Question Number 54376 by Abdo msup. last updated on 02/Feb/19 | ||
$$\left.\mathrm{1}\right)\:{calculate}\:\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{ax}\:\:+\mathrm{1}} \\ $$ $${with}\:\:\:\mid{a}\mid<\mathrm{2} \\ $$ $$\left.\mathrm{2}\right)\:{calculate}\:{g}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{x}}{\left({x}^{\mathrm{2}} \:+{ax}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ $$\left.\mathrm{3}\right){find}\:{values}\:{of}\:{integrals}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}} \\ $$ $${and}\:\int_{−\infty} ^{+\infty} \:\frac{{x}}{\left({x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ $$\left.\mathrm{4}\right)\:{calculate}\:{A}\left(\theta\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{cos}\theta\:+\mathrm{1}} \\ $$ $$\theta\:{is}\:{a}\:{given}\:{real}. \\ $$ | ||
Commented bymaxmathsup by imad last updated on 03/Feb/19 | ||
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{ax}\:+\mathrm{1}}\:\:\Rightarrow{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\frac{{a}}{\mathrm{2}}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$ $$=\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{\left({x}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{4}−{a}^{\mathrm{2}} }{\mathrm{4}}}\:\:=_{{x}+\frac{{a}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }{t}} \:\:\:\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{1}}{\frac{\mathrm{4}−{a}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\frac{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }{dt}}{\mathrm{2}} \\ $$ $$=\mathrm{2}\frac{\mathrm{1}}{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\frac{\mathrm{2}\pi}{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\:\Rightarrow{f}\left({a}\right)\:=\frac{\mathrm{2}\pi}{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\:. \\ $$ $$\left.\mathrm{2}\right){we}\:{have}\:{f}^{'} \left({a}\right)\:=−\int_{−\infty} ^{+\infty} \:\:\frac{{xdx}}{\left({x}^{\mathrm{2}} +{ax}+\mathrm{1}\right)^{\mathrm{2}} }\:=−{g}\left({a}\right)\:\Rightarrow \\ $$ $${g}\left({a}\right)\:=−{f}^{'} \left({a}\right)\:=−\mathrm{2}\pi\:\left(\left(\mathrm{4}−{a}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{'} =−\mathrm{2}\pi\:.\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\mathrm{2}{a}\right)\left(\mathrm{4}−{a}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$ $$=−\mathrm{2}\pi{a}\:\frac{\mathrm{1}}{\left(\mathrm{4}−{a}^{\mathrm{2}} \right)\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\:=\frac{−\mathrm{2}\pi{a}}{\left(\mathrm{4}−{a}^{\mathrm{2}} \right)\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }} \\ $$ | ||
Commented bymaxmathsup by imad last updated on 03/Feb/19 | ||
$$\left.\mathrm{3}\right)\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:={f}\left(\sqrt{\mathrm{2}}\right)\:=\frac{\mathrm{2}\pi}{\sqrt{\mathrm{4}−\mathrm{2}}}\:=\frac{\mathrm{2}\pi}{\sqrt{\mathrm{2}}}\:=\pi\sqrt{\mathrm{2}}. \\ $$ $$\int_{−\infty} ^{+\infty} \:\:\:\frac{{xdx}}{\left({x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}\right)^{\mathrm{2}} }\:={g}\left(\sqrt{\mathrm{2}}\right)\:=\:\frac{−\mathrm{2}\pi\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}}\:=−\pi\:. \\ $$ | ||
Commented bymaxmathsup by imad last updated on 03/Feb/19 | ||
$$\left.\mathrm{4}\right)\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{cos}\theta\:{x}\:+\mathrm{1}}\:={f}\left(\mathrm{2}{cos}\theta\right)\:=\:\frac{\mathrm{2}\pi}{\sqrt{\mathrm{4}−\mathrm{4}{cos}^{\mathrm{2}} \theta}}\:=\frac{\mathrm{2}\pi}{\mathrm{2}\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} \theta}}\:=\frac{\pi}{\mid{sin}\theta\mid} \\ $$ $$\left(\:{we}\:{suppose}\:{here}\:\theta\:\neq\:{k}\pi\:\:{kfrom}\:{Z}\right). \\ $$ $$ \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19 | ||
