How does the integral $\int_0^{\infty} \ln\left( 1+\frac{1}{x^2} \right) \,\text{d}x$ converge?

Question

I tried using the fact that $\ln(f(x)) < f(x)$ but that doesn't seem to work. It's an improper integral. $$ \int_0^{\infty} \ln\left( 1+\frac{1}{x^2} \right) \,\text{d}x $$

Answer 1

Note that $$\int \ln(1+\frac{1}{x^2}) dx = x \log \left(\frac{1}{x^2}+1\right)+2 \arctan (x) + C$$ so $$\begin{align} \int_0^{\infty} \ln(1+\frac{1}{x^2}) dx \\ &= \lim_{x \to \infty} \left( x \log \left(\frac{1}{x^2}+1\right)+2 \arctan (x) \right) - \lim_{x \to 0} \left(x \log \left(\frac{1}{x^2}+1\right)+2 \arctan (x) \right) \\ &= \left( 0 + 2 \cdot \frac{\pi}{2} \right) - \left( 0 + 0 \right) \\ &= \pi \end{align} $$

Answer 2

To prove convergence, you should estimate behavior of the integrand near zero and near infinity.

Near zero $\ln(1 + x^{-2}) \approx \ln(x^{-2}) = -2 \ln x $. It is known that $\int \ln x \; dx$ converges near zero.

Near infinity $\ln(1 + x^{-2}) \approx x^{-2}$. It is known that $\int x^{-2} \; dx$ converges near infinity.

So the integral converges properly. In fact, notation $f(x) \approx g(x)$ should be replaced with more strict $f(x) = \Theta(g(x))$ everywhere.

Answer 3

As $x\to 0$, the integrand behaves as $-2 \log{x}$, which is an integrable singularity.

As $x \to \infty$, the integrand behaves as $1/x^2$, which is integrable in this limit.

There are no other singular points in the integration interval.