# Relation between Riemann integral and Lebesgue integral on an unbounded interval

The original problem is to compute $$\lim_{n \to \infty} \int_{\mathbb{R}}\frac{\cos^n x}{1+x^2} dm(x),$$ where $$m$$ is the Lebesgue measure. I think this can be done by observing that $$f_n(x):= \frac{\cos^n x}{1+x^2}$$ is bounded above by $$g(x) := \frac{1}{1+x^2}$$, which is integrable on $$\mathbb{R}$$, and also that $$f_n(x)$$ converges to $$0$$ a.e. Hence, by the dominated convergence theorem, the answer is $$0$$.

My question is whether the following limit of Riemann integral (instead of Lebesgue integral) also makes sense and is equal to the above : $$\lim_{n \to \infty} \int_{-\infty}^{\infty}\frac{\cos^n x}{1+x^2} dx.$$

Here is my attempt: Using $$f_n(x) \leq g(x)$$ for all $$x$$ and $$\int_{-\infty}^{\infty}g(x) dx = \pi$$, we can say that the improper (Riemann) integral exists (and is less than or equal to $$\pi$$), which depends on $$n$$. But it is not clear to me why the limit exists.

Another thing I know is the following result:

Let $$f$$ be a bounded function on a compact interval $$[a, b]$$. If $$f$$ is Riemann integrable on $$[a, b]$$, then $$f$$ is Lebesgue integrable on $$[a, b]$$, and $$\int_a^b f(x) dx = \int_{[a, b]} f(x) dm(x).$$

But this result requires a compact interval, which is not the case here.

To summarize, could anyone give a rigorous explanation of how to show $$\lim_{n \to \infty} \int_{\mathbb{R}}\frac{\cos^n x}{1+x^2} dm(x) = \lim_{n \to \infty} \int_{-\infty}^{\infty}\frac{\cos^n x}{1+x^2} dx,$$ if it is indeed true? Thank you.

We can easily extend the theorem you quoted to handle unbounded intervals:

Theorem $$\star$$

Suppose $$f:\Bbb{R}\to\Bbb{C}$$ is a function with the following properties:

• $$f$$ is improperly Riemann-integrable on $$\Bbb{R}$$ (i.e for every $$-\infty, the restricted map $$f|_{[a,b]}:[a,b]\to\Bbb{R}$$ is properly Riemann-integrable, and the limit $$\lim\limits_{\substack{b\to\infty\\a\to-\infty}}\int^b_af(x)\,dx$$ exists in $$\Bbb{R}$$; this limit being denoted $$\int_{-\infty}^{\infty}f(x)\,dx$$).
• $$f$$ is Lebesgue-integrable on $$\Bbb{R}$$: $$\int_{\Bbb{R}}|f|\,dm<\infty$$ (Lebesgue-measurability of $$f$$ is guaranteed by the previous condition)

Then, the two integrals coincide: \begin{align} \int_{\Bbb{R}}f\,dm&=\int_{-\infty}^{\infty}f(x)\,dx. \end{align}

The proof is immediate from the theorem you quoted and dominated convergence: \begin{align} \int_{\Bbb{R}}f\,dm&=\lim_{n\to\infty}\int_{[-n,n]}f\,dm\tag{DCT}\\ &=\lim_{n\to\infty}\int_{-n}^nf(x)\,dx\tag{your quoted theorem}\\ &=\int_{-\infty}^{\infty}f(x)\,dx, \end{align} where the final equality is because $$f$$ is assumed to be improperly Riemann-integrable.

Next, you write

Here is my attempt: Using $$f_n(x) \leq g(x)$$ for all $$x$$ and $$\int_{-\infty}^{\infty}g(x) dx = \pi$$, we can say that...

Well, this by itself is not good enough. You need to say $$|f_n(x)|\leq g(x)$$ for all $$x$$; the absolute values are important (and I assume you know why this inequality is sufficient to guarantee the improper Riemann-integrability of $$f_n$$ on $$\Bbb{R}$$).

Finally, we know that for each $$n\in\Bbb{N}$$, the function $$f_n$$ satisfies the conditions of theorem $$(\star)$$. So, $$\int_{-\infty}^{\infty}f_n(x)\,dx=\int_{\Bbb{R}}f_n\,dm$$. By the dominated convergence theorem (applied to the right side), we know the limit exists and equals $$0$$. Therefore, \begin{align} \lim_{n\to\infty}\int_{-\infty}^{\infty}\frac{\cos^n(x)}{1+x^2}\,dx&=\lim_{n\to\infty}\int_{\Bbb{R}}\frac{\cos^n(x)}{1+x^2}\,dm(x)=0. \end{align} Hopefully it's clear which are (improper)Riemann integrals and which are Lebesgue integrals.

• Thank you for your detailed answer. Commented Jun 8, 2022 at 22:34