Prove of refute uniform convergence of integral.

Question

I can't prove (or refute) uniform convergence of

$$ \int_{1}^{+\infty} \arctan\frac{2y}{x^2+y^2}dx,\;y\in\mathbb{R} $$

I tried Weierstrass M-test, but failed. Can't find the solution to this.

Answer 1

Note that $\arctan(\cdot)$ is increasing on $[0,1]$ and for $x \geqslant 1$ and $y > 0$ we have

$$0 \leqslant \frac{2y}{x^2 + y^2} \leqslant \frac{2xy}{x^2 + y^2} \leqslant 1$$

Thus,

$$\sup_{y \in \mathbb{R^+}}\int_n^\infty \arctan\frac{2y}{x^2 + y^2} \, dx > \sup_{y \in \mathbb{R^+}}\int_n^{2n} \arctan\frac{2y}{x^2 + y^2} \, dx \\ > \sup_{y \in \mathbb{R^+}}\left(n\arctan\frac{2y}{(2n)^2 + y^2}\right) \\ \geqslant n\arctan\frac{2n}{4n^2 + n^2 } \\= n\arctan\frac{2}{5n } $$

Using the Taylor expansion $\arctan x = x + \mathcal{O}(x^3)$ for $x \in (-1,1]$ we have

$$\sup_{y \in \mathbb{R^+}}\int_n^\infty \arctan\frac{2y}{x^2 + y^2} \, dx > n\left[\frac{2}{5n} + \mathcal{O}\left(\frac{1}{n^3} \right)\right] \\ = \frac{2}{5} + \mathcal{O} \left(\frac{1}{n^2} \right)$$

Since the RHS does not converge to $0$ as $n \to \infty$, the convergence of the improper integral is not uniform for $y \in [0,\infty)$ and, consequently, for $y \in \mathbb{R}$.

Answer 2

$$\DeclareMathOperator\sgn{sgn} f(A,y)=\int_1^A \frac {2y\, \mathrm dx}{x^2+y^2} = \int_1^A \frac {2y\cdot y}{y^2} \cdot \frac {\mathrm d(x/y)}{(x/y)^2 + 1} = 2\int_{1/y}^{A/y} \frac {\mathrm d u}{1+u^2} = 2\arctan (A/y) - 2\arctan (1/y) \to \pi \sgn y - 2\arctan (1/y) [A\to +\infty, y \neq 0], $$ and $f(A,0) = 0$. So the limit function $g(y) = 2\cdot 1_{\mathbb R \setminus \{0\}} (y)(\pi \sgn y - \arctan (1/y))$.

Now consider $$ \sup_{y \in \mathbb R^*} \vert f(A,y) - g(y) \vert = \sup_{\mathbb R^*} \vert 2 \arctan (A/y) - \pi \sgn y\vert = \sgn_{\mathbb R^+} \vert 2\arctan (A/y) - \pi \vert \geqslant \vert 2\arctan (A/A) -\pi \vert =\frac\pi 2, $$ thus the integral does not converge uniformly.