To test the convergence of $\int_0^{1} \frac{x^p \log x}{1+ x^2} dx$

Question

To test the convergence of the improper integral:

$$\int_0^{1} \frac{x^p \log x}{1+ x^2} dx$$

Here we see that $0$ is the point of infinite discontinuty for $p<0$.

Let $f(x) = \frac{x^p \log x}{1+ x^2}$ and $g(x) = x^p$.

Now as we take..

$$\lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{\log x}{1+ x^2}$$

Will this $g(x)$ work? Not getting clue to solve it. Thanks in advance.

Answer 1

You may observe that $$ \frac{x^p \log x}{1+ x^2} \sim x^p \log x, \quad \quad x \to 0^+. $$ The integral $$ \int_0^1x^p \log x \: dx $$ is convergent if and only if $p>-1$, with (using integration by parts) the value $$ \int_0^1x^p \log x \: dx =-\frac{1}{(p+1)^2}, \quad \quad p>-1. $$ Then $$ \int_0^1\frac{x^p \log x}{1+ x^2} \:dx $$ is convergent if and only if $p>-1$.