I) For witch value of α the integral C=∫_0 ^( ∞) ((1/(√(1+2x^2 )))−(1/(x+1)))dx conveege ? And in this case calculate α. II) Let Δ={(x; y)/ ∣x∣+∣y∣≤2} a) Calculate I_1 = ∫∫_Δ dxdy and ∫∫


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Question Number 79730 by Henri Boucatchou last updated on 27/Jan/20
$I) For witch value of α the integral C=∫_0 ^( ∞) ((1/(√(1+2x^2 )))−(1/(x+1)))dx conveege ? And in this case calculate α. II) Let Δ={(x; y)/ ∣x∣+∣y∣≤2} a) Calculate I_1 = ∫∫_Δ dxdy and ∫∫_Δ ((dxdy)/((∣x∣+∣y∣)^2 +4))$
$I) F o r w i t c h v a l u e o f α t h e i n t e g r a l$ $C = \int_{0}^{\infty} (\frac{1}{\sqrt{1 + 2 x^{2}}} - \frac{1}{x + 1}) d x c o n v e e g e ?$ $A n d i n t h i s c a s e c a l c u l a t e α .$ $I I) L e t Δ = {(x; y) / ∣ x ∣ + ∣ y ∣⩽ 2}$ $a) C a l c u l a t e I_{1} = \int \int_{Δ} d x d y a n d \int \int_{Δ} \frac{d x d y}{{(∣ x ∣ + ∣ y ∣)}^{2} + 4}$
Commented by Henri Boucatchou last updated on 27/Jan/20

$P l e a s e α i s h e r e : C = \int_{0}^{\infty} (\frac{1}{\sqrt{1 + 2 x^{2}}} - \frac{α}{1 + x}) d x . . .$
	Answered by mind is power last updated on 28/Jan/20
	$f(x)=(1/(√(1+2x^2 )))−(1/(1+x)) continus x→∞ f(x)=(((x+1)−(√((1+2x^2 ))))/((x+1)(√(1+2x^2 ))))=((2x−x^2 )/((x+1)(√((1+2x^2 )))(x+1+(√(1+2x^2 ))))) ∼((−x^2 )/(x^3 (√2).(1+(√2))))=−(1/(x(√2)(1+(√2)))) not integrabl in+∞ C diverge a)∫∫_Δ dxdy x∈[−2,2] ∣y∣<2−∣x∣ ⇒∣x∣−2≤y≤2−∣x∣ =∫_(−2) ^2 ∫_(∣x∣−2) ^(2−∣x∣) dxdy =∫_(−2) ^2 [4−2∣x∣]dx =2∫_0 ^2 [4−2x]dx =16−2(4)=8 ∫∫_Δ ((dxdy)/((∣x∣+∣y∣)^2 +4)) we haveΔ=∪_(i=1) ^4 D_i D_1 =(x,y)∣ x+y<2,D_2 x−y<2,D_3 =−x+y<2,D_4 −x−y<2∣ g(x,y)=(1/((∣x∣+∣y∣)^2 +4)) is invsriant in sense g(_− ^+ x,_− ^+ y)=g(x,y) D_1 =ϕ(D_2 )=ϕ′(D_3 )=ϕ′′(D_4 ) ϕ(x,y)=(x,−y) ϕ′(x,y)=(−x,y) ϕ′′(x,y)=(−x,−y) goϕ^i (x,y)=g(x,h),i∈{1,2,3} ⇒∫∫_D_i g(x,y)dxdy=∫∫_D_j g(x,y)dxdy,∀i,j∈{1,2,3,4} ⇒∫∫_Δ g(x,y)dxdy=4∫∫_D_1 g(x,y)dxdy =4∫_0 ^2 ∫_0 ^(2−x) ((dxdy)/((x+y)^2 +4)) ∫_0 ^2 ∫_0 ^(2−x) (dy/(((((x+y)/2))^2 +1)))dx=∫_0 ^2 2∫_0 ^(2−x) (1/2)(dy/(((((x+y)/2))^2 +1))) =∫_0 ^2 2.[tan^(−1) (((x+y)/2))]_0 ^(2−x) ]dx =2∫_0 ^2 ((π/4)−tan^− ((x/2))] by part =π−2∫_0 ^2 tan^(−1) ((x/2))dx=π−2[[_0 ^2 xtan^(−1) ((x/2))]−∫_0 ^2 ((2x)/(1+x^2 ))]dx π−4tan^− (1)+2ln(5)=2ln(5)$
