For witch value of α the integral I=∫_0 ^∞ ((1/(√(1+2x^2 )))−(α/(1+x)))dx converge; and in this case calculate α


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Question Number 79869 by Henri Boucatchou last updated on 28/Jan/20

$F o r w i t c h v a l u e o f α t h e i n t e g r a l$ $I = \int_{0}^{\infty} (\frac{1}{\sqrt{1 + 2 x^{2}}} - \frac{α}{1 + x}) d x c o n v e r g e;$ $a n d i n t h i s c a s e c a l c u l a t e α$
Commented by mathmax by abdo last updated on 29/Jan/20
$let a>0 we have I =∫_0 ^a ((1/(√(1+2x^2 )))−(α/(1+x)))dx +∫_a ^(+∞) ((1/(√(1+2x^2 )))−(α/(1+x)))dx the gunction x→(1/(√(1+2x^2 )))−(α/(1+x)) is continue on [0,a] ⇒ integrable on[0,a] ∀α now let look what happend on +∞! we have (1/(√(1+2x^2 )))−(α/(1+x)) =(1/(x(√2)((√((1/(2x^2 ))+1))))) −(α/(x(1+(1/x)))) =(1/(x(√2)))(1+(1/(2x^2 )))^(−(1/2)) −(α/(x(1+(1/x)))) ∼(1/(x(√2))){1−(1/(4x^2 ))}−(α/x)(1−(1/x)) ∼(1/x)((1/(√2))−α)−(1/(4(√2)x^3 )) +(α/x^2 ) so I converges ⇔(1/(√2))−α=0 ⇒α=(1/(√2)) if α≠(1/(√2)) this integral diverges!$
$l e t a > 0 w e h a v e I = \int_{0}^{a} (\frac{1}{\sqrt{1 + 2 x^{2}}} - \frac{α}{1 + x}) d x + \int_{a}^{+ \infty} (\frac{1}{\sqrt{1 + 2 x^{2}}} - \frac{α}{1 + x}) d x$ $t h e g u n c t i o n x \to \frac{1}{\sqrt{1 + 2 x^{2}}} - \frac{α}{1 + x} i s c o n t i n u e o n [0, a] \Rightarrow$ $i n t e g r a b l e o n [0, a] \forall α n o w l e t l o o k w h a t h a p p e n d o n + \infty!$ $w e h a v e \frac{1}{\sqrt{1 + 2 x^{2}}} - \frac{α}{1 + x} = \frac{1}{x \sqrt{2} (\sqrt{\frac{1}{2 x^{2}} + 1})} - \frac{α}{x (1 + \frac{1}{x})}$ $= \frac{1}{x \sqrt{2}} {(1 + \frac{1}{2 x^{2}})}^{- \frac{1}{2}} - \frac{α}{x (1 + \frac{1}{x})} \sim \frac{1}{x \sqrt{2}} {1 - \frac{1}{4 x^{2}}} - \frac{α}{x} (1 - \frac{1}{x})$ $\sim \frac{1}{x} (\frac{1}{\sqrt{2}} - α) - \frac{1}{4 \sqrt{2} x^{3}} + \frac{α}{x^{2}} s o I c o n v e r g e s \Leftrightarrow \frac{1}{\sqrt{2}} - α = 0 \Rightarrow α = \frac{1}{\sqrt{2}}$ $i f α \neq \frac{1}{\sqrt{2}} t h i s i n t e g r a l d i v e r g e s!$
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