Can somebody explain to me the convergence of $\int_0^\infty t^{2k}e^{-xt^2}dt $ being uniform on $[\delta,\infty)$ for $\delta\gt 0$ ?
Also according to the theorem above shouldn't be $t^{2k}e^{-xt^2}\leq t^{2k}e^{-\delta t^2}$ for $t\geq \delta$ instead of $x\geq \delta$ ? And how do we show $\int_\delta^{\infty} t^{2k}e^{-\delta t^2}dt \lt \infty$?
The theorem you quote is sometimes called 'dominated convergence,' and the idea is that if the family of functions lives within a region cut out by an integrable function, then the family of functions cannot escape and be wild.
The observation is that as a function in $x$, we have $f(x, \cdot) = t^{2k}e^{-xt^2}$ which is monotonically decreasing as $x$ increases and $t$ is held constant, but not integrable at $0$, as now you're just integrating an increasing polynomial over a ray to $\infty$. All one has to do then is check that this function is integrable, and then the claim follows on any ray $[a, \infty)$ which does not contain $0$, since the function is not defined there. Thus, taking $\delta > 0$, no matter how small, suffices.
The value of that last integral is not easy to find by hand, but if you know about the Gamma function, then you can show that it can be expressed in those terms. There's a calculation available here, if you enter the integral.