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

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 $$K$$.

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?

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

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 ?

• 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 Commented Feb 10, 2019 at 9:26
• @Anvit. Make $x=\frac 1t$ and expand series around $t=0$. This is still Taylor expansion. Sorry for not being very talkative ! Cheers Commented Feb 10, 2019 at 9:52