Question Number 31097 by abdo imad last updated on 02/Mar/18
$${calculate}\:{interms}\:{of}\:{a}\:{and}\:{b}\:{the}\:{integral} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({bt}\right)\:−{arctan}\left({at}\right)}{{t}}{dt}\:\:{with}\:{a}\:{and}\:{b}>\mathrm{0}. \\ $$
Commented byabdo imad last updated on 05/Mar/18
![let put I=∫_0 ^∞ ((arctan(bt) −arctan(at))/t)dt and for ξ>0 I(ξ)= ∫_0 ^ξ ((arctan(bt)−arctan(at))/t)dt we have I=lim_(ξ→+∞) I(ξ) but I(ξ)=∫_0 ^ξ ((arctan(bt))/t)dt −∫_0 ^ξ ((arctan(at))/t)dt but ∫_0 ^ξ ((arctan(bt))/t)dt=_(bt=x) ∫_0 ^(bξ) ((arctanx)/x)dx ∫_0 ^ξ ((artan(at))/t)dt= _(at=x) ∫_0 ^(aξ) ((arctanx)/x)dx ⇒ I(ξ)=∫_0 ^(bξ) ((arctanx)/x)dx −∫_0 ^(aξ) ((arctanx)/x)dx=∫_(aξ) ^(bξ) ((arctanx)/x)dx ∃ c∈]aξ,bξ[ / I(ξ)=arctanξ ∫_(aξ) ^(bξ) (dx/x)=ln((b/a))arctanξ ⇒ lim _(ξ→+∞) I(ξ)=(π/2)ln((b/a)) ⇒ I=(π/2)(ln(b)−ln(a)) .](Q31308.png)
$${let}\:{put}\:{I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({bt}\right)\:−{arctan}\left({at}\right)}{{t}}{dt}\:{and}\:{for}\:\xi>\mathrm{0} \\ $$ $${I}\left(\xi\right)=\:\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left({bt}\right)−{arctan}\left({at}\right)}{{t}}{dt}\:{we}\:{have} \\ $$ $${I}={lim}_{\xi\rightarrow+\infty} \:{I}\left(\xi\right)\:{but}\:{I}\left(\xi\right)=\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left({bt}\right)}{{t}}{dt}\:−\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left({at}\right)}{{t}}{dt} \\ $$ $${but}\:\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left({bt}\right)}{{t}}{dt}=_{{bt}={x}} \:\int_{\mathrm{0}} ^{{b}\xi} \:\frac{{arctanx}}{{x}}{dx} \\ $$ $$\int_{\mathrm{0}} ^{\xi} \:\frac{{artan}\left({at}\right)}{{t}}{dt}=\:_{{at}={x}} \int_{\mathrm{0}} ^{{a}\xi} \:\frac{{arctanx}}{{x}}{dx}\:\Rightarrow \\ $$ $${I}\left(\xi\right)=\int_{\mathrm{0}} ^{{b}\xi} \:\frac{{arctanx}}{{x}}{dx}\:−\int_{\mathrm{0}} ^{{a}\xi} \:\frac{{arctanx}}{{x}}{dx}=\int_{{a}\xi} ^{{b}\xi} \:\:\frac{{arctanx}}{{x}}{dx} \\ $$ $$\left.\exists\:{c}\in\right]{a}\xi,{b}\xi\left[\:/\:{I}\left(\xi\right)={arctan}\xi\:\int_{{a}\xi} ^{{b}\xi} \:\frac{{dx}}{{x}}={ln}\left(\frac{{b}}{{a}}\right){arctan}\xi\:\Rightarrow\right. \\ $$ $${lim}\:_{\xi\rightarrow+\infty} {I}\left(\xi\right)=\frac{\pi}{\mathrm{2}}{ln}\left(\frac{{b}}{{a}}\right)\:\Rightarrow\:{I}=\frac{\pi}{\mathrm{2}}\left({ln}\left({b}\right)−{ln}\left({a}\right)\right)\:. \\ $$