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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.

saddysaw
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  • 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. – macton Jan 29 '21 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. – macton Jan 29 '21 at 11:37
  • @macton , there is a thing called uniform convergence of improper integrals. – saddysaw Jan 29 '21 at 11:37
  • Related: https://math.stackexchange.com/questions/2505047/how-do-you-evaluate-int-infty-infty-frac-cosxx21dx-without-usin , https://math.stackexchange.com/questions/2372949/evaluate-the-integral-int-limits-infty-infty-frac-cosxx21dx , https://math.stackexchange.com/questions/3755888/solution-to-a-general-integral-int-0-infty-frac-costxx2k2e-sxdx – V.G Jan 29 '21 at 11:39
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    https://math.stackexchange.com/questions/3942333/study-the-uniform-convergence-of-the-integral-int-0-infty-dfrac-sin-alpha-x – V.G Jan 29 '21 at 11:40
  • My fault for not understand the topic correctly. Still the limit should be fixed like $$\lim_{N\to\infty} 2\int_{0}^{N}\displaystyle{\frac{1}{4+x^2}},dx$$ – macton Jan 29 '21 at 11:41

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