Algebra Questions

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

$${f}\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:{dx}\:\:\:\left(\mathrm{1}\right) \\$$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{ax}^{\mathrm{2}} }{\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)}\:{dx} \\$$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left(\frac{{a}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{b}^{\mathrm{2}} +{x}^{\mathrm{2}} }\right)\:{dx} \\$$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}}\left[{a}.\mathrm{arctan}\left({ax}\right)−\frac{\mathrm{1}}{{b}}\mathrm{arctan}\left(\frac{{x}}{{b}}\right)\right]_{\mathrm{0}} ^{\infty} \\$$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}}.\frac{\pi}{\mathrm{2}}\left({a}−\frac{\mathrm{1}}{{b}}\right) \\$$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\pi\left({a}^{\mathrm{2}} −\frac{{a}}{{b}}\right)}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}} \\$$$$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\pi}{{b}^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}{b}^{\mathrm{3}} }.\frac{\mathrm{2}{ab}^{\mathrm{2}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}}+\frac{\pi}{{b}^{\mathrm{2}} }.\frac{\mathrm{1}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}} \\$$$${f}\left({a},{b}\right)\:=\:\frac{\pi{a}}{{b}^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}{b}^{\mathrm{3}} }\mathrm{ln}\mid{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}\mid+\frac{\pi}{\mathrm{2}{b}^{\mathrm{3}} }\mathrm{ln}\mid\frac{{ab}−\mathrm{1}}{{ab}+\mathrm{1}}\mid+\mathrm{C}\left({b}\right)\:\:\left(\mathrm{2}\right) \\$$$$\mathrm{with}\:\left(\mathrm{1}\right)\::\:{f}\left(\mathrm{0},\mathrm{1}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{0}{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\:\mathrm{0} \\$$$$\mathrm{with}\:\left(\mathrm{2}\right)\::\:{f}\left(\mathrm{0},\mathrm{1}\right)\:=\:\mathrm{C}\left({b}\right) \\$$$$\Rightarrow\:\mathrm{C}\left({b}\right)\:=\:\mathrm{0} \\$$$$\\$$$${f}\left({a},{b}\right)\:=\:\frac{\pi{a}}{{b}^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}{b}^{\mathrm{3}} }\mathrm{ln}\mid{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{1}\mid+\frac{\pi}{\mathrm{2}{b}^{\mathrm{3}} }\mathrm{ln}\mid\frac{{ab}−\mathrm{1}}{{ab}+\mathrm{1}}\mid \\$$$${f}\left({a},{b}\right)\:=\:\frac{\pi{a}}{{b}^{\mathrm{2}} }−\frac{\pi}{{b}^{\mathrm{3}} }\mathrm{ln}\mid{ab}+\mathrm{1}\mid \\$$

Commented by mathdanisur last updated on 23/Aug/21

$$\mathrm{cool}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{Ser} \\$$