$$\left.\mathrm{2}\right)\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+{a}−{a}}{\left({x}^{\mathrm{2}} +{ax}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}^{\mathrm{2}} +{ax}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +{ax}+\mathrm{1}\right)^{\mathrm{2}} }{dx}−\frac{{a}}{\mathrm{2}}\int\frac{{dx}}{\left[\left({x}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\:\right)^{\mathrm{2}} \right]^{\mathrm{2}} } \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}×\frac{−\mathrm{1}}{\left({x}^{\mathrm{2}} +{ax}+\mathrm{1}\right)}−\frac{{a}}{\mathrm{2}}{I}_{\mathrm{2}} \\ $$ $${I}_{\mathrm{2}} \:\:{let}\:\left({x}+\frac{{a}}{\mathrm{2}}\right)=\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\:{tan}\theta={ktan}\theta \\ $$ $${dx}={k}\:{sec}^{\mathrm{2}} \theta{d}\theta \\ $$ $${so}\:{I}_{\mathrm{2}} =\int\frac{{ksec}^{\mathrm{2}} \theta{d}\theta}{\left[{k}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta+{k}^{\mathrm{2}} \right]^{\mathrm{2}} } \\ $$ $$\int\frac{{ksec}^{\mathrm{2}} \theta}{{k}^{\mathrm{4}} {sec}^{\mathrm{4}} \theta}{d}\theta \\ $$ $$\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\int\left(\frac{\mathrm{1}+{cos}\mathrm{2}\theta}{\mathrm{2}}\right){d}\theta \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{3}} }\theta+\frac{\mathrm{1}}{\mathrm{4}{k}^{\mathrm{3}} }{sin}\mathrm{2}\theta+{c} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{3}} }{tan}^{−\mathrm{1}} \left(\frac{{x}+\frac{{a}}{\mathrm{2}}}{{k}}\right)+\frac{\mathrm{1}}{\mathrm{4}{k}^{\mathrm{3}} }×\frac{\mathrm{2}\left(\frac{{x}+\frac{{a}}{\mathrm{2}}}{{k}}\right)}{\mathrm{1}+\left(\frac{{x}+\frac{{a}}{\mathrm{2}}}{{k}}\right)^{\mathrm{2}} }+{c} \\ $$ $${so}\: \\ $$ $$\frac{−\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +{ax}+\mathrm{1}\right)}−\frac{{a}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{3}} }{tan}^{−\mathrm{1}} \left(\frac{{x}+\frac{{a}}{\mathrm{2}}}{{k}}\right)+\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{3}} }×\frac{\left(\frac{{x}+\frac{{a}}{\mathrm{2}}}{{k}}\right)}{\mathrm{1}+\left(\frac{{x}+\frac{{a}}{\mathrm{2}}}{{k}}\right)^{\mathrm{2}} }\right] \\ $$ $$\mid\frac{−\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +{ax}+\mathrm{1}\right)}−\frac{{a}}{\mathrm{4}{k}^{\mathrm{3}} }\left[{tan}^{−\mathrm{1}} \left(\frac{{x}+\frac{{a}}{\mathrm{2}}}{{k}}\right)+\frac{\frac{{x}+\frac{{a}}{\mathrm{2}}}{{k}}}{\mathrm{1}+\left(\frac{{x}+\frac{{a}}{\mathrm{2}}}{{k}}\right)^{\mathrm{2}} }\right]\mid_{−\infty} ^{\infty} \\ $$ $$=\mathrm{0}−\frac{{a}}{\mathrm{4}{k}^{\mathrm{3}} }\left[\left\{{tan}^{−\mathrm{1}} \left(\infty\right)−{tan}^{−\mathrm{1}} \left(−\infty\right)\right\}+\mathrm{0}\right] \\ $$ $$=−\frac{{a}}{\mathrm{4}{k}^{\mathrm{3}} }×\frac{\pi}{\mathrm{1}}=\frac{−{a}\pi}{\mathrm{4}\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$ $$ \\ $$ $$\left.\mathrm{3}\right){second}\:{part}\:=\frac{−\sqrt{\mathrm{2}}\:\pi}{\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\frac{−\sqrt{\mathrm{2}}\:\pi}{\mathrm{4}}×\sqrt{\mathrm{2}}\:×\sqrt{\mathrm{2}}\:×\sqrt{\mathrm{2}}\:=−\pi\:{answer} \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19 | ||
$$\left.\mathrm{1}\right)\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}\frac{{a}}{\mathrm{2}}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}{dx} \\ $$ $$\int\frac{{dx}}{\left(\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\:\right)^{\mathrm{2}} +\left({x}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$ $${so}\:{answer}\:{is} \\ $$ $$\mid\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\:}{tan}^{−\mathrm{1}} \left(\frac{{x}+\frac{{a}}{\mathrm{2}}}{\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}}\right)\mid_{−\infty} ^{\infty} \:\left[\frac{{a}^{\mathrm{2}} }{\mathrm{4}}<\mathrm{1}\right] \\ $$ $$=\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}}×\left\{\frac{\pi}{\mathrm{2}}−\left(−\frac{\pi}{\mathrm{2}}\right)\right\}=\frac{\pi}{\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}} \\ $$ $$\left.\mathrm{3}\right)\frac{\pi}{\sqrt{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{4}}}}=\pi\sqrt{\mathrm{2}} \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19 | ||
$$\left.\mathrm{4}\right)\int_{−\infty} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\left[\left[\left[{b}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{cos}\theta\right]\right.\right. \\ $$ $$\frac{\mathrm{1}}{{b}}\mid{tan}^{−\mathrm{1}} \left(\frac{{x}}{{b}}\right)\mid_{−\infty} ^{\infty} \\ $$ $$\frac{\mathrm{1}}{{b}}\left[{tan}^{−\mathrm{1}} \left(\infty\right)−{tan}^{−\mathrm{1}} \left(−\infty\right)\right] \\ $$ $$=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{2}{cos}\theta}}×\left\{\frac{\pi}{\mathrm{2}}−\frac{−\pi}{\mathrm{2}}\right\} \\ $$ $$=\frac{\pi}{\sqrt{\mathrm{1}+\mathrm{2}{cos}\theta}} \\ $$ $$ \\ $$ | ||