Question Number 30773 by abdo imad last updated on 25/Feb/18
$${let}\:{a}>\mathrm{0}\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$ $$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}. \\ $$
Commented byabdo imad last updated on 26/Feb/18
![1)the ch. x=a tant give I=∫_0 ^∞ (dx/((x^2 +a^2 )^2 )) = ∫_0 ^(π/2) (1/(a^4 (1+tan^2 t)^2 )) a(1+tan^2 t)dt =(1/a^3 ) ∫_0 ^(π/2) (dt/(1+tan^2 t))⇒a^3 I= ∫_0 ^(π/2) cos^2 tdt = ∫_0 ^(π/2) ((1+cos(2t))/2)dt=(π/4) +[(1/4) sin(2t)]_0 ^(π/2) =(π/4) ⇒ I=(π/(4 a^3 )) 2) let put J= ∫_(−∞) ^(+∞) (x^2 /((x^2 +a^2 )^2 ))dx J= 2∫_0 ^∞ ((x^2 +a^(2 ) −a^2 )/((x^2 +a^2 )^2 ))dx =2∫_0 ^∞ (dx/((x^2 +a^2 )))dx −2a^2 ∫_0 ^∞ (dx/((x^2 +a^2 )^2 ))dx ch x=at ⇒ ∫_0 ^∞ (dx/((x^2 +a^2 )))= ∫_0 ^∞ ((adt)/(a^2 (1+t^2 )))=(1/a)(π/2)= (π/(2a)) J=(π/a) −2a^2 I = (π/a) −2a^2 (π/(4a^3 ))= (π/a) −(π/(2a))⇒ J=(π/(2a)) .](Q30847.png)
$$\left.\mathrm{1}\right){the}\:{ch}.\:{x}={a}\:{tant}\:{give} \\ $$ $${I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{1}}{{a}^{\mathrm{4}} \left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }\:{a}\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$ $$=\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}\Rightarrow{a}^{\mathrm{3}} \:{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}} {tdt} \\ $$ $$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt}=\frac{\pi}{\mathrm{4}}\:+\left[\frac{\mathrm{1}}{\mathrm{4}}\:{sin}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:=\frac{\pi}{\mathrm{4}}\:\Rightarrow \\ $$ $${I}=\frac{\pi}{\mathrm{4}\:{a}^{\mathrm{3}} } \\ $$ $$\left.\mathrm{2}\right)\:{let}\:{put}\:{J}=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$ $${J}=\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}\:} \:−{a}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)}{dx}\:−\mathrm{2}{a}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$ $${ch}\:{x}={at}\:\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)}=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{adt}}{{a}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{{a}}\frac{\pi}{\mathrm{2}}=\:\frac{\pi}{\mathrm{2}{a}} \\ $$ $${J}=\frac{\pi}{{a}}\:−\mathrm{2}{a}^{\mathrm{2}} \:{I}\:=\:\frac{\pi}{{a}}\:−\mathrm{2}{a}^{\mathrm{2}} \:\frac{\pi}{\mathrm{4}{a}^{\mathrm{3}} }=\:\frac{\pi}{{a}}\:−\frac{\pi}{\mathrm{2}{a}}\Rightarrow\:{J}=\frac{\pi}{\mathrm{2}{a}}\:. \\ $$