Family of Generalized Integrals ${I}(a,b,p)=\int_0^{ab} \left( \left\{\frac{x}{a}\right\}-p\right) \left( \left\{\frac{x}{b}\right\}-p\right) \; dx$

Question

Background:

I came across the following family of generalized Franel integrals, and found them quite interesting. I don't think I've seen anything about these integrals before, at least not generalized, and I want to know if this is a known family of generalized integrals. I am also interested behind the behavior of this family of integrals, specifically when $p=\frac{1}{2}$. I want to find out if there's a simple algebraic closed form expression for this family of integrals. Wolfram does not do a good job factoring the integrands, but to be fair it didn't factor $I(a,b,p)$ as I have done in this post. Also, how would you approach cases where $\gcd{(a,b,c,\dots)} \neq 1$?

Consider the family of generalized integrals as the following:

$${I}(a,b,p)=\int_0^{ab} \left( \bigg\{\frac{x}{a}\bigg\}-p\right) \left( \bigg\{\frac{x}{b}\bigg\}-p\right) \; dx$$ $${I}(a,b,c,p)=\int_0^{abc} \left( \bigg\{\frac{x}{a}\bigg\}-p\right) \left( \bigg\{\frac{x}{b}\bigg\}-p\right) \left( \bigg\{\frac{x}{c}\bigg\}-p\right)\; dx$$ $${I}(a,b,c,d,p)=\int_0^{abcd} \left( \bigg\{\frac{x}{a}\bigg\}-p\right) \left( \bigg\{\frac{x}{b}\bigg\}-p\right) \left( \bigg\{\frac{x}{c}\bigg\}-p\right) \left( \bigg\{\frac{x}{d}\bigg\}-p\right)\; dx$$ $$\ldots$$ Where $a,b,c,\ldots \in \mathbb{N}$, $p \in \mathbb{Q}^+$, and $\gcd{(a,b,c,\ldots)}=1$.

Calculations:

Express the integral as the following: $${I}(a,b,p)=\sum_{i=0}^{a-1} \sum_{k=0}^{b-1} \int_0^1 \left(\frac{t+i}{a}-p\right)\left(\frac{t+k}{b}-p\right) \; dt$$ Changing the order of the summations and integral and using some algebra: $${I}(a,b,p)=\int_0^1 \left(\frac{a-1}{2}+t-ap\right)\left(\frac{b-1}{2}+t-bp\right) \; dt$$ Expanding the integrand out and factoring yields: $${I}(a,b,p)=\int_0^1 \frac{ab}{4}{\left(2p-1\right)}^2+\frac{at}{2}\left(1-2p\right)+\frac{bt}{2}\left(1-2p\right)+\frac{(a+b)}{4}\left(2p-1\right)+{\left(t-\frac{1}{2}\right)}^2 \; dt$$ $${I}(a,b,p)=\int_0^1 \frac{ab}{4}{\left(2p-1\right)}^2+{\left(t-\frac{1}{2}\right)}^2 \; dt$$ And so: $$\boxed{{I}(a,b,p)= \frac{ab}{4}{\left(1-2p\right)}^2+\frac{1}{12}}$$ Calculated similarly,I got the following: $$I(a,b,c,p)=\frac{abc{\left(1-2p\right)}^3}{8}+\frac{c}{24}\left(1-2p\right)$$ $$I(a,b,c,d,p)= \frac{abcd}{16}{\left(1-2p\right)}^4+\frac{{(1-2p)}^2}{48}\left(ab+cd\right)+\frac{1}{80} $$ However, as @Varun Vejalla and @OliverDiaz pointed out in the comments, these results are illogical, and there actually is no closed form for $I(a,b,c,d,p)$.

