# Uniform convergence of the Integral depending on a parameter

Prove uniform convergence of the following Integral for every $$\alpha\in E=\left[-\infty;+\infty\right]$$:

$$I(\alpha)=\int_{-\infty}^{+\infty}\displaystyle{\frac{\cos(\alpha x)}{4+x^2}}\,dx$$

Here's my approach: Since $$f(x,\alpha)=\displaystyle{\frac{\cos(\alpha x)}{4+x^2}}$$ is even function we can write our integral as follows $$I(\alpha)=2\int_{0}^{+\infty}\displaystyle{\frac{\cos(\alpha x)}{4+x^2}}\,dx.$$

From Weierstrass's test we have $$\left|f(x,\alpha)\right|\le\left|\displaystyle{\frac{\cos(\alpha x)}{4+x^2}}\right|\le\displaystyle{\frac{1}{4+x^2}}$$ and $$\displaystyle { \lim_{N\to\infty} 2\int_{0}^{N}\displaystyle{\frac{1}{4+x^2}}\,dx } = 2\int_{0}^{+\infty}\displaystyle{\frac{1}{4+x^2}}\,dx = 4\arctan(\displaystyle{\frac{x}{2}})\bigg\rvert_{0}^{\infty}=4(\displaystyle{\frac{\pi}{2}} - 0) = 2\pi$$. Thus, this integral converges uniformly.

Am I allowed to bound integrand by $$\displaystyle{\frac{1}{4+x^2}}$$, cause there are some problems of $$\cos(\alpha x)$$ in E. I guess it's not that simple, any hint will be helpful.

• Um, if I understand correctly, uniform convergence is used to describe the convergence behavior of a sequence of functions $f_n \to f$. What you here is calculating the improper integral given a parameter. Jan 29, 2021 at 11:34
• Also $$\lim_{x\to\infty} 2\int_{0}^{+\infty}\displaystyle{\frac{1}{4+x^2}}\,dx$$ is very questionable: you don't take limit with respect to the dummy variable in the integrand. Jan 29, 2021 at 11:37
• @macton , there is a thing called uniform convergence of improper integrals. Jan 29, 2021 at 11:37
• – V.G
Jan 29, 2021 at 11:39
• math.stackexchange.com/questions/3942333/…
– V.G
Jan 29, 2021 at 11:40