Question Number 151768 by mathdanisur last updated on 22/Aug/21
$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{ln}\left(\mathrm{1}\:+\:\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \right)}{\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 22/Aug/21
![f(a,b) = ∫_0 ^∞ ((ln(1+a^2 x^2 ))/(b^2 +x^2 )) dx (1) (∂f/∂a)(a,b) = ∫_0 ^∞ ((2ax^2 )/((1+a^2 x^2 )(b^2 +x^2 ))) dx (∂f/∂a)(a,b) = ((2a)/(a^2 b^2 −1))∫_0 ^∞ ((a^2 /(1+a^2 x^2 ))−(1/(b^2 +x^2 ))) dx (∂f/∂a)(a,b) = ((2a)/(a^2 b^2 −1))[a.arctan(ax)−(1/b)arctan((x/b))]_0 ^∞ (∂f/∂a)(a,b) = ((2a)/(a^2 b^2 −1)).(π/2)(a−(1/b)) (∂f/∂a)(a,b) = ((π(a^2 −(a/b)))/(a^2 b^2 −1)) (∂f/∂a)(a,b) = (π/b^2 )−(π/(2b^3 )).((2ab^2 )/(a^2 b^2 −1))+(π/b^2 ).(1/(a^2 b^2 −1)) f(a,b) = ((πa)/b^2 )−(π/(2b^3 ))ln∣a^2 b^2 −1∣+(π/(2b^3 ))ln∣((ab−1)/(ab+1))∣+C(b) (2) with (1) : f(0,1) = ∫_0 ^∞ ((ln(1+0x^2 ))/(1+x^2 ))dx = 0 with (2) : f(0,1) = C(b) ⇒ C(b) = 0 f(a,b) = ((πa)/b^2 )−(π/(2b^3 ))ln∣a^2 b^2 −1∣+(π/(2b^3 ))ln∣((ab−1)/(ab+1))∣ f(a,b) = ((πa)/b^2 )−(π/b^3 )ln∣ab+1∣](Q151775.png)
$${f}\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:{dx}\:\:\:\left(\mathrm{1}\right) \\ $$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{ax}^{\mathrm{2}} }{\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)}\:{dx} \\ $$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left(\frac{{a}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{b}^{\mathrm{2}} +{x}^{\mathrm{2}} }\right)\:{dx} \\ $$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}}\left[{a}.\mathrm{arctan}\left({ax}\right)−\frac{\mathrm{1}}{{b}}\mathrm{arctan}\left(\frac{{x}}{{b}}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}}.\frac{\pi}{\mathrm{2}}\left({a}−\frac{\mathrm{1}}{{b}}\right) \\ $$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\pi\left({a}^{\mathrm{2}} −\frac{{a}}{{b}}\right)}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\pi}{{b}^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}{b}^{\mathrm{3}} }.\frac{\mathrm{2}{ab}^{\mathrm{2}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}}+\frac{\pi}{{b}^{\mathrm{2}} }.\frac{\mathrm{1}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}} \\ $$$${f}\left({a},{b}\right)\:=\:\frac{\pi{a}}{{b}^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}{b}^{\mathrm{3}} }\mathrm{ln}\mid{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}\mid+\frac{\pi}{\mathrm{2}{b}^{\mathrm{3}} }\mathrm{ln}\mid\frac{{ab}−\mathrm{1}}{{ab}+\mathrm{1}}\mid+\mathrm{C}\left({b}\right)\:\:\left(\mathrm{2}\right) \\ $$$$\mathrm{with}\:\left(\mathrm{1}\right)\::\:{f}\left(\mathrm{0},\mathrm{1}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{0}{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\:\mathrm{0} \\ $$$$\mathrm{with}\:\left(\mathrm{2}\right)\::\:{f}\left(\mathrm{0},\mathrm{1}\right)\:=\:\mathrm{C}\left({b}\right) \\ $$$$\Rightarrow\:\mathrm{C}\left({b}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$$${f}\left({a},{b}\right)\:=\:\frac{\pi{a}}{{b}^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}{b}^{\mathrm{3}} }\mathrm{ln}\mid{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}\mid+\frac{\pi}{\mathrm{2}{b}^{\mathrm{3}} }\mathrm{ln}\mid\frac{{ab}−\mathrm{1}}{{ab}+\mathrm{1}}\mid \\ $$$${f}\left({a},{b}\right)\:=\:\frac{\pi{a}}{{b}^{\mathrm{2}} }−\frac{\pi}{{b}^{\mathrm{3}} }\mathrm{ln}\mid{ab}+\mathrm{1}\mid \\ $$
Commented by mathdanisur last updated on 23/Aug/21
$$\mathrm{cool}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$