Question Number 79730 by Henri Boucatchou last updated on 27/Jan/20
$$\left.{I}\right)\:\:{For}\:{witch}\:{value}\:{of}\:\alpha\:{the}\:{integral} \\ $$$$\:{C}=\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){dx}\:\:{conveege}\:\:? \\ $$$${And}\:{in}\:{this}\:{case}\:{calculate}\:\alpha. \\ $$$$\left.{II}\right)\:\:{Let}\:\Delta=\left\{\left({x};\:{y}\right)/\:\mid{x}\mid+\mid{y}\mid\leqslant\mathrm{2}\right\} \\ $$$$\left.\:\:\:\:\:{a}\right)\:{Calculate}\:{I}_{\mathrm{1}} =\:\int\int_{\Delta} {dxdy}\:\:\:{and}\:\:\int\int_{\Delta} \frac{{dxdy}}{\left(\mid{x}\mid+\mid{y}\mid\right)^{\mathrm{2}} +\mathrm{4}} \\ $$
Commented by Henri Boucatchou last updated on 27/Jan/20
$$\:\boldsymbol{{P}}{lease}\:\alpha\:{is}\:{here}\::\:\:{C}=\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }}−\frac{\alpha}{\mathrm{1}+{x}}\right){dx}... \\ $$
Answered by mind is power last updated on 28/Jan/20
![f(x)=(1/(√(1+2x^2 )))−(1/(1+x)) continus x→∞ f(x)=(((x+1)−(√((1+2x^2 ))))/((x+1)(√(1+2x^2 ))))=((2x−x^2 )/((x+1)(√((1+2x^2 )))(x+1+(√(1+2x^2 ))))) ∼((−x^2 )/(x^3 (√2).(1+(√2))))=−(1/(x(√2)(1+(√2)))) not integrabl in+∞ C diverge a)∫∫_Δ dxdy x∈[−2,2] ∣y∣<2−∣x∣ ⇒∣x∣−2≤y≤2−∣x∣ =∫_(−2) ^2 ∫_(∣x∣−2) ^(2−∣x∣) dxdy =∫_(−2) ^2 [4−2∣x∣]dx =2∫_0 ^2 [4−2x]dx =16−2(4)=8 ∫∫_Δ ((dxdy)/((∣x∣+∣y∣)^2 +4)) we haveΔ=∪_(i=1) ^4 D_i D_1 =(x,y)∣ x+y<2,D_2 x−y<2,D_3 =−x+y<2,D_4 −x−y<2∣ g(x,y)=(1/((∣x∣+∣y∣)^2 +4)) is invsriant in sense g(_− ^+ x,_− ^+ y)=g(x,y) D_1 =ϕ(D_2 )=ϕ′(D_3 )=ϕ′′(D_4 ) ϕ(x,y)=(x,−y) ϕ′(x,y)=(−x,y) ϕ′′(x,y)=(−x,−y) goϕ^i (x,y)=g(x,h),i∈{1,2,3} ⇒∫∫_D_i g(x,y)dxdy=∫∫_D_j g(x,y)dxdy,∀i,j∈{1,2,3,4} ⇒∫∫_Δ g(x,y)dxdy=4∫∫_D_1 g(x,y)dxdy =4∫_0 ^2 ∫_0 ^(2−x) ((dxdy)/((x+y)^2 +4)) ∫_0 ^2 ∫_0 ^(2−x) (dy/(((((x+y)/2))^2 +1)))dx=∫_0 ^2 2∫_0 ^(2−x) (1/2)(dy/(((((x+y)/2))^2 +1))) =∫_0 ^2 2.[tan^(−1) (((x+y)/2))]_0 ^(2−x) ]dx =2∫_0 ^2 ((π/4)−tan^− ((x/2))] by part =π−2∫_0 ^2 tan^(−1) ((x/2))dx=π−2[[_0 ^2 xtan^(−1) ((x/2))]−∫_0 ^2 ((2x)/(1+x^2 ))]dx π−4tan^− (1)+2ln(5)=2ln(5)](Q79796.png)
$${f}\left({x}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\mathrm{1}+{x}} \\ $$$${continus}\:{x}\rightarrow\infty \\ $$$${f}\left({x}\right)=\frac{\left({x}+\mathrm{1}\right)−\sqrt{\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right)}}{\left({x}+\mathrm{1}\right)\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }}=\frac{\mathrm{2}{x}−{x}^{\mathrm{2}} }{\left({x}+\mathrm{1}\right)\sqrt{\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right)}\left({x}+\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }\right)} \\ $$$$\sim\frac{−{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} \sqrt{\mathrm{2}}.\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}=−\frac{\mathrm{1}}{{x}\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}\:{not}\:{integrabl}\:{in}+\infty \\ $$$${C}\:{diverge} \\ $$$$\left.