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Find positive $K$ such that $$\int_0^\infty\left(\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}\right)dx$$ converges

I used the fact that $\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}>\frac{-K+1/\sqrt2}{1+x}$ for $x>1$ to prove it diverges for $K < 1/\sqrt2$.
And the fact that $\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}<\frac{1-K}{x+1}$ to prove it diverges for $K>1$.

However I am not sure how to prove it converges (or not) in $[\frac1{\sqrt2},1]$.

I think it would not converge for any $K$ becase the denomiantor is sorta linear and this would never be identically $0$ for any $K$.

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2 Answers 2

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Your bounds are correct but too weak. Here we need a more precise asymptotic analysis.

Note that as $x\to +\infty$, $$\begin{align*}\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}&=\frac{x+1-\sqrt{2}Kx\sqrt{1+\frac{1}{2x^2}}}{(x+1)\sqrt{2x^2+1}}\\ &=\frac{x+1-\sqrt{2}Kx(1+\frac{1}{4x^2}+o(\frac{1}{x^2}))}{\sqrt{2}x^2+O(x)}\\ &=\frac{(1-\sqrt{2}K)x+1+O(1/x)}{\sqrt{2}x^2+O(x)}.\end{align*}$$ What may we conclude?

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  • $\begingroup$ So K must be equal to $1/\sqrt2$? $\endgroup$
    – Anvit
    Commented Feb 10, 2019 at 9:15
  • $\begingroup$ Yes, you are correct! $\endgroup$
    – Robert Z
    Commented Feb 10, 2019 at 9:16
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Hint

I suggest to have a look at what happens at the bounds.

For $x$ close to $0$, the Taylor expansion of the integrand is $$\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}=(1-K)+K x-(K+1) x^2+O\left(x^3\right)$$

Now, for infinitely large values of $x$, $$\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}=\frac{\frac{1}{\sqrt{2}}-K}{x}+\frac{K}{x^2}+O\left(\frac{1}{x^3}\right)$$

Does this tell you something ?

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    $\begingroup$ Could you name the process you have used to obtain the second series? I identify the first one as taylor expansion with $0$ as center $\endgroup$
    – Anvit
    Commented Feb 10, 2019 at 9:26
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    $\begingroup$ @Anvit. Make $x=\frac 1t$ and expand series around $t=0$. This is still Taylor expansion. Sorry for not being very talkative ! Cheers $\endgroup$ Commented Feb 10, 2019 at 9:52

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