Interchanging two limits in front of an integral.

Question

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.

Answer 1

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.

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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.

Answer 2

It seems that you are thinking to complicate. This follows by the multiplicativity of the integral and the limes. for constants $a,b,c$ : $$\int_a^b cf(x)dx=c \int_a^b f(x)dx\tag{1}$$ $$\lim_{x\to a} c f(x) = c\lim_{x\to a} f(x)\tag{2}$$

We have $$\lim_{\alpha\to0^+}\biggl(\lim_{M\to+\infty}\int_{-M}^{M}\dfrac{t\sin(tx)}{\alpha^2+x^2}dt\biggr)=\lim_{\alpha\to0^+}\biggl(\lim_{M\to+\infty}\int_{-M}^{M}\dfrac{1}{\alpha^2+x^2}t\sin(tx)dt\biggr)$$ From $(1)$ we get

$$=\lim_{\alpha\to0^+}\biggl(\lim_{M\to+\infty}\dfrac{1}{\alpha^2+x^2}\int_{-M}^{M}t\sin(tx)dt\biggr)$$ and from $(2)$ we get

$$=\lim_{\alpha\to0^+}\biggl(\dfrac{1}{\alpha^2+x^2}\lim_{M\to+\infty}\int_{-M}^{M}t\sin(tx)dt\biggr)$$

and again from $(2)$ we get $$=\biggl(\lim_{\alpha\to0^+}\dfrac{1}{\alpha^2+x^2}\biggr)\lim_{M\to+\infty}\int_{-M}^{M}t\sin(tx)dt$$

Now we use $(2)$ to interchange the limits by viewing the other limit expression as constant: $$=\lim_{M\to+\infty}\biggl(\lim_{\alpha\to0^+}\dfrac{1}{\alpha^2+x^2}\biggr)\int_{-M}^{M}t\sin(tx)dt$$ and so on until we arrive at

$$\lim_{M\to+\infty}\biggl(\lim_{\alpha\to0^+}\int_{-M}^{M}\dfrac{t\sin(tx)}{\alpha^2+x^2}dt\biggr)$$