The Definite Integral Problem (with a twist)?

The Definite Integral Problem (with a twist)

In the Riemann integral one essentially calculates the area by splitting the area into $N$ rectangular strips and then taking $N \to \infty$.

Here’s something I asked myself related to the Riemann integral.

Let’s say I split the area into say $3$ strips however I recount the first strip $a_1$ times, the $2$‘nd strip $a_2$ times and the third strip $a_3$ times.

N is 3

Similarly we ask about $N= 4$ and recount the first strip $a_1$ times, the $2$‘nd strip $a_2$ times, the third strip $a_3$ times and the fourth strip $a_4$ times:

N is 4

Now while this all is doable (but hardwork ?) is there any way to make it work for the case $N \to \infty$ after which I take the limit $k \to \infty$

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Notice in the above pictures the area beneath the curve (in the Riemann integration sense where $a_r=1$ for all $r$) is obviously infinite. Let’s add the conditions that the curve $f(x)$ is a smooth continuous function whose integral $\int_0^\infty f(x) d x $ is absolutely convergent.

Conjectured solution

I discovered the following relation for arbitrary $a_r$:

$$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n a_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{a_r}{r^s}}_{\text{removable singularity}} \int_0^\infty f(x) \, dx $$

Where $f(x)$ is a smooth continuous function whose integral $\int_0^\infty f(x) d x $ is absolutely convergent. $a_r$ is the $r$‘th number of a sequence.

Heuristic Proof (a lot of cheating involved)

Consider an integral such that $$ \int_0^\infty f(x) \, dx = C,$$where, $f(x)$ is a smooth and continuous function and absolutely converges.

Now we raise both sides to the power s:

$$\left(\int_0^\infty f(x) \, dx\right)^s = C^s $$

We substitute $x$ with $rx$ to get:

$$\left(\int_0^\infty f(rx) \, dx\right)^s = (C/r)^s $$

Multiplying both sides by an arbitrary coefficient:

$$ (b_r)\left(\int_0^\infty f(rx) \, dx\right)^s = (b_r)( C/r)^s $$

Taking their sum:

$$ \sum_{r=1}^\infty b_r \left(\int_0^\infty f(rx) \, dx\right)^s = C^s \underbrace{\sum_{r=1}^\infty \frac{b_r}{r^s}}_{\text{dirichlet series}} $$

We write the integral as a limit of a Riemann sum:

$$ \sum_{r=1}^\infty \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right)^s = C^s \sum_{r=1}^\infty \frac{b_r}{r^s} $$

Using the mobius inversion formula:

$$ \sum_{r=1}^\infty \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right)^s = C^s \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_r}{r^s} $$

We define $ a_r = \sum_{e|r} b_e $

Note:

$$ (\frac{b_1}{1^s} + \frac{b_2}{2^s} + \frac{b_3}{3^s} + \frac{b_4}{4^s} + \dots) \times (\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \dots) = \frac{b_1}{1^s} + \frac{b_1 + b_2}{2^s} + \frac{b_1 + b_3}{3^s} + \frac{b_1 + b_2 + b_4}{4^s} + \dots $$

Now focusing on the L.H.S ($s \nearrow 1 $):

$$ \sum_{r=1}^\infty \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right)^s = \sum_{r=1}^\infty \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right) $$

Focusing on the L.H.S (and vertically summing):

$$ \lim_{k \to \infty }\lim_{n \to \infty} b_1 ((f(\frac{k}{n}) + f(2 \frac{k}{n}) + f(3 \frac{k}{n}) +f(4 \frac{k}{n}) + \cdots)\frac{k}{n} $$
$$+$$
$$ \lim_{n \to \infty} b_2 (0.f(\frac{k}{n}) + f(2 \frac{k}{n}) + 0.f(3 \frac{k}{n}) +f(4 \frac{k}{n}) +\cdots) \frac{k}{n}$$
$$+$$
$$ \vdots $$

$$ = \lim_{n \to \infty} (\underbrace{b_1}_{a_1} (f(\frac{k}{n}) + \underbrace{(b_1 + b_2)}_{a_2}f(2 \frac{k}{n}) + \underbrace{(b_1 + b_3)}_{a_3}f(3 \frac{k}{n}) +\underbrace{(b_1 + b_2 + b_4)}_{a_4}f(4 \frac{k}{n}) + \cdots)\frac{k}{n} $$

Note: this resummation trick can be only done for special functions ($f$ must absolutely converge)

Hence, for special $a_r$ the L.H.S converges:

$$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n a_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{a_r}{r^s}}_{\text{removable singularity}} \times \int_0^\infty f(x) \, dx $$

Example:

Let $f(x) = e^{-x}$

$$ a_{2r} = 1$$
$$ a_{2r+1} = 0$$

Hence,

Let us compute the R.H.S

$$\lim_{s \to 1} \frac{1}{\zeta(s)} (\frac{2^{(-s)}}{\zeta(s)}) . 1 = \frac{1}{2}$$

Looking at the L.H.S:

We are essentially adding all the even strips! This can be computed also by doing:

$$ \int_{0}^\infty e^{-x} dx = 1 $$
$$ \implies \int_{0}^\infty e^{-2x} d(2x) = 1 $$
$$ \implies \int_{0}^\infty e^{-2x} d(x) = 1/2 $$

Hence both answers match!

Here’s a crazier example with non-periodic $a_r$ but the notation there is ($a_r = d_r$) What is the limit of this Dirichlet series?

