I am taking a course in real analysis (materials include Measure theory, Lebesgue integral, etc.) and I couldn't justify a statement in the course material. Consider
$$\lim_{\alpha\to0^+}\biggl(\lim_{M\to+\infty}\int_{-M}^{M}\dfrac{t\sin(tx)}{\alpha^2+t^2}dt\biggr)$$
My instructor asserted that this is equal to
$$\lim_{M\to+\infty}\biggl(\lim_{\alpha\to0^+}\int_{-M}^{M}\dfrac{t\sin(tx)}{\alpha^2+t^2}dt\biggr)$$
i.e. we can interchange the two limits infront of the integral. But I don't see why this is true. I have checked out this answer, Here is what I did: set
$a_{m,n}=\int_{-M_m}^{M_m}\dfrac{t\sin(tx)}{\alpha_n^2+t^2}dt$ where $M_m\uparrow+\infty$ and $\alpha\downarrow0$. Then
$$\lim_{\alpha\to0^+}\biggl(\lim_{M\to+\infty}\int_{-M}^{M}\dfrac{t\sin(tx)}{\alpha^2+t^2}dt\biggr)=\lim_{n\to\infty}(\lim_{m\to\infty}a_{m,n})=\lim_{n\to\infty}\sum_{m=1}^\infty c_{m,n}$$
if we put $c_{m,n}=a_{m,n}-a_{m-1,n}$ and take $a_{0,n}=0$ for every n. But then its easy to see that $\{c_{m,n}\}$ is neither nonnegative nor absolutely summable (it is not nonnegative because $\sin(tx)$ could be negative, its not absolutely summable because $\frac{sinx}{x}$ is not integrable), so I can't interchange the infinite sum and the limit using Monotone convergence theorem or Dominated convergence theorem. Hence I am stuck.
Any help is appreciated, thank you for all your help in advance.
Edit: sorry I made a typo (the $x^2$ on the denominator should really be $t$). But I think proving uniform convergence in $M$ would still work.
Best Answer
Note: There was a mistake in the answer, I was integrating with respect to $x$...
If the integral is with respect to $t$ instead, since $\lim_{t \to \infty} t \sin(tx) \neq 0$ for all $x \neq 0$, the integral $$ \lim_{M\to+\infty}\int_{-M}^{M}\dfrac{t\sin(tx)}{\alpha^2+x^2}dt=2\lim_{M\to+\infty}\int_{0}^{M}\dfrac{t\sin(tx)}{\alpha^2+x^2}dt $$ is divergent, so both sides are divergent in this case.
Note that $$ \left| \int_{-M}^{M}\dfrac{t\sin(tx)}{\alpha^2+x^2}dx -\int_{-M}^{M}\dfrac{t\sin(tx)}{x^2}dx \right|= \alpha^2\left| \int_{-M}^{M}\dfrac{t\sin(tx)}{(\alpha^2+x^2)x^2}dx \right| \\ \leq \alpha^2\left| \int_{-M}^{M}\dfrac{t\sin(tx)}{x^4}dx \right| \leq \alpha^2 \int_{-\infty}^{\infty}\dfrac{|t\sin(tx)|}{x^4}dx $$
This shows that $\lim_{\alpha\to0^+}\int_{-M}^{M}\dfrac{t\sin(tx)}{\alpha^2+x^2}dx=\int_{-M}^{M}\dfrac{t\sin(tx)}{x^2}dx$ uniformly in $M$
You can use this uniformity to show that the limits can be interchanged, it is an instructive and relatively easy exercise.
This idea is usually called the Moore-Osgood theorem.