Question Number 220972 by SdC355 last updated on 21/May/25
$$\int_{−\infty} ^{+\infty} \int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\:\mathrm{d}{x}\mathrm{d}{y} \\ $$$${x}={r}\mathrm{cos}\left(\theta\right) \\ $$$${y}={r}\mathrm{sin}\left(\theta\right) \\ $$$$\mid\mid\boldsymbol{{J}}\mid\mid={r}\mathrm{d}{r}\mathrm{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{{r}}{\left({r}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\mathrm{d}{r}\mathrm{d}\theta=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\mathrm{d}\theta=\frac{\pi}{\mathrm{4}} \\ $$$${Q}.\:\mathrm{if}\:\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\:\mathrm{d}{x}\mathrm{d}{y} \\ $$$$\mathrm{0}\leq{r}<\infty\:,\:\mathrm{0}\leq\theta\leq\frac{\pi}{\mathrm{2}}…..??? \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{confuse}\:\mathrm{how}\:\mathrm{could}\:\:\mathrm{i}\:\mathrm{select}\:\mathrm{interval}\:\mathrm{of}\:\mathrm{integral} \\ $$
Commented by Frix last updated on 21/May/25
$$\underset{−\infty} {\overset{+\infty} {\int}}\:\underset{−\infty} {\overset{+\infty} {\int}}\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dxdy}= \\ $$$$=\mathrm{4}\underset{\mathrm{0}} {\overset{+\infty} {\int}}\:\underset{\mathrm{0}} {\overset{+\infty} {\int}}\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dxdy}= \\ $$$$=\mathrm{4}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\underset{\mathrm{0}} {\overset{+\infty} {\int}}\frac{{r}}{\left({r}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{drd}\theta \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{of}\:\mathrm{it}\:\mathrm{as}\:\mathrm{a}\:\mathrm{quarter}\:\mathrm{circle}… \\ $$
Commented by SdC355 last updated on 22/May/25
$$\mathrm{hmmm}\:\mathrm{thx}\sim \\ $$
Answered by Mathspace last updated on 22/May/25
![la fonction f(x,y)=(1/((x^2 +y^2 +4)^2 )) verifie f(−x,−y)=f(x,y) ⇒ ∫∫_R^2 ((dxdy)/((x^2 +y^2 +4)^2 ))dx=4∫∫_([0,+∞[^2 ) ((dxdy)/((x^2 +y^2 +4)^2 )) le changement x=rcosθ et y=rsinθ donne r>0 et θ∈[0,(π/2)] ⇒ I=4∫_0 ^(π/2) ∫_0 ^(+∞) (1/((r^2 +4)^2 )) r dr dθ =2π∫_0 ^∞ ((rdr)/((r^2 +4)^2 ))=2π×[−(1/2)(1/(r^2 +4))]_0 ^(+∞) =π((1/4)−0)=(π/4)](https://www.tinkutara.com/question/Q220996.png)
$${la}\:{fonction}\:{f}\left({x},{y}\right)=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} } \\ $$$${verifie}\:{f}\left(−{x},−{y}\right)={f}\left({x},{y}\right)\:\Rightarrow \\ $$$$\int\int_{{R}^{\mathrm{2}} } \:\frac{{dxdy}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}=\mathrm{4}\int\int_{\left[\mathrm{0},+\infty\left[^{\mathrm{2}} \right.\right.} \frac{{dxdy}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} } \\ $$$${le}\:{changement}\:{x}={rcos}\theta\:{et}\:{y}={rsin}\theta\:{donne} \\ $$$${r}>\mathrm{0}\:{et}\:\theta\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\:\Rightarrow \\ $$$${I}=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{\left({r}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\:{r}\:{dr}\:{d}\theta \\ $$$$=\mathrm{2}\pi\int_{\mathrm{0}} ^{\infty} \frac{{rdr}}{\left({r}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{2}\pi×\left[−\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{1}}{{r}^{\mathrm{2}} +\mathrm{4}}\right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\pi\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{0}\right)=\frac{\pi}{\mathrm{4}} \\ $$