$$I=\int_0^\infty \frac {x} {x^2+1} – \frac {C} {3x+1} dx$$
Find the value of C for which the integral converges.
I've gotten as far as solving the integral
$\frac 1 2 \ln|x^2+1| – \frac 1 3 c\ln|3x+1|$ but not sure where to go from here as evaluating it gives $\infty – 0 – c \infty + 0$. Any help would be appreciated!
Best Answer
You can do this without finding the actual integral. The basic question is what is the behaviour of the integrand as $x \to \infty$. We have $$ \frac{x}{x^2+1} = \frac{1}{x (1+1/x^2)} = \frac{1}{x} \left(1 - \frac{1}{x^2} + \ldots \right) = \frac{1}{x} + O\left(\frac{1}{x^3}\right)\ \text{as}\ x \to \infty$$ while $$\frac{C}{3x+1} = \frac{C}{3 x (1 + 1/(3x))} = \frac{C}{3x} + O\left(\frac{1}{x^2}\right) $$ So if $C = 3$, $$ \frac{x}{x^2+1} - \frac{3}{3x+1} = O\left(\frac{1}{x^2}\right) $$ which is integrable as $x \to \infty$. If $C \ne 3$, you have a term in $1/x$ which is not integrable.