Question Number 184925 by cortano1 last updated on 14/Jan/23
Commented by Frix last updated on 14/Jan/23
$$\mathrm{I}\:\mathrm{think}\:\mathrm{that} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{tan}^{−\mathrm{1}} \:{ax}\:−\mathrm{tan}^{−\mathrm{1}} \:{bx}}{{x}}{dx}=\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\frac{{a}}{{b}} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 14/Jan/23
![I=∫_0 ^∞ ((tan^(−1) (ax)−tan^(−1) (bx))/x)dx =∫_0 ^∞ ((tan^(−1) (ax))/x)dx−∫_0 ^∞ ((tan^(−1) (bx))/x)dx=I(a)−I(b) ⇒I ′(a)−I ′(b)=∫_0 ^∞ (dx/(1+(ax)^2 ))−∫_0 ^∞ (dx/(1+(bx)^2 )) =(1/a)[tan^(−1) (ax)]_0 ^∞ −(1/b)[tan^(−1) (bx)]_0 ^∞ =(π/(2a))−(π/(2b)) ⇒I(a)−I(b)=(π/2)lna−(π/2)lnb+C I(1)−I(1)=0=C ⇒I=I(a)−I(b)=(π/2)ln((a/b)) ∫_0 ^∞ ((tan^(−1) (πx)−tan^(−1) (2x))/x)dx=I(π)−I(2)=(π/2)ln((π/2))★](Q184930.png)
$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left({ax}\right)−\mathrm{tan}^{−\mathrm{1}} \left({bx}\right)}{{x}}{dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left({ax}\right)}{{x}}{dx}−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left({bx}\right)}{{x}}{dx}={I}\left({a}\right)−{I}\left({b}\right) \\ $$$$\Rightarrow{I}\:'\left({a}\right)−{I}\:'\left({b}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+\left({ax}\right)^{\mathrm{2}} }−\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+\left({bx}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{a}}\left[\mathrm{tan}^{−\mathrm{1}} \left({ax}\right)\right]_{\mathrm{0}} ^{\infty} −\frac{\mathrm{1}}{{b}}\left[\mathrm{tan}^{−\mathrm{1}} \left({bx}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{2}{a}}−\frac{\pi}{\mathrm{2}{b}}\:\Rightarrow{I}\left({a}\right)−{I}\left({b}\right)=\frac{\pi}{\mathrm{2}}\mathrm{ln}{a}−\frac{\pi}{\mathrm{2}}\mathrm{ln}{b}+{C} \\ $$$${I}\left(\mathrm{1}\right)−{I}\left(\mathrm{1}\right)=\mathrm{0}={C}\:\Rightarrow{I}={I}\left({a}\right)−{I}\left({b}\right)=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\frac{{a}}{{b}}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left(\pi{x}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{{x}}{dx}={I}\left(\pi\right)−{I}\left(\mathrm{2}\right)=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\frac{\pi}{\mathrm{2}}\right)\bigstar \\ $$