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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?

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

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

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

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  • $\begingroup$ That’s really a very nice way! $\endgroup$ – user Aug 24 '18 at 9:03
  • $\begingroup$ Still there is another great answer. Thank you. $\endgroup$ – nik Aug 24 '18 at 9:06
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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 $$\begin{eqnarray*} \int_{0}^{+\infty}\left[\frac{1}{\sqrt{x^2+4}}-\frac{1}{x+2}\right]\,\text dx &\stackrel{x\mapsto 2z}{=}& \int_{0}^{+\infty}\left[\frac{1}{\sqrt{z^2+1}}-\frac{1}{z+1}\right]\,\text dz\\[0.3cm]&=&\left[\text{arcsinh}(z)-\log(z+1)\right]_{0}^{+\infty}\\[0.3cm]&=&\lim_{z\to +\infty}\text{arcsinh}(z)-\log(z+1)\\&\stackrel{z\mapsto\sinh t}{=}&\lim_{t\to +\infty} \log\left(\frac{e^t}{\sinh t+1}\right)=\color{red}{\log 2}.\end{eqnarray*}$$

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