Question Number 32339 by abdo imad last updated on 23/Mar/18
$${calculate}\:\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{th}\left(\mathrm{3}{x}\right)\:−{th}\left(\mathrm{2}{x}\right)}{{x}}\:{dx}\:. \\ $$
Commented by abdo imad last updated on 24/Mar/18
![I =lim _(ξ→+∞) I(ξ) with I(ξ) = ∫_0 ^(ξ ) ((th(3x)−th(2x))/x)dx I(ξ) = ∫_0 ^ξ ((th(3x))/x) dx −∫_0 ^ξ ((th(2x))/x)dx but ch.3x=t give ∫_0 ^ξ ((th(3x))/x)dx = ∫_0 ^(3ξ) ((th(t))/(t/3)) (dt/3) =∫_0 ^(3ξ) ((th(t))/t)dt also we have ∫_0 ^ξ ((th(2x))/x)dx = ∫_0 ^(2ξ) ((th(t))/t)dt ⇒ I(ξ) = ∫_0 ^(3ξ) ((th(t))/t)dt −∫_0 ^(2ξ) ((th(t))/t)dt =∫_(2ξ) ^(3ξ) ((th(t))/t)dt but ∃ c ∈]2ξ,3ξ[ / I(ξ) =th(c) ∫_(2ξ) ^(3ξ) (dt/t)=ln((3/2))th(c) ⇒ lim_(ξ→+∞) I(ξ) =ln(3)−ln(2) .( look that lim_(c→+∞) thc=1)](Q32434.png)
$${I}\:={lim}\:_{\xi\rightarrow+\infty} {I}\left(\xi\right)\:\:{with}\:{I}\left(\xi\right)\:=\:\int_{\mathrm{0}} ^{\xi\:\:} \:\:\frac{{th}\left(\mathrm{3}{x}\right)−{th}\left(\mathrm{2}{x}\right)}{{x}}{dx} \\ $$$${I}\left(\xi\right)\:=\:\int_{\mathrm{0}} ^{\xi} \:\:\:\frac{{th}\left(\mathrm{3}{x}\right)}{{x}}\:{dx}\:−\int_{\mathrm{0}} ^{\xi} \:\frac{{th}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:\:{but}\:{ch}.\mathrm{3}{x}={t}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\xi} \:\:\frac{{th}\left(\mathrm{3}{x}\right)}{{x}}{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{3}\xi} \:\:\frac{{th}\left({t}\right)}{\frac{{t}}{\mathrm{3}}}\:\frac{{dt}}{\mathrm{3}}\:=\int_{\mathrm{0}} ^{\mathrm{3}\xi} \:\:\frac{{th}\left({t}\right)}{{t}}{dt}\:{also}\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\xi} \:\frac{{th}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\xi} \:\:\frac{{th}\left({t}\right)}{{t}}{dt}\:\Rightarrow \\ $$$${I}\left(\xi\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{3}\xi} \:\frac{{th}\left({t}\right)}{{t}}{dt}\:−\int_{\mathrm{0}} ^{\mathrm{2}\xi} \:\:\frac{{th}\left({t}\right)}{{t}}{dt}\:=\int_{\mathrm{2}\xi} ^{\mathrm{3}\xi} \:\:\:\frac{{th}\left({t}\right)}{{t}}{dt}\:{but} \\ $$$$\left.\exists\:{c}\:\in\right]\mathrm{2}\xi,\mathrm{3}\xi\left[\:/\:{I}\left(\xi\right)\:={th}\left({c}\right)\:\int_{\mathrm{2}\xi} ^{\mathrm{3}\xi} \:\:\frac{{dt}}{{t}}={ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right){th}\left({c}\right)\:\:\Rightarrow\right. \\ $$$${lim}_{\xi\rightarrow+\infty} \:{I}\left(\xi\right)\:={ln}\left(\mathrm{3}\right)−{ln}\left(\mathrm{2}\right)\:.\left(\:{look}\:{that}\:{lim}_{{c}\rightarrow+\infty} {thc}=\mathrm{1}\right) \\ $$$$ \\ $$