# Improper integral $\int_{0}^{\infty}\left(\frac{1}{\sqrt{x^2+4}}-\frac{k}{x+2}\right)\text dx$

Given improper integral $$\int \limits_{0}^{\infty}\left(\frac{1}{\sqrt{x^2+4}}-\frac{k}{x+2}\right)\text dx \, ,$$
there exists $$k$$ that makes this integral convergent.
Find its integration value.

Choices are $$\ln 2$$, $$\ln 3$$, $$\ln 4$$, and $$\ln 5$$.

I've written every information from the problem.
Yet I'm not sure whether I should find the integration value from the given integral or $$k$$.

What I've tried so far is,
$$\int_{0}^{\infty} \frac{1}{\sqrt{x^2+4}} \, \text dx= \left[\sinh^{-1}{\frac{x}{2}}\right]_{0}^{\infty}$$

How should I proceed?

• They don't let you say "none of the above"? – Angina Seng Aug 24 '18 at 8:34

We have that for $$x\to \infty$$

$$\frac{1}{\sqrt{x^2+4}}=\frac1x(1+4/x^2)^{-1/2}\sim \frac1x-\frac2{x^3}$$

and

$$\frac k {x+2}\sim \frac k x$$

therefore in order to have convergence we need $$k=1$$ in such way that the $$\frac1x$$ term cancels out.

Then we need to solve and evaluate

$$\int_{0}^{\infty}\left (\frac{1}{\sqrt{x^2+4}}-\frac{1}{x+2}\right)\text dx=\left[\sinh^{-1}\frac x 2 -\log (x+2)\right]_{0}^{\infty}$$

and to evaluate the value at $$\infty$$ recall that

$$\sinh^{-1}\frac x 2=\log \left(\frac x 2 + \sqrt{\frac{x^2}{4}+1}\right) \sim \log x$$

• Thanks! the last calculation part still bothers me. I'll give it a try. – nik Aug 24 '18 at 8:53
• You're still writing it. Awsome. – nik Aug 24 '18 at 8:54
• @nik If you use the log expression for arcsinh it becomes trivial – user Aug 24 '18 at 8:54

As shown in gimusi's answer, you do not need to compute the integral.

However, if you want to integrate, consider $$I=\int_{0}^{p}\left(\frac{1}{\sqrt{x^2+4}}-\frac{k}{x+2}\right)\,\text dx=\sinh ^{-1}\left(\frac{p}{2}\right)-k \log (p+2)+k \log (2)$$ and use series expansion for large $$p$$. This should give $$I=(1-k) \log \left({p}\right)+k \log (2)-\frac{2 k}{p}+\frac{2 k+1}{p^2}+O\left(\frac{1}{p^3}\right)$$ and look at the limit when $$p\to \infty$$.

• That’s really a very nice way! – user Aug 24 '18 at 9:03
• Still there is another great answer. Thank you. – nik Aug 24 '18 at 9:06

There are no integration issue in a right neighbourhood of the origin, but when $$x\to +\infty$$ we have that the integrand function behaves like $$\frac{1-k}{x}+O\left(\frac{1}{x^2}\right)$$, so a necessary and sufficient condition for the integrability is $$k=1$$. In such a case $$\begin{eqnarray*} \int_{0}^{+\infty}\left[\frac{1}{\sqrt{x^2+4}}-\frac{1}{x+2}\right]\,\text dx &\stackrel{x\mapsto 2z}{=}& \int_{0}^{+\infty}\left[\frac{1}{\sqrt{z^2+1}}-\frac{1}{z+1}\right]\,\text dz\\[0.3cm]&=&\left[\text{arcsinh}(z)-\log(z+1)\right]_{0}^{+\infty}\\[0.3cm]&=&\lim_{z\to +\infty}\text{arcsinh}(z)-\log(z+1)\\&\stackrel{z\mapsto\sinh t}{=}&\lim_{t\to +\infty} \log\left(\frac{e^t}{\sinh t+1}\right)=\color{red}{\log 2}.\end{eqnarray*}$$