Uniform convergence for improper integrals dependig on a parameter

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I'm having troubles to understand how apply certain covergence criteria for improper integrals. In the book "Elements of real analysis" of Robert G. Bartle we have at the problem set 25.M:

"For wich values of t do the following infinite integrals converges uniformly?" $$a)\int_{0}^{\infty}\frac{dx}{x^2+t^2}$$ and $$b)\int_{0}^{\infty}\frac{dx}{x^2+t}$$

My atemmp to "a)" is: Since $\dfrac{1}{x^2+t^2}\leq \dfrac{1}{x^2}$ we can use the Weirestrass M-Test but the integral depending only in $x$ diverges when the lower limit is $0$. In "b)" we can argue for $t>0$. However in several post on internet and in the "hints for the selected excercises" section of the book said that "a)" and "b)" converges uniformly for every value of $t$. Could anyone explain me this?

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The integrals diverge for any real $t \leqslant 0$, so we only need to consider uniform convergence for values of $t$ in the interval $(0,\infty)$.

For any $\alpha > 0$, both integrals are uniformly convergent for all $t \in [\alpha,\infty)$. This follows easily from the Weierstrass M-test, since $x \mapsto (x^2 +c)^{-1}$ is integrable for any $c > 0$ and

$$\left|\frac{1}{x^2+t^2}\right| \leqslant \frac{1}{x^2 + \alpha^2}, \quad \left|\frac{1}{x^2+t}\right| \leqslant \frac{1}{x^2 + \alpha}$$

This leaves the question of uniform convergence for $t \in (0,\infty)$. If the interval of integration were $[a,\infty)$ with $a > 0$, then the convergence of both improper integrals would be unequivocally uniform by the Weierstrass M_Test since for $m = 1,2$ and all $t \in (0,\infty)$

$$\left|\frac{1}{x^2 + t^m}\right|\leqslant \frac{1}{x^2}, \quad \int_a^\infty \frac{dx}{x^2} = \frac{1}{a} < \infty$$

Furthermore, we can also argue that convergence is uniform when the interval of integration is $[0,\infty)$ despite the singular behavior of the integrand as both $x,t \to 0$.

This follows from the Cauchy criterion for uniform convergence which states that the improper integral of $x \mapsto f(x,t)$ over $[0,\infty)$ is uniformly convergent for $t \in I$ if and only if the integrand is Riemann integrable over any finite interval and for every $\epsilon > 0$ and there exists $K > 0$ such that for all $c_2 > c_1 > K$ and $t \in I$ we have

$$\tag{*}\left|\int_{c_1}^{c_2} f(x,t) \, dx \right|< \epsilon$$

We have already shown that for both integrands (a) and (b), the improper integrals taken over an interval such as $[1, \infty)$ are uniformly convergent for $t \in [0,\infty)$. Hence, there exists $K > 1$ for which (*) is satisfied for any $\epsilon > 0$ and all $t \in (0,\infty)$.