Further Observations:

Interestingly enough, $p=\frac{1}{2}$ is a special case for this entire family generalized integrals. Why is this? Assuming the aforementioned conditions are met:

$$I\left(a,b,\frac{1}{2}\right)=\frac{1}{12}$$ $$I\left(a,b,c,\frac{1}{2}\right)=\int_0^1 {\left(t-\frac{1}{2}\right)}^3 \; dt=0$$ $$I\left(a,b,c,d,\frac{1}{2}\right)=\int_0^1 {\left(t-\frac{1}{2}\right)}^4 \; dt=\frac{1}{80} $$ $$I\left(a,b,c,d,e,\frac{1}{2}\right)=\int_0^1 {\left(t-\frac{1}{2}\right)}^5 \; dt=0$$ And so it seems that the following statement is true: $$I\left(a_1,a_2,\ldots,a_n,\frac{1}{2}\right)=\int_0^1 {\left(t-\frac{1}{2}\right)}^n \; dt=\cases{ 0 & $n \; \text{is odd}$ \cr \frac{1}{2^n\left(n+1\right)} & $n \; \text{is even}$ }$$ However, Wolfram Alpha computed $I\left(a,b,c,d,\frac{1}{2}\right)=0$ for valid $a,b,c,d$ values.

Final Remarks:

I wonder what other interesting observations can be made about this family of generalized integrals. Specifically, are there other interesting special cases, and if so why are they so special?

Answer 1

This just addresses the simplest questions in the OP, namely the estimation of $I(a,b;p)$ when $d:=g.c.d(a,b)>1$. In such case, there are integers $q,r$ such that $a=qd$, $b=rd$ and $g.c.d(q,r)=1$

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The change of variable $u=x/d$ and the periodicity of $x\mapsto\{x\}$ results in

$$ \begin{align} I(a,b;p)&=d\int^{dqr}_0\Big(\big\{\tfrac{x}{q}\big\}-p\Big)\Big(\big\{\tfrac{x}{r}\big\}-p\Big)\,dx=d\sum^{d-1}_{j=0}\int^{(j+1)qr}_{jqr}\Big(\big\{\tfrac{x}{q}\big\}-p\Big)\Big(\big\{\tfrac{x}{r}\big\}-p\Big)\,dx\\ &= d\sum^{d-1}_{j=0}\int^{qr}_0\Big(\big\{\tfrac{x+jqr}{q}\big\}-p\Big)\Big(\big\{\tfrac{x+jqr}{r}\big\}-p\Big)\,dx\\ &=d^2\int^{qr}_0\Big(\big\{\tfrac{x}{q}\big\}-p\Big)\Big(\big\{\tfrac{x}{r}\big\}-p\Big)\,dx \end{align}$$

Form the case of relative prime one obtained (the OP provides a sketch of the proof for this case):

$$\begin{align} \frac{1}{ab}I(a,b;p)= \frac{1}{4}{\left(1-2p\right)}^2+\frac{1}{12}\frac{g.c.d(a,b)}{l.c.m(a,b)} \end{align}$$

where $l.c.m(a,b)$ is the lowest common multiple of $a$ and $b$. When $p=1/2$ we recover Franel's formula.

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For higher order integrals, I have a few references that may be useful to those interested in this question:

Franel integrals of order three

Franel Integrals of over four

Multiple Franel integrals

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Here I provide further details for the expression $I(a,b;p)$ when $g.c.d(a, b)=1$.

Without loss of generality assume $a>b$. The $a=bq+r$ for $q\in\mathbb{N}$ and $1\leq r<b$

