Improper integrals

For sufficiently large $x$ , the integrand can be approximated as

$\dfrac{k x^2 + 4x - 3x \sqrt{x^2 + 9} } { (x^2 + 4) \sqrt{x^2 + 9} } \approx \dfrac{ kx^2 - 3x^2}{ x^2 \cdot x} = \dfrac{x^2(k-3)}{x^3} = \dfrac{k-3}x .$

This means that for some gigantic number $N$ , $\begin{aligned} \int_0^\infty \left( \dfrac k{\sqrt{x^2+9}} - \dfrac{3x}{x^2 + 4} \right) \, dx &=& \int_0^N \left( \dfrac k{\sqrt{x^2+9}} - \dfrac{3x}{x^2 + 4} \right) \, dx + \int_N^\infty \left( \dfrac k{\sqrt{x^2+9}} - \dfrac{3x}{x^2 + 4} \right) \, dx \\ &\approx & \int_0^N \left( \dfrac k{\sqrt{x^2+9}} - \dfrac{3x}{x^2 + 4} \right) \, dx + \int_N^\infty \dfrac {k-3}x \, dx \end{aligned}$

The latter integral diverges when $k - 3 \ne 0$ .

What's left to do is to prove that the integral in question converges when $k=3$ .

$\int_0^\infty \left( \dfrac 3{\sqrt{x^2+9}} - \dfrac{3x}{x^2 + 4} \right) \, dx = 3 \int_0^\infty \left( \dfrac 1{\sqrt{x^2+9}} - \dfrac{x}{x^2 + 4} \right) \, dx \\$

Recall that $\dfrac d{dx} \text{sinh}^{-1} x = \dfrac1{\sqrt{x^2 + 1}}$ and $\dfrac d{dx} \ln x= \dfrac 1x$ . The integral can be expressed as $3 \left [ \text{sinh}^{-1} \left( \frac x3 \right) - \frac12 \ln(x^2 + 4) \right]_{x=0}^{x\to\infty} = 3 \left [ \lim_{y\to\infty} f(y) - f(0) \right ] ,$ where $f(z) = \text{sinh}^{-1} \left( \frac z3 \right) - \frac12 \ln(z^2 + 4)$ .

$f(0)$ can easily be evaluated to be $-\ln 2$ .

We are left to evaluate $\displaystyle \lim_{y\to\infty} \text{sinh}^{-1} \left( \frac y3 \right) - \frac12 \ln(y^2 + 4)$ .

By definition, $\text{sinh}^{-1} x = \ln\left(x + \sqrt{x^2 + 1} \right)$ . The limit becomes $\begin{aligned} \displaystyle \lim_{y\to\infty} \ln \left(\frac y3 + \sqrt{\left( \frac y3\right)^2 + 1} \right) - \frac12 \ln(y^2 + 4) &=& \displaystyle \lim_{y\to\infty} \left [ \ln\left(\frac13\right) + \ln\left(y + \sqrt{y^2 + 9}\right) - \frac12\ln(y^2 + 4)\right ] \\ &=& \displaystyle \lim_{y\to\infty} \left [ -\ln3 + \frac12 \ln\left(y + \sqrt{y^2 + 9}\right)^2 - \frac12\ln(y^2 + 4) \right ] \\ &=& \displaystyle -\ln3 + \frac12 \lim_{y\to\infty} \ln\dfrac{\left(y + \sqrt{y^2 + 9}\right)^2}{y^2 + 4} \\ &=& \displaystyle -\ln3 + \frac12 \ln \left( \lim_{y\to\infty} \dfrac{2y^2 + 9 + 2y \sqrt{y^2+9}}{y^2 + 4} \right) \\ &=& \displaystyle -\ln3 + \frac12 \ln \left( \lim_{y\to\infty} \dfrac{2 + 9/y^2 + 2 \sqrt{1 +9/y^2}}{1 + 4/y^2} \right) = \ln2 - \ln3 \\ \end{aligned}$ Putting it all together, the integral is equal to $3 \ln\left ( \frac43\right)$ when $k=3$ .

Hence, the integral only converges when $k=\boxed3$ .

The integral is simply $k\ln\frac{x+\sqrt{x^2+9}}{3}-\frac{3}{2}\ln\left(x^2+4\right)$ evaluated between the limits $x = \infty$ and $x = 0$ .

It's value at $x = 0$ is $-3 \ln2$ which is convergent, therefore, we now just need to find the value of $k$ such that its value at $x = \infty$ is convergent.

This is $\displaystyle \lim_{x \to \infty} \left(k\ln\frac{x+\sqrt{x^2+9}}{3}-\frac{3}{2}\ln\left(x^2+4\right)\right)$ .

Substituting $x=3\tan\theta$ and simplifying the limit, we get:

$\displaystyle \lim_{\theta \to \frac{\pi}{2}} \ln\left(\frac{\left(1+\sin\theta\right)^k}{\left(5\sin^2\theta+4\right)^{\frac{3}{2}}}\cdot\cos^{3-k}\theta\right)$

$\displaystyle = \ln\left(\lim_{\theta \to \frac{\pi}{2}} \frac{\left(1+\sin\theta\right)^k}{\left(5\sin^2\theta+4\right)^{\frac{3}{2}}}\cdot\lim_{\theta \to \frac{\pi}{2}} \cos^{3-k}\theta\right)$

$\displaystyle = \ln\left(\frac{2^k}{27}\cdot \lim_{\theta \to \frac{\pi}{2}} \cos^{3-k}\theta\right)$

Now, let's look at the following cases:

Case 1: k > 3

In this case, the expression inside $\ln$ is of the form $\frac{nonzero}{0}$ , which certainly accounts for a divergent value.

Case 2: k < 3

In this case, the expression inside $\ln$ is always going to be $0$ , but this leads us to $\ln 0$ , which is again a sign of divergence.

Case 3: k = 3

In this case, the expression is equal to $\ln(\frac{8}{27})$ and is convergent. Hence, for $\boxed{k = 3}$ the integral will have a convergent value.

Also, the integral then equals to $\ln(\frac{8}{27} - (- 3 \ln 2)) = \ln(\frac{64}{27}) = 3 \ln(\frac{4}{3})$

We can evaluate:

$\displaystyle\int_{0}^{\infty}{\dfrac{k}{\sqrt{x^2+9}}-\dfrac{3x}{x^2+4}}dx=k\sinh^{-1}\left(\dfrac{x}{3}\right)-\dfrac{3}{2}\ln|x^2+4|$

With a few calculations, we get

$\ln\left|\dfrac{(x+\sqrt{x^2+9})^k}{(x^2+4)\sqrt{x^2+4}}\right|$

To make it converge when $x \rightarrow \infty$ , $k=\boxed{3}$ .