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Question Number 79730 by Henri Boucatchou last updated on 27/Jan/20

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 \dots

Answered by mind is power last updated on 28/Jan/20

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)