Prove the continuity and differentiability of parametric integration

Question

$$F(\alpha )=\int_{0}^{+ \infty } \frac{\cos x}{1+(x+\alpha )^{2} } dx$$ Prove the function F is continuous and differentiable on the interval $[0, +\infty )$

Answer 1

$|F(\alpha) - F(\beta)| \leq \displaystyle \int_{0}^\infty \left|\dfrac{cosx}{1+(x+\alpha)^2} - \dfrac{cosx}{1+(x+\beta)^2}\right|dx \leq |\beta - \alpha|\cdot \displaystyle \int_{0}^\infty \dfrac{(x+\alpha)+(x+\beta)}{(1+(x+\alpha)^2)(1+(x+\beta)^2)}dx \leq |\alpha - \beta|\cdot \left(\displaystyle \int_{0}^\infty \dfrac{(x+\alpha)}{(1+(x+\alpha)^2)(1+(x+\beta)^2)}dx + \displaystyle \int_{0}^\infty \dfrac{(x+\beta)}{(1+(x+\alpha)^2)(1+(x+\beta)^2)}dx\right) \leq \dfrac{|\alpha - \beta|}{2}\cdot \left(\displaystyle \int_{0}^\infty \dfrac{1}{1+(x+\beta)^2}dx + \displaystyle \int_{0}^\infty \dfrac{1}{1+(x+\alpha)^2}dx\right) = \dfrac{|\alpha - \beta|}{2}\cdot \left(\displaystyle \int_{\beta}^\infty \dfrac{1}{1+x^2}dx + \displaystyle \int_{\alpha}^\infty \dfrac{1}{1+x^2}dx\right) = \dfrac{|\alpha - \beta|}{2}\cdot \left(\dfrac{\pi}{2} - tan^{-1}\beta + \dfrac{\pi}{2} - tan^{-1}\alpha\right) \leq \dfrac{|\alpha - \beta|}{2}\cdot \left(\dfrac{\pi}{2} + \dfrac{\pi}{2} + \dfrac{\pi}{2} + \dfrac{\pi}{2}\right) = \pi\cdot |\alpha - \beta|$. This shows $F$ is Lipschitz and therefore is uniformly continuous, hence continuous on the indicated interval. You can easily show it is differentiable there.