{a}\right)\int\int_{\Delta} {dxdy} \\ $$$${x}\in\left[−\mathrm{2},\mathrm{2}\right] \\ $$$$\mid{y}\mid<\mathrm{2}−\mid{x}\mid \\ $$$$\Rightarrow\mid{x}\mid−\mathrm{2}\leqslant{y}\leqslant\mathrm{2}−\mid{x}\mid \\ $$$$=\int_{−\mathrm{2}} ^{\mathrm{2}} \int_{\mid{x}\mid−\mathrm{2}} ^{\mathrm{2}−\mid{x}\mid} {dxdy} \\ $$$$=\int_{−\mathrm{2}} ^{\mathrm{2}} \left[\mathrm{4}−\mathrm{2}\mid{x}\mid\right]{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}} \left[\mathrm{4}−\mathrm{2}{x}\right]{dx} \\ $$$$=\mathrm{16}−\mathrm{2}\left(\mathrm{4}\right)=\mathrm{8} \\ $$$$\int\int_{\Delta} \frac{{dxdy}}{\left(\mid{x}\mid+\mid{y}\mid\right)^{\mathrm{2}} +\mathrm{4}} \\ $$$${we}\:{have}\Delta=\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\cup}}{D}_{{i}} \\ $$$${D}_{\mathrm{1}} =\left({x},{y}\right)\mid\:\:\:{x}+{y}<\mathrm{2},{D}_{\mathrm{2}} {x}−{y}<\mathrm{2},{D}_{\mathrm{3}} =−{x}+{y}<\mathrm{2},{D}_{\mathrm{4}} −{x}−{y}<\mathrm{2}\mid \\ $$$${g}\left({x},{y}\right)=\frac{\mathrm{1}}{\left(\mid{x}\mid+\mid{y}\mid\right)^{\mathrm{2}} +\mathrm{4}}\:\:{is}\:{invsriant}\:{in}\:{sense} \\ $$$${g}\left(_{−} ^{+} {x},_{−} ^{+} {y}\right)={g}\left({x},{y}\right) \\ $$$${D}_{\mathrm{1}} =\varphi\left({D}_{\mathrm{2}} \right)=\varphi'\left({D}_{\mathrm{3}} \right)=\varphi''\left({D}_{\mathrm{4}} \right) \\ $$$$\varphi\left({x},{y}\right)=\left({x},−{y}\right) \\ $$$$\varphi'\left({x},{y}\right)=\left(−{x},{y}\right) \\ $$$$\varphi''\left({x},{y}\right)=\left(−{x},−{y}\right) \\ $$$${go}\varphi^{{i}} \left({x},{y}\right)={g}\left({x},{h}\right),{i}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3}\right\} \\ $$$$\Rightarrow\int\int_{{D}_{{i}} } {g}\left({x},{y}\right){dxdy}=\int\int_{{D}_{{j}} } {g}\left({x},{y}\right){dxdy},\forall{i},{j}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\right\} \\ $$$$\Rightarrow\int\int_{\Delta} {g}\left({x},{y}\right){dxdy}=\mathrm{4}\int\int_{{D}_{\mathrm{1}} } {g}\left({x},{y}\right){dxdy} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{2}−{x}} \frac{{dxdy}}{\left({x}+{y}\right)^{\mathrm{2}} +\mathrm{4}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{2}−{x}} \frac{{dy}}{\left(\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}\right)}{dx}=\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}−{x}} \frac{\mathrm{1}}{\mathrm{2}}\frac{{dy}}{\left(\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\left.=\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{2}.\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}+{y}}{\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\mathrm{2}−{x}} \right]{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−{tan}^{−} \left(\frac{{x}}{\mathrm{2}}\right)\right] \\ $$$${by}\:{part} \\ $$$$=\pi−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{2}}\right){dx}=\pi−\mathrm{2}\left[\left[_{\mathrm{0}} ^{\mathrm{2}} {x}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{2}}\right)\right]−\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right]{dx} \\ $$$$\pi−\mathrm{4}{tan}^{−} \left(\mathrm{1}\right)+\mathrm{2}{ln}\left(\mathrm{5}\right)=\mathrm{2}{ln}\left(\mathrm{5}\right) \\ $$$$ \\ $$