For-witch-value-of-the-integral-I-0-1-1-2x-2-1-x-dx-converge-and-in-this-case-calculate-

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

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!