The uniform convergence of $\int_0^\infty \mathrm{e}^{-ax^2}x^{2n}\mathrm{d}x \quad w.r.t \ a$

Question

For fixed $n\in \mathbb{N}$, study the uniform convergence of $$\int_0^\infty \mathrm{e}^{-ax^2}x^{2n}\mathrm{d}x \quad w.r.t \ a\in(0,\infty)$$.

Actually I want to compute the integral by differentiating $$ \int_0^\infty \mathrm{e}^{-ax^2}\mathrm{d}x=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}, $$ but we need to prove $$\int_0^\infty \mathrm{e}^{-ax^2}x^{2n+2}\mathrm{d}x $$ converges uniformly w.r.t $a>0$. Then the procedure is legal. The Weierstrass, Dirichlet and Abel criterion all seem do not work.

Appreciate any help!

Answer 1

The convergence of the improper integral is uniform for $a$ in any interval $[A,\infty)$ where $A > 0$.

This follows from the Weierstrass M-test since $|e^{-ax^2}x^{2n}| \leqslant e^{-Ax^2}x^{2n}$ for all $a \geqslant A$ and for any fixed $n$ the improper integral $\int_0^\infty e^{-Ax^2}x^{2n} \, dx$ converges.

However, for any $c > 0$ and $a_c = 1/c^2 \in (0,\infty)$,

$$\sup_{a \in (0,\infty)}\left|\int_c^{\infty} e^{-ax^2}x^{2n} \, dx\right| \geqslant \sup_{a \in (0,\infty)}\int_c^{2c} e^{-ax^2}x^{2n} \, dx\geqslant \sup_{a \in (0,\infty)}c \cdot e^{-a (2c)^2} c^{2n}\geqslant e^{-4a_c c^2} c^{2n+1}\\ = e^{-4}c^{2n+1} \underset{c \to \infty}\longrightarrow \infty$$

Thus, the improper integral fails to converge uniformly (with respect to $a$) on $(0,\infty)$.

Answer 2

Hint

$$I_n=\int_0^\infty {e}^{-ax^2}x^{2n}\,dx$$ $$x=\sqrt t \implies I_n= \frac{1}{2}\int_0^\infty e^{-a t} t^{n-\frac{1}{2}}\,dt=\frac{1}{2} a^{-(n+\frac{1}{2})} \Gamma \left(n+\frac{1}{2}\right)$$