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 $k \to \infty$

enter image description here

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 $


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

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:

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


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

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


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

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    $\begingroup$ I know in the end the information is redundant, but it'd still be nice if you'd state all implicit prerequisites for your statement to hold in the beginning $\endgroup$
    – Sudix
    Commented Aug 20, 2018 at 16:54
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    $\begingroup$ Done ... There may be further requirements never been too good a rigour (Sorry I'm a physicist in training) $\endgroup$ Commented Aug 20, 2018 at 17:01
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    $\begingroup$ Your question is little bit unclear to me. What's $d_r$ actually? I think you have to state that.! $\endgroup$
    – Empty
    Commented Aug 20, 2018 at 17:18
  • $\begingroup$ @Empty Stated ... also if u look at the (long) proof one can see you get them by the specific sum of particular $b_r$ coefficients $\endgroup$ Commented Aug 20, 2018 at 17:21
  • $\begingroup$ @MoreAnonymous very nice question! see my answer below $\endgroup$ Commented Sep 17, 2019 at 9:22

2 Answers 2


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. $\square$

  • $\begingroup$ Thank you so much!!! A couple of quick questions: which part of the proof made required you to write the condition $\lim_{n \to \infty} \frac{\log^2(n)}{n}\sum_{r=1}^n |b_r| = 0$. I'm not even sure how you arrived on it let alone why the proof would break beyond that? Also as previously mentioned in the comment (does the "sufficent condition" include the example in that link - feel free to comment on that post too)? Also does this proof enable a measure theorist to call the L.H.S as some notion of area? Like L.H.S = Riemann measure ? $\endgroup$ Commented Sep 17, 2019 at 10:18
  • $\begingroup$ Either way this is a spectacular answer ... Perhaps I'm commenting too early but I'm so excited my 1 year old question is answered! $\endgroup$ Commented Sep 17, 2019 at 10:18
  • $\begingroup$ Also your proof seems to suggest absolute convergence is an overkill? $\endgroup$ Commented Sep 17, 2019 at 10:50
  • $\begingroup$ "where the last equality used the decay assumption on the $b_d$'s". We want convergence of the integral $\int_1^\infty \frac{\frac{1}{t}\sum_{d \le t} |b_d|}{t}dt$. Having $\frac{1}{t}\sum_{d \le t} |b_d| = o(\frac{1}{\log^2t})$ suffices. As I commented above, I don't fully know the definition of $d_r$. Specifically, I don't know what $d'_{r(i,j)}$ is. Also, I don't understand your last question. The LHS is equal to the RHS, and the RHS is some quantity times the Riemann integral of $f$, so I really don't know what you want. $\endgroup$ Commented Sep 17, 2019 at 10:54
  • $\begingroup$ your right "The LHS is equal to the RHS, and the RHS is some quantity times the Riemann integral of $f$" ... I see also $d_r$ plays the same role as $a_r$ here (in fact I might as well edit that post for consistent notation ) ... Maybe it might be easier to say I'm using the formula to say there should be a relation with $(\int e^{-x^2} dx)^2 = \text{constant} \times \int e^{-x} dx$. Please give me a moment to edit that post? $\endgroup$ Commented Sep 17, 2019 at 11:03

(I have no answer but a comment which is too long for the comment section:)

The r.h.s in your first equation $$ \lim_{s \to 1} \! \frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s} \int_0^\infty f(x) \, dx $$

can be identified as a term containing the limit of the quotient of a Dirichlet series ($F(s)$) and the Riemann zeta function when approaching 1:

$$ \lim_{s \to 1} \! \frac{F(s)}{\zeta(s)} \int_0^\infty f(x) \, dx $$

(here $F(s)=\sum_{n=1}^\infty d_n/n^s$). You maybe can think about the convergence and evaluation of that using known convergence conditions on $F(s)$. E.g in case F(1) would be finite the whole thing evaluates to $0$. The special case $d_n = 1$ which as you point out should break down to the Riemann integral unfortunately is just about the limit case $\sigma_0=1$ where we cannot say anything about convergence or divergence from that criteria alone.

  • $\begingroup$ Using the Möbius transform you also use, here we get $$ F(s)/\zeta(s) = \sum_n n^{-s} \sum_{r|n} \mu(n/r) d_r ,$$ with your $d_r$. Not sure if we can do any double sum transformation gymnastics here ... ? $\endgroup$ Commented Aug 21, 2018 at 10:53
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    $\begingroup$ It seems to work in other cases too... Check math.stackexchange.com/questions/3173284/… $\endgroup$ Commented Apr 3, 2019 at 16:44
  • $\begingroup$ Also for those reading this comment/answer the previous version of the question used ($a_r = d_r$) $\endgroup$ Commented Sep 16, 2019 at 20:14

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