# Turning sum into integral when the widths of the intervals are not uniform

I have a sum of the form $$S = \sum_\lambda f(\lambda),$$ where the "index" of the sum solves the equation $$\tan(C \lambda) = \lambda.$$

This equation has a countably infinite number of solutions. One can see that as $$C$$ gets large, these solutions get closer and closer together, so it should be possible to express $$S$$ as an integral as $$C \rightarrow \infty$$, but I am unclear how to achieve this.

How can I introduce the vanishing width of the intervals into the sum to make a Riemann-type sum for $$S$$?

Attempt:

Set $$S = \lim_{N\rightarrow \infty}\sum_{n=-N}^N f(\lambda_n)$$ where $$\tan(C\lambda_n) = \lambda_n$$ and the $$\lambda_n$$ are sorted ($$\lambda_{n-1}< \lambda_n$$ for all $$n$$). Then $$S = \lim_{N\rightarrow \infty} 2N\sum_{n=-N}^N f(2N\frac{\lambda_n}{2N})\frac{1}{2N} \sim \lim_{N\rightarrow \infty} 2N\int_{-N}^N f(2Nz)dz$$

Is this correct? Or am I missing something here? I suppose there are some strong constraints required on $$f$$ such that $$S$$ converges.

• The Riemann sum requires that the intervals all tend to zero in the limit. It does not matter if their relative size varies; the key thing is that as $n\to\infty$, $\Delta x$ must tend to $0$ Aug 22, 2021 at 17:45
• In that case @Fshrike I suppose I'm left wondering how to convert a sum of the form $S$ with an unspecified $f(\lambda)$ in it into a Riemann sum. I wonder if my last line seems appropriate Aug 22, 2021 at 18:11
• I am not sure. I do know that one can take integrals over discrete measure spaces (someone did this in an answer to one of my questions, converting a sum into an integral) Aug 22, 2021 at 18:51

We prove:

Claim. Consider $$f : \mathbb{R} \to \mathbb{R}$$ such that

1. $$f$$ is locally Riemann integrable on $$\mathbb{R}$$, that is, $$f$$ is Riemann integrable on each closed bounded subinterval of $$\mathbb{R}$$,

2. $$\phi(r) = \sup_{|x| \geq r} |f(x)|$$ is integrable on $$[0, \infty)$$.

Then we have

$$\lim_{C \to \infty} \frac{1}{C} \sum_{\lambda : \tan(C\lambda) = \lambda} f(\lambda) = \frac{1}{\pi} \int_{-\infty}^{\infty} f(x) \, \mathrm{d}x.$$

For the proof, if $$C > 1$$ then we note that the equation $$\tan(C\lambda) = \lambda$$ has a unique solution $$\lambda_n$$ in each interval $$I_n = [\frac{n\pi}{C}-\frac{\pi}{2C}, \frac{n\pi}{C}+\frac{\pi}{2C}]$$. This allows us to write

$$\sum_{\lambda : \tan(C\lambda) = \lambda} f(\lambda) = \sum_{n\in\mathbb{Z}} f(\lambda_n),$$

provided the sum converges absolutely.

1. From this, for each fixed $$r > 0$$,

$$S_{C,r} := \sum_{n\in\mathbb{Z}} f(\lambda_n) \operatorname{length}(I_n \cap [-r, r])$$

is a Riemann of $$f$$ over $$[-r, r]$$. (Even though the index $$n$$ runs over the infinite set $$\mathbb{Z}$$, only finitely many terms are non-zero and hence $$S_{n,r}$$ is essentially a finite sum.) Also, the mesh of the associated tagged partition decreases to $$0$$ as $$C \to \infty$$, and so, we have

$$\lim_{C\to\infty} S_{C,r} = \int_{-r}^{r} f(x) \, \mathrm{d}x.$$

2. On the other hand, for $$|n| \geq 2$$,

$$\Bigl( \sup_{I_n}|f| \Bigr) \operatorname{length}(I_n) \leq \phi\biggl( \frac{|n|\pi}{C}-\frac{\pi}{2C} \biggr) \operatorname{length}(I_n) \leq \int_{\frac{|n|\pi}{C}-\frac{3\pi}{2C}}^{\frac{|n|\pi}{C}-\frac{\pi}{2C}} \phi (x) \, \mathrm{d}x,$$

and so, if $$C$$ is sufficiently large so that $$r > \frac{5\pi}{2C}$$, then

$$\sum_{n\in\mathbb{Z}} | f(\lambda_n) | \operatorname{length}(I_n \setminus [-r, r]) \leq 2\int_{r-\frac{5\pi}{2C}}^{\infty} \phi (x) \, \mathrm{d}x.$$

This in particular shows that the sum $$\sum_n f(\lambda_n)$$ converges absolutely.

3. On the other hand, since $$|f(x)| \leq \varphi(|x|)$$ and $$\varphi(|x|)$$ is integrable on $$\mathbb{R}$$, $$\int_{-\infty}^{\infty} f(x) \, \mathrm{d}x$$ converges absolutely. Moreover, it is clear that

$$\left| \int_{|x|\geq r} f(x) \, \mathrm{d}x \right| \leq 2 \int_{r}^{\infty} \varphi(x) \, \mathrm{d}x.$$

4. Finally, combining all the observations, we find that

$$\limsup_{C\to\infty} \, \Biggl| \frac{\pi}{C} \sum_{\lambda : \tan(C\lambda) = \lambda} f(\lambda) - \int_{-\infty}^{\infty} f(x) \, \mathrm{d}x \Biggr| \leq 4\int_{r}^{\infty} \phi (x) \, \mathrm{d}x$$

Since this is true for any $$r > 0$$, letting $$r\to \infty$$ proves the desired claim.