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.



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