# Non-uniform convergence of improper integrals II

I'm looking at approaches for proving or disproving uniform convergence of improper integrals. I asked this question:

Techniques for proving non-uniform-convergence of improper integrals

but I realized later it was a trivial example since the integral does not depend on non-zero values of the parameter.

A better example is $$y \in \mathbb{R}$$ and

$$\int_2^\infty \frac{y}{y^2 + x^2} \frac{dx}{\ln(x)}$$

I first tried to show it is not uniformly convergent for $$y \in \mathbb{R}$$ because the Cauchy condition isn’t satisfied:

$$\int_a^b\frac{y}{y^2 + x^2} \frac{dx}{\ln(x)} > \frac{y}{y^2 + b^2}\frac{b - a}{\ln(b)}$$

but whatever I try for $$y$$ such as $$y=b$$ doesn’t make the right-side not go to $$0$$ as $$a,b \to \infty$$.

So I then tried to show the Cauchy condition is satisfied with

$$\int_a^b\frac{y}{y^2 + x^2} \frac{dx}{\ln(x)} < \frac{y}{y^2 + a^2}\frac{b - a}{\ln(a)} < \frac{1}{2a}\frac{b - a}{\ln(a)}$$

This seems to work insofar as the right side going to $$0$$, but this must be true for all $$b > a$$ and this is where I think it breaks down. How can I prove or disprove the uniform convergence?

The convergence is uniform, which is why your first attempt failed. Also, your second argument is insufficient because you could take $$b = a^2$$ and find
$$\frac{1}{2a}\frac{b - a}{\ln(a)} = \frac{1}{2}\frac{a - 1}{\ln(a)} \xrightarrow[a \to \infty]{} \infty$$
To prove it more directly in detail, note that with $$b > a > 2$$ we have
$$\left|\int_a^b\frac{y}{y^2 + x^2} \frac{dx}{\ln x}\right| < \frac{1}{\ln a}\int_a^b\frac{|y|}{y^2 + x^2} \, dx = \frac{1}{\ln a}\int_{a/|y|}^{b/|y|}\frac{du}{1 + u^2} < \frac{\pi}{2\ln a}$$
Since the RHS converges to $$0$$ as $$a \to \infty$$ independently of $$y \in \mathbb{R}$$, the Cauchy criterion for uniform convergence of the improper integral is satisfied.