Question Number 174024 by Michaelfaraday last updated on 23/Jul/22
Answered by behi834171 last updated on 23/Jul/22
![(1/((a^2 +t^2 )(b^2 +t^2 )))=(1/(a^2 −b^2 )).(((a^2 +t^2 )−(b^2 +t^2 ))/((a^2 +t^2 )(b^2 +t^2 )))⇒ I=(1/(a^2 −b^2 ))∫_0 ^∞ (((a^2 +t^2 )−(b^2 +t^2 ))/((a^2 +t^2 )(b^2 +t^2 )))dt= =(1/(a^2 −b^2 ))[∫_( 0) ^( ∞) (1/(b^2 +t^2 ))dt−∫_( 0) ^( ∞) (1/(a^2 +t^2 ))dt]= =(1/(a^2 −b^2 ))[(1/b)tg^(−1) ((t/b))−(1/a)tg^(−1) ((t/a))]_0 ^( ∞) = =(1/(a^2 −b^2 ))[(π/(2b))−(π/(2a))]=(π/(2ab)).(1/(a^2 −b^2 )).(a−b)= =(𝛑/(2ab(a+b))) .■](https://www.tinkutara.com/question/Q174027.png)
$$\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }.\frac{\left({a}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)−\left({b}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)}\Rightarrow \\ $$$$\boldsymbol{\mathrm{I}}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\:\frac{\left({a}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)−\left({b}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)}\mathrm{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }\left[\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\frac{\mathrm{1}}{{b}^{\mathrm{2}} +{t}^{\mathrm{2}} }{dt}−\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{t}^{\mathrm{2}} }{dt}\right]= \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\left[\frac{\mathrm{1}}{{b}}{tg}^{−\mathrm{1}} \left(\frac{{t}}{{b}}\right)−\frac{\mathrm{1}}{{a}}{tg}^{−\mathrm{1}} \left(\frac{{t}}{{a}}\right)\underset{\mathrm{0}} {\overset{\:\infty} {\right]}}= \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\left[\frac{\pi}{\mathrm{2}{b}}−\frac{\pi}{\mathrm{2}{a}}\right]=\frac{\pi}{\mathrm{2}{ab}}.\frac{\mathrm{1}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }.\left({a}−{b}\right)= \\ $$$$=\frac{\boldsymbol{\pi}}{\mathrm{2}\boldsymbol{\mathrm{ab}}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)}\:\:\:\:\:\:.\blacksquare \\ $$
Commented by Tawa11 last updated on 23/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by peter frank last updated on 24/Jul/22
$$\mathrm{thanks} \\ $$
Answered by CElcedricjunior last updated on 25/Jul/22
![∫_0 ^∞ (dt/((a^2 +t^2 )(b^2 +t^2 )))=k decomposition en element simple (1/((a^2 +t^2 )(b^2 +t^2 )))=(c/(a^2 +t^2 )) +(d/(b^2 +t^2 )) ((t^2 (c+d)+a^2 d+b^2 c)/((a^2 +t^2 )(b^2 +t^2 ))) par identification { ((c+d=0)),((a^2 d+b^2 c=1)) :}=> { ((d=−c)),((c=(1/(b^2 −a^2 )))) :} d=(1/(a^2 −b^2 )) k=∫_0 ^∞ (((1/(b^2 −a^2 ))/(a^2 +t^2 ))+((1/(a^2 −b))/(b^2 +t^2 )))dt k=(1/(a(b^2 −a^2 )))[arctan((t/a))]_0 ^(+∞) +(1/(b(a^2 −b^2 )))[arctan((t/b))]_0 ^(+∞) k=(𝛑/(2a(b^2 −a^2 )))+(𝛑/(2b(a^2 −b^2 ))) k=(𝛑/(2(b^2 −a^2 )))((1/a)−(1/b)) k=(𝛑/(2(b^2 −a^2 )))(((b−a)/(ab)))=((𝛑(b−a))/(2ab(b−a)(b+a))) k=(𝛑/(2ab(a+b))) ............le celebre cedric junior...........](https://www.tinkutara.com/question/Q174126.png)
$$\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{dt}}}{\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} \right)\left(\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} \right)}=\boldsymbol{\mathrm{k}} \\ $$$$\boldsymbol{\mathrm{decomposition}}\:\boldsymbol{\mathrm{en}}\:\boldsymbol{\mathrm{element}}\:\boldsymbol{\mathrm{simple}}\: \\ $$$$\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} \right)\left(\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} \right)}=\frac{\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} }\:+\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} } \\ $$$$\frac{\boldsymbol{\mathrm{t}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{d}}\right)+\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{d}}+\boldsymbol{\mathrm{b}}^{\mathrm{2}} \boldsymbol{\mathrm{c}}}{\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} \right)\left(\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} \right)} \\ $$$$\boldsymbol{\mathrm{par}}\:\boldsymbol{\mathrm{identification}} \\ $$$$\begin{cases}{\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{d}}=\mathrm{0}}\\{\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{d}}+\boldsymbol{\mathrm{b}}^{\mathrm{2}} \boldsymbol{\mathrm{c}}=\mathrm{1}}\end{cases}=>\begin{cases}{\boldsymbol{\mathrm{d}}=−\boldsymbol{\mathrm{c}}}\\{\boldsymbol{\mathrm{c}}=\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}^{\mathrm{2}} }}\end{cases} \\ $$$$\boldsymbol{\mathrm{d}}=\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} } \\ $$$$\boldsymbol{\mathrm{k}}=\int_{\mathrm{0}} ^{\infty} \left(\frac{\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}^{\mathrm{2}} }}{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} }+\frac{\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}}}{\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} }\right)\boldsymbol{\mathrm{dt}} \\ $$$$\boldsymbol{\mathrm{k}}=\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}\left(\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)}\left[\boldsymbol{\mathrm{arctan}}\left(\frac{\boldsymbol{\mathrm{t}}}{\boldsymbol{\mathrm{a}}}\right)\right]_{\mathrm{0}} ^{+\infty} +\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} \right)}\left[\boldsymbol{\mathrm{arctan}}\left(\frac{\boldsymbol{\mathrm{t}}}{\boldsymbol{\mathrm{b}}}\right)\right]_{\mathrm{0}} ^{+\infty} \\ $$$$\boldsymbol{\mathrm{k}}=\frac{\boldsymbol{\pi}}{\mathrm{2}\boldsymbol{{a}}\left(\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right)}+\frac{\boldsymbol{\pi}}{\mathrm{2}\boldsymbol{\mathrm{b}}\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} \right)} \\ $$$$\boldsymbol{\mathrm{k}}=\frac{\boldsymbol{\pi}}{\mathrm{2}\left(\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}}\right) \\ $$$$\boldsymbol{\mathrm{k}}=\frac{\boldsymbol{\pi}}{\mathrm{2}\left(\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)}\left(\frac{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{ab}}}\right)=\frac{\boldsymbol{\pi}\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{a}}\right)}{\mathrm{2}\boldsymbol{\mathrm{ab}}\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{a}}\right)\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}\right)} \\ $$$$\boldsymbol{\mathrm{k}}=\frac{\boldsymbol{\pi}}{\mathrm{2}\boldsymbol{\mathrm{ab}}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)} \\ $$$$ \\ $$$$ \\ $$$$\:…………{le}\:{celebre}\:{cedric}\:{junior}……….. \\ $$