Questions from Measure Theory

Is it possible to prove the formula (without cheating :P)?
(edit: answered with brilliance https://math.stackexchange.com/a/3359525/430082)

When all $a_r=1$ for all $r$ then we have a Riemann integral formula. Is it possible to associate the conjectured formula with a measure (the LHS in variable form) rigorously ?

$$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n a_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{a_r}{r^s}}_{\text{removable singularity}} \int_0^\infty f(x) \, dx $$

where the curve $f(x)$ is a smooth continuous function whose integral $\int_0^\infty f(x) d x $ is absolutely convergent

Answer

Let $b_r = \sum_{d \mid r} a_d\mu(\frac{r}{d})$. We prove that if the $b_r$‘s are small enough, the result is true.

Claim: If $\lim_{n \to \infty} \frac{\log^2(n)}{n}\sum_{r=1}^n |b_r| = 0$ and $f$ is smooth, then $$\lim_{k \to \infty} \lim_{n \to \infty} \sum_{r=1}^n a_rf\left(\frac{kr}{n}\right)\frac{k}{n} = \left(\lim_{s \to 1} \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_s}{r^s}\right)\int_0^\infty f(x)dx.$$

Proof: It suffices to show, for any smooth $f$, that $$\lim_{n \to \infty} \sum_{r=1}^n a_rf\left(\frac{rk}{n}\right)\frac{k}{n} = \left(\lim_{s \to 1} \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_r}{r^s}\right)\int_0^k f(x)dx$$ for each $k$. By replacing $f(\cdot)$ with $f(k\cdot)$, it suffices to show $$\lim_{n \to \infty} \sum_{r=1}^n a_rf\left(\frac{r}{n}\right)\frac{1}{n} = \left(\lim_{s \to 1}\frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_r}{r^s}\right)\int_0^1 f(x)dx$$ for any smooth $f$. By mobius inversion, $a_r = \sum_{d \mid r} b_d$, so $$\sum_{r=1}^n a_rf(\frac{r}{n})\frac{1}{n} = \sum_{r=1}^n \sum_{d \mid r} b_d f(\frac{r}{n})\frac{1}{n} = \sum_{d=1}^n \sum_{d \mid r \le n} b_df(\frac{r}{n})\frac{1}{n} = \sum_{d=1}^n b_d\frac{1}{n}\sum_{m \le n/d} f(\frac{md}{n}).$$ Also, $$\lim_{s \to 1} \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_r}{r^s} = \lim_{s \to 1} \sum_{d=1}^\infty \frac{b_d}{d^s} = \sum_{d=1}^\infty \frac{b_d}{d},$$ where the last equality used the decay assumption on the $b_d$‘s (precisely, we used dominated convergence theorem, which is justified, since $\sum_{d=1}^\infty \frac{|b_d|}{d} < \infty$, since, by partial summation, $\sum_{d=1}^\infty \frac{|b_d|}{d} = \int_1^\infty \frac{\frac{1}{t}\sum_{d \le t} |b_d|}{t}dt$). We therefore wish to prove $$\lim_{n \to \infty} \sum_{d=1}^n \frac{b_d}{d}\frac{1}{n/d}\sum_{m \le n/d} f(\frac{md}{n}) = \lim_{n \to \infty} \sum_{d=1}^n \frac{b_d}{d}\int_0^1 f(x)dx.$$ Take $n \ge 1$. Then, $$\left|\sum_{d=1}^n \frac{b_d}{d}\frac{1}{n/d}\sum_{m \le n/d} f(\frac{md}{n})-\sum_{d=1}^n \frac{b_d}{d}\int_0^1 f(x)dx\right| \le \sum_{d=1}^n \frac{|b_d|}{d}\left|\frac{1}{n/d}\sum_{m \le n/d} f(\frac{md}{n})-\int_0^1 f(x)dx\right|.$$ Now, using $$\int_0^1 f(x)dx = \sum_{m \le n/d} \int_{(m-1)d/n}^{md/n} f(x)dx + \int_{\lfloor \frac{n}{d}\rfloor \frac{d}{n}}^1 f(x)dx,$$ we get that, for any $d \le n$, $$\left|\frac{1}{n/d}\sum_{m \le n/d} f(\frac{md}{n})-\int_0^1 f(x)dx\right| \le \sum_{m \le n/d} \int_{(m-1)d/n}^{md/n} \left|f(\frac{md}{n})-f(x)\right|dx + \int_{\lfloor \frac{n}{d}\rfloor\frac{d}{n}}^1 |f(x)|dx.$$ By the mean value theorem, for any $x \in [(m-1)\frac{d}{n},m\frac{d}{n}]$, $|f(\frac{md}{n})-f(x)| \le ||f’||_\infty \frac{d}{n}$, so we bound $$\int_{(m-1)d/n}^{md/n} \left|f(\frac{md}{n})-f(x)\right|dx \le ||f’||_\infty \frac{d}{n}\frac{d}{n}.$$ We also have, using $\lfloor \frac{n}{d}\rfloor\frac{d}{n} \ge 1-\frac{d}{n}$, the bound $$\int_{\lfloor \frac{n}{d}\rfloor\frac{d}{n}}^1 |f(x)|dx \le ||f||_\infty \frac{d}{n}.$$ Therefore, $$\left|\frac{1}{n/d}\sum_{m \le n/d} f(\frac{md}{n})-\int_0^1 f(x)dx\right| \le (||f’||_\infty+||f||_\infty)\frac{d}{n}.$$ Combining everything, we obtain $$\left|\sum_{d=1}^n \frac{b_d}{d}\frac{1}{n/d}\sum_{m \le n/d} f(\frac{md}{n})-\sum_{d=1}^n \frac{b_d}{d}\int_0^1 f(x)dx\right| \le \frac{1}{n}\sum_{d=1}^n \frac{|b_d|}{d},$$ which does indeed tend to $0$ as $n \to \infty$, as desired. $\square$

Attribution
Source : Link , Question Author : More Anonymous , Answer Author : Alex Ravsky

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