What to do when an improper integral returns infinity-infinity

Question

I have to solve this following question:

Find the value(s) of the (real) constant C such that the following integral converges:

$$\int_0^\infty \left(\frac{x}{{x^2+1}} - \frac{C}{{3x+1}}\right) \, dx$$

However, when I integrate it, the end result is $$\lim_{{A \to \infty}} \frac{{\ln(A^2+1)}}{2} - \frac{{C \ln(3A+1)}}{3}$$

Which gives $\infty - C\infty$, which is indeterminate. What should I do in this case?

The only thing that came to my mind is that the expression may be rewritten as $$\frac{\ln(\infty)}{2} - \frac{C\ln(\infty)}{3}$$ so then if $C = \frac{3}{2}$ $$\frac{\ln(\infty)}{2} - \frac{\ln(\infty)}{2} = 0$$

Which in theory is the answer to the question, but I don't know if rewritting the equation as I did above is something that is allowable.

Thanks in advance for the help.

Answer 1

We have \begin{align} &\int_0^\infty\left(\frac{x}{{x^2+1}} - \frac{C}{{3x+1}}\right) \, dx\\ =&\ \ln\frac{(1+x^2)^{1/2}}{(3x+1)^{C/3}}\bigg|_0^{\infty} = \lim_{x\to\infty}\ln\frac{(1+x^2)^{1/2}}{(3x+1)^{C/3}}=\lim_{x\to\infty}\ln \frac{x^{1-\frac C3}}{3^{C/3}} \end{align} which is finite (converging) if $C=3$.

Answer 2

When you have an indeterminate form of the limit, all it means is that there is some arithmetic property of the limit that cannot be applied. In this case the property that

If $\lim f$ and $\lim g$ exist, then $\lim (f-g)=\lim f-\lim g$

What you do in those cases, is to determine the limit in some other way. You need to make the operation in question (the $-$ in you case) cancel "all it can", such that the behavior of the result becomes apparent.

So, "compute" the difference.

$$\frac{\ln(A^2+1)}{2}-\frac{C\ln(3A+1)}{3}=\ln\left(\frac{(A^2+1)^{1/2}}{(3A+1)^{C/3}}\right)$$

Now, inside the logarithm, there is still an indeterminate case $\infty/\infty$.

To make the analysis simpler, consider the case $C=3$ first and divide numerator and denominator by $A$. We get

$$\ln\left(\frac{(1+1/A^2)^{1/2}}{(3+1/A)}\right)$$

Now the limit is apparent.

Try next the case $C\neq 3$.

Answer 3

In order to determine $C$ there is no need to use antiderivatives, The first function behaves like $1/x$ at $\infty.$ Therefore we should take $C$ so that the second function behaves like $1/x,$ i.e. $C=3.$ Next $$\left |{x\over x^2+1}-{3x\over 3x+1}\right |={2x\over (x^2+1)(3x+1)}\\ \le {1\over x^2},\quad x\ge 1$$ Therefore the integral is absolutely convergent.

If $C\neq 3$ then $${x\over x^2+1}-{Cx\over 3x+1}\\ =\left [{x\over x^2+1}-{3x\over 3x+1}\right ]+{3-C\over 3x+1}$$ The integral of the function in square brackets is absolutely convergent, while the integral of the remaining term is divergent.