	$f (x) = \frac{1}{\sqrt{1 + 2 x^{2}}} - \frac{1}{1 + x}$ $c o n t i n u s x \to \infty$ $f (x) = \frac{(x + 1) - \sqrt{(1 + 2 x^{2})}}{(x + 1) \sqrt{1 + 2 x^{2}}} = \frac{2 x - x^{2}}{(x + 1) \sqrt{(1 + 2 x^{2})} (x + 1 + \sqrt{1 + 2 x^{2}})}$ $\sim \frac{- x^{2}}{x^{3} \sqrt{2} . (1 + \sqrt{2})} = - \frac{1}{x \sqrt{2} (1 + \sqrt{2})} n o t i n t e g r a b l i n + \infty$ $C d i v e r g e$ $a) \int \int_{Δ} d x d y$ $x \in [- 2, 2]$ $∣ y ∣< 2 - ∣ x ∣$ $\Rightarrow∣ x ∣ - 2 ⩽ y ⩽ 2 - ∣ x ∣$ $= \int_{- 2}^{2} \int_{∣ x ∣ - 2}^{2 - ∣ x ∣} d x d y$ $= \int_{- 2}^{2} [4 - 2 ∣ x ∣] d x$ $= 2 \int_{0}^{2} [4 - 2 x] d x$ $= 16 - 2 (4) = 8$ $\int \int_{Δ} \frac{d x d y}{{(∣ x ∣ + ∣ y ∣)}^{2} + 4}$ $w e h a v e Δ = \underset{i = 1}{\overset{4}{\cup}} D_{i}$ $D_{1} = (x, y) ∣ x + y < 2, D_{2} x - y < 2, D_{3} = - x + y < 2, D_{4} - x - y < 2 ∣$ $g (x, y) = \frac{1}{{(∣ x ∣ + ∣ y ∣)}^{2} + 4} i s i n v s r i a n t i n s e n s e$ $g (_{-}^{+} x,_{-}^{+} y) = g (x, y)$ $D_{1} = φ (D_{2}) = φ^{'} (D_{3}) = φ^{″} (D_{4})$ $φ (x, y) = (x, - y)$ $φ^{'} (x, y) = (- x, y)$ $φ^{″} (x, y) = (- x, - y)$ $g o φ^{i} (x, y) = g (x, h), i \in {1, 2, 3}$ $\Rightarrow \int \int_{D_{i}} g (x, y) d x d y = \int \int_{D_{j}} g (x, y) d x d y, \forall i, j \in {1, 2, 3, 4}$ $\Rightarrow \int \int_{Δ} g (x, y) d x d y = 4 \int \int_{D_{1}} g (x, y) d x d y$ $= 4 \int_{0}^{2} \int_{0}^{2 - x} \frac{d x d y}{{(x + y)}^{2} + 4}$ $\int_{0}^{2} \int_{0}^{2 - x} \frac{d y}{({(\frac{x + y}{2})}^{2} + 1)} d x = \int_{0}^{2} 2 \int_{0}^{2 - x} \frac{1}{2} \frac{d y}{({(\frac{x + y}{2})}^{2} + 1)}$ $= \int_{0}^{2} 2 . {[\tan^{- 1} (\frac{x + y}{2})]}_{0}^{2 - x}] d x$ $= 2 \int_{0}^{2} (\frac{π}{4} - {t a n}^{-} (\frac{x}{2})]$ $b y p a r t$ $= π - 2 \int_{0}^{2} \tan^{- 1} (\frac{x}{2}) d x = π - 2 [[_{0}^{2} x \tan^{- 1} (\frac{x}{2})] - \int_{0}^{2} \frac{2 x}{1 + x^{2}}] d x$ $π - 4 {t a n}^{-} (1) + 2 l n (5) = 2 l n (5)$
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