Question Number 78276 by msup trace by abdo last updated on 15/Jan/20
![find I_n =∫∫_([1,n]^2 ) (√(x^2 +y^2 ))ln(x^2 +y^2 )dxdy](https://www.tinkutara.com/question/Q78276.png)
$${find}\:{I}_{{n}} =\int\int_{\left[\mathrm{1},{n}\right]^{\mathrm{2}} } \:\:\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dxdy} \\ $$
Commented by mathmax by abdo last updated on 17/Jan/20
![let consider the diffeomorphism (r,θ)→(x,y)=(rcosθ,rsinθ) we have 1≤x≤n and 1≤y≤n ⇒2≤x^2 +y^2 ≤2n^2 ⇒(√2)≤r≤n(√2) and 0≤θ≤(π/2) ⇒ I_n =∫_0 ^(π/2) ∫_(√2) ^(n(√2)) rln(r^2 )rdr dθ =π ∫_(√2) ^(n(√2)) r^2 ln(r)dr and by parts ∫_(√2) ^(n(√2)) r^2 ln(r)dr =[(r^3 /3)ln(r)]_(√2) ^(n(√2)) −∫_(√2) ^(n(√2)) (r^3 /3)(dr/r) =(1/3)(2(√2)n^3 ln(n(√2))−((2(√2))/3)ln((√2))) −(1/3)∫_(√2) ^(n(√2)) r^2 dr =((2(√2))/3){ n^3 ln(n(√2))−ln((√2))}−(1/9)[r^3 ]_(√2) ^(n(√2)) ⇒ I_n =((2(√2))/3){n^3 ln(n(√2))−ln((√2))}−(1/9){2(√2)n^3 −2(√2)}](https://www.tinkutara.com/question/Q78436.png)
$${let}\:{consider}\:{the}\:{diffeomorphism}\:\:\left({r},\theta\right)\rightarrow\left({x},{y}\right)=\left({rcos}\theta,{rsin}\theta\right) \\ $$$${we}\:{have}\:\mathrm{1}\leqslant{x}\leqslant{n}\:{and}\:\mathrm{1}\leqslant{y}\leqslant{n}\:\Rightarrow\mathrm{2}\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow\sqrt{\mathrm{2}}\leqslant{r}\leqslant{n}\sqrt{\mathrm{2}} \\ $$$${and}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\sqrt{\mathrm{2}}} ^{{n}\sqrt{\mathrm{2}}} {rln}\left({r}^{\mathrm{2}} \right){rdr}\:{d}\theta \\ $$$$=\pi\:\int_{\sqrt{\mathrm{2}}} ^{{n}\sqrt{\mathrm{2}}} {r}^{\mathrm{2}} {ln}\left({r}\right){dr}\:\:{and}\:{by}\:{parts}\:\int_{\sqrt{\mathrm{2}}} ^{{n}\sqrt{\mathrm{2}}} {r}^{\mathrm{2}} {ln}\left({r}\right){dr} \\ $$$$=\left[\frac{{r}^{\mathrm{3}} }{\mathrm{3}}{ln}\left({r}\right)\right]_{\sqrt{\mathrm{2}}} ^{{n}\sqrt{\mathrm{2}}} \:−\int_{\sqrt{\mathrm{2}}} ^{{n}\sqrt{\mathrm{2}}} \frac{{r}^{\mathrm{3}} }{\mathrm{3}}\frac{{dr}}{{r}}\:=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{2}\sqrt{\mathrm{2}}{n}^{\mathrm{3}} {ln}\left({n}\sqrt{\mathrm{2}}\right)−\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}{ln}\left(\sqrt{\mathrm{2}}\right)\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}\int_{\sqrt{\mathrm{2}}} ^{{n}\sqrt{\mathrm{2}}} \:{r}^{\mathrm{2}} {dr}\:=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}\left\{\:{n}^{\mathrm{3}} {ln}\left({n}\sqrt{\mathrm{2}}\right)−{ln}\left(\sqrt{\mathrm{2}}\right)\right\}−\frac{\mathrm{1}}{\mathrm{9}}\left[{r}^{\mathrm{3}} \right]_{\sqrt{\mathrm{2}}} ^{{n}\sqrt{\mathrm{2}}} \:\Rightarrow \\ $$$${I}_{{n}} =\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}\left\{{n}^{\mathrm{3}} {ln}\left({n}\sqrt{\mathrm{2}}\right)−{ln}\left(\sqrt{\mathrm{2}}\right)\right\}−\frac{\mathrm{1}}{\mathrm{9}}\left\{\mathrm{2}\sqrt{\mathrm{2}}{n}^{\mathrm{3}} −\mathrm{2}\sqrt{\mathrm{2}}\right\} \\ $$