pi^{{s}−\mathrm{1}} \mathrm{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−{s}\right)\zeta\left(\mathrm{1}−{s}\right) \\ $$$${g}\left({s}\right)−\mathrm{2}^{{s}} \pi^{{s}−\mathrm{1}} \Gamma\left(\mathrm{1}−{s}\right)\zeta\left(\mathrm{1}−{s}\right) \\ $$$${g}\left(−\mathrm{2}\right)=\mathrm{2}^{−\mathrm{2}} \pi^{−\mathrm{3}} \Gamma\left(\mathrm{3}\right)\zeta\left(\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{4}}\pi^{−\mathrm{3}} \centerdot\mathrm{2}\centerdot\zeta\left(\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{2}}\pi^{−\mathrm{3}} \zeta\left(\mathrm{3}\right) \\ $$$${g}'\left({s}\right)\mathrm{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)+{g}\left({s}\right)\mathrm{cos}\left(\frac{\pi{s}}{\mathrm{2}}\right)\left(\frac{\pi}{\mathrm{2}}\right) \\ $$$${s}=−\mathrm{2}: \\ $$$$\mathrm{sin}\left(\frac{\pi\left(−\mathrm{2}\right)}{\mathrm{2}}\right)=\mathrm{sin}\left(−\pi\right)=\mathrm{0},\mathrm{cos}\left(\frac{\pi\left(−\mathrm{2}\right)}{\mathrm{2}}\right)=\mathrm{cos}\left(−\pi\right)=−\mathrm{1} \\ $$$$\zeta'\left(−\mathrm{2}\right)={g}'\left(−\mathrm{2}\right)\mathrm{sin}\left(\mathrm{0}\right)+{g}\left(−\mathrm{2}\right)\mathrm{cos}\left(−\pi\right)\frac{\pi}{\mathrm{2}}=\mathrm{0}+\left(\frac{\mathrm{1}}{\mathrm{2}}\pi^{−\mathrm{3}} \zeta\left(\mathrm{3}\right)\right)\centerdot\left(−\mathrm{1}\right)\centerdot\frac{\pi}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{4}}\pi^{−\mathrm{2}} \zeta\left(\mathrm{3}\right) \\ $$$$\int_{\epsilon} ^{\infty} \frac{\mathrm{tanh}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{tanh}^{\mathrm{2}} {x}}{\epsilon}−\mathrm{8}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}{m}^{\mathrm{2}} \left(−\mathrm{1}\right)^{{m}} {E}_{\mathrm{1}} \left(\mathrm{2}{m}\epsilon\right) \\ $$$$\frac{\mathrm{tanh}^{\mathrm{2}} \epsilon}{\epsilon}\sim\epsilon\rightarrow\mathrm{0},\epsilon\rightarrow\mathrm{0}^{+} \\ $$$$−\mathrm{8}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}{m}^{\mathrm{2}} \left(−\mathrm{1}\right)^{{m}} {E}_{\mathrm{1}} \left(\mathrm{2}{m}\epsilon\right)\sim\mathrm{8}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}{m}^{\mathrm{2}} \left(−\mathrm{1}\right)^{{m}} \left(−\gamma−\mathrm{ln}\left(\mathrm{2}{m}\epsilon\right)\right)=\mathrm{8}\eta'\left(−\mathrm{2}\right)+\mathrm{const}.\centerdot\mathrm{0} \\ $$$$\eta^{'} \left(−\mathrm{2}\right)=−\mathrm{7}\zeta'\left(−\mathrm{2}\right)=−\mathrm{7}\left(−\frac{\mathrm{1}}{\mathrm{4}}\pi^{−\mathrm{2}} \zeta\left(\mathrm{3}\right)\right)=\frac{\mathrm{7}}{\mathrm{4}}\pi^{−\mathrm{2}} \zeta\left(\mathrm{3}\right) \\ $$$$\mathrm{8}\eta'\left(−\mathrm{2}\right)=\mathrm{8}\centerdot\frac{\mathrm{7}}{\mathrm{4}}\pi^{−\mathrm{2}} \zeta\left(\mathrm{3}\right)=\mathrm{14}\pi^{−\mathrm{2}} \zeta\left(\mathrm{3}\right) \\ $$$$\frac{\mathrm{14}\zeta\left(\mathrm{3}\right)}{\pi^{\mathrm{2}} } \\ $$
Commented by Nicholas666 last updated on 28/Jun/25
$$\mathrm{you}\:\mathrm{are}\:\mathrm{really}\:\mathrm{a}\:\mathrm{great}\:\mathrm{mathematicians}\: \\ $$