$$ \begin{align} I(a,b;p)&=\sum^{b-1}_{k=1}\int^{(k+1)a}_{ka}\Big(\big\{\tfrac{x}{a}\big\}-p\Big)\Big(\big\{\tfrac{x}{b}\big\}-p\Big)\,dx = \sum^{b-1}_{k=0}\int^a_0\Big(\big\{\tfrac{x+ak}{a}\big\}-p\Big)\Big(\big\{\tfrac{x+ak}{b}\big\}-p\Big)\,dx\\ &=\sum^{b-1}_{k=0}\int^a_0\Big(\big\{\tfrac{x}{a}\big\}-p\Big)\Big(\big\{\tfrac{x+rk}{b}\big\}-p\Big)\,dx=\sum^{b-1}_{k=0}\sum^{a-1}_{\ell=0}\int^{\ell+1}_{\ell}\Big(\big\{\tfrac{x}{a}\big\}-p\Big)\Big(\big\{\tfrac{x+rk}{b}\big\}-p\Big)\,dx\\ &=\sum^{b-1}_{k=0}\sum^{a-1}_{\ell=0}\int^1_0\Big(\big\{\tfrac{x+\ell}{a}\big\}-p\Big)\Big(\big\{\tfrac{x+\ell+rk}{b}\big\}-p\Big)\,dx\ \end{align} $$ Since $g.c.d(a,b)=g.c.d(b,r)=1$, for each $0\leq\ell<a$ fixed, $\ell+kr$ ranges over $\{0,\ldots,b-1\}\mod \,b$ as $k$ over $\{0,\ldots,b-1\}$. Hence $$ I(a, b;p)=\sum^{b-1}_{j=0}\sum^{a-1}_{\ell=0} \int^1_0\Big(\big\{\tfrac{x+\ell}{a}\big\}-p\Big)\Big(\big\{\tfrac{x+j}{b}\big\}-p\Big)\,dx=\sum^{b-1}_{j=0}\sum^{a-1}_{\ell=0}\int^1_0 \Big(\tfrac{x+\ell}{a}-p\Big)\Big(\tfrac{x+j}{b}-p\Big)\,dx $$

The rest is as the OP indicated.

I ignore thus far whether a similar argument carries over for higher orders under the assumption that $g.c.d(a_1,\ldots,a_n)=1$.

Answer 2

This answer is only for the case $\gcd(a_1, a_2), \gcd(a_1, a_3),... = 1$ (i.e. the $\gcd$ of any pair of $a_i$ is $1$).

Starting from what you have already done: $$I(a_1, a_2, ..., a_n, p) = \int_0^1 \prod_{i=1}^n \left(\frac{a_i-1}{2} +t-a_ip \right)dt$$

Let $c_i = \frac{a_i-1}{2}-a_ip$. Then the task is to find $$\int_0^1 \prod_{i=1}^n\left( t+c_i \right)dt$$

The integrand can be expanded as $$\int_0^1 \sum_{k=0}^nS_{k, n}x^{n-k} dt$$

where $S_{k, n}$ is the sum of the product of all $\binom{n}{k}$ "$k$-tuples" from $c_1, c_2, ..., c_n$ (except for $S_{0, n} = 1$). For example, $S_{2, 4} = c_1c_2+c_1c_3+c_1c_4 + c_2c_3+c_2c_4+c_3c_4$ and $S_{3, 4} = c_1c_2c_3+c_1c_2c_4+c_1c_3c_4+c_2c_3c_4$. Let $s_{k, n}$ be defined in a similar way, but for $a_i$ instead of $c_i$

Then the integral is $$\sum_{k=0}^{n} \frac{S_{k, n}}{n+1-k}$$

For $n = 3$, this gives $$\frac{(1-2p)^3a_1a_2a_3}{8}+\frac{(1-2p)(a_1+a_2+a_3)}{24} = $$ $$\frac{(1-2p)^3 s_{3, 3}}{8} + \frac{(1-2p) s_{1, 3}}{24}$$

For $n = 4$, this gives $$\frac{(1-2p)^4 s_{4, 4}}{16}+\frac{s_{2, 4}}{48}(1-2p)^2 + \frac{1}{80}$$

For $n = 5$, this gives $$\frac{(1-2p)^5 s_{5, 5}}{32} + \frac{(1-2p)^3 s_{3, 5}}{96} + \frac{(1-2p)s_{1, 5}}{160}$$

In general it seems like $$I(a_1, ..., a_n, p) = \sum_{1 \le k \le n+1, k\pmod2 = 1} \frac{(1-2p)^{n+1-k} s_{n+1-k, n}}{k\cdot 2^n}$$

although I have not confirmed this.