Improper Integrals Value of a Constant

Question

I have the integral: $$\int ^\infty _0\left(\frac{1}{\sqrt[2]{x^2+4}}-\frac{\Phi}{x+2}\right)dx$$ Where $\Phi$ is a constant. The question is to find the value of this constant, so this integral converges. I know that first I should split the Integrals and then apply trig. substitution with $\tan(\theta)$ to the first. But mmy question is how to proceed after that in order to find $\Phi$.

Answer 1

Just as mickep commented, the problem is for an infinite value of the upper bound. So, consider $$A=\frac{1}{\sqrt{x^2+4}}-\frac{\Phi}{x+2}=\frac 1x\Big(\frac{1}{\sqrt{1+\frac{4}{x^2}}}-\frac \Phi{1+\frac 2x} \Big)$$ and use Taylor series for each term and combine them to get $$A=\frac{1-\Phi }{x}+\frac{2 \Phi }{x^2}-\frac{4 \Phi +2}{x^3}+O\left(\frac{1}{x^4}\right)$$ Now, if you integrate $$\int A\,dx=(1-\Phi) \log (x)-\frac{2 \Phi }{x}+\frac{2 \Phi +1}{x^2}+O\left(\frac{1}{x^3}\right)$$

I am sure that you see where is the problem.

Answer 2

Hint: I will call the constant $c$. Rewrite the expression as $$\frac{(x+2)-c\sqrt{x^2+4}}{(x+2)\sqrt{x^2+4}}.$$ Multiply top and bottom by $(x+2)+c\sqrt{x^2+4}$. Choose $c$ so as to kill the $x^2$ terms you get on top.