Examine $\int \limits_{0}^{\infty} \frac{\ln x}{1+x^{2}} d x$ for convergence by using the direct comparison test

Question

Examine the following improper integral for convergence and determine the value if necessary.

$$\int_{0}^{\infty} \frac{\ln x}{1+x^{2}} dx$$

My solution approach:

I am not sure how to use the comparison test properly. This is how I determined the value of the integral.

$$ \begin{split} I &= \int_{0}^{+\infty} \frac{\ln x}{x^{2}+1} dx \quad \text{substitute } x=\frac{1}{t}, d x=\frac{-1}{t^{2}} d t\\ &=\int_{+\infty}^{0} \frac{\ln \frac{1}{t}}{\frac{1}{t^{2}}+1} \frac{-1}{t^{2}} dt \\ &=\int_{+\infty}^{0} \frac{\ln t^{-1}}{\frac{1}{t^{2}}+1} \frac{-1}{t^{2}} dt\\ &= \int_{+\infty}^{0} \frac{-\ln t}{t^{2}+1} \frac{-t^{2}}{t^{2}} d t\\ &= \int_{+\infty}^{0} \frac{\ln t}{t^{2}+1} d t \\ &= -\int_{0}^{+\infty} \frac{\ln t}{t^{2}+1} d t\\ &= -\int_{0}^{+\infty} \frac{\ln x}{x^{2}+1} d x\\ &\Longrightarrow I=-I \Longrightarrow I+I=0 \Longrightarrow 2 I=0 \Longrightarrow I=0 \end{split} $$

I still need to show that it converges with the comparison test though. Thanks in advance!

Answer 1

We can use your idea: $$\int\limits_0^{+\infty}\frac{\ln{x}}{1+x^2}dx=\int\limits_0^{1}\frac{\ln{x}}{1+x^2}dx+\int\limits_1^{+\infty}\frac{\ln{x}}{1+x^2}dx=$$ $$=\int\limits_0^{1}\frac{-\ln{\frac{1}{x}}}{1+\frac{1}{x^2}}\left(-d\frac{1}{x}\right)+\int\limits_1^{+\infty}\frac{\ln{x}}{1+x^2}dx=$$ $$=\int\limits_{+\infty}^1\frac{\ln{x}}{1+x^2}dx+\int\limits_1^{+\infty}\frac{\ln{x}}{1+x^2}dx=0.$$ $\int\limits_1^{+\infty}\frac{\ln{x}}{1+x^2}dx$ converges because there is $a>1$ for which: $$\int\limits_a^{+\infty}\frac{\ln{x}}{1+x^2}dx<\int\limits_a^{+\infty}\frac{1}{x^{1.5}}dx.$$

Answer 2

As a side-effect of your analysis, it's enough to check only one of $x\to0$ and $x\to\infty$ as they contribute equally.

For small $x$, then, what is an obviously good approximation for $\frac{1}{1+x^2}$ which happens to be an upper bound for it?

Answer 3

For large $x$ is well known that $\ln x\leq\sqrt x$, then $\dfrac{\ln x}{1+x^2}\leq\dfrac{1}{x^{\frac{3}{2}}}$, so at infinity the integral converges.

For small $x$ we have $-\dfrac{\ln x}{1+x^2}\sim-\ln x$. Integral $-\displaystyle\int\limits_0^\delta\ln xdx$ can be calculated by integration by parts and it is finite.