Integral involving fractional part.

[mathisfun1]

View attachment 3249

I started with taking three cases,
1. a=b
since a and b are co primes, both will have to be equal to 1 and then we can easily get lhs = rhs.
2. a>b
3.a<b
I have no idea as to how to proceed for the second and third cases. Any help would be greatly appreciated.

 

[Opalg]

Divide the unit interval into $ab$ subintervals $\bigl[\frac r{ab},\,\frac{r+1}{ab}\bigr]$ of length $\frac1{ab}$, for $0\leqslant r <ab.$ In each subinterval, $x = \frac r{ab} + t$ (where $0\leqslant t\leqslant \frac1{ab}$), and $ax = \frac rb + at$. Thus $\{ax\} = \bigl\{ \frac rb \bigr\} + at = \frac kb + at$ for some $k$ with $0\leqslant k <b.$ Similarly, $\{bx\} = \bigl\{ \frac ra \bigr\} + bt = \frac la + bt$ for some $l$ with $0\leqslant l <a.$

You then need to show that as $r$ goes from $0$ to $ab-1$, each of the possible values of the pair $k,\: l$ (with $0\leqslant k <b,\ 0\leqslant l <a$) occurs exactly once. To prove that, you will need to use the fact that $a$ and $b$ are relatively prime.

It will then follow that $$\begin{aligned} \int_0^1\bigl(\{ax\} - \tfrac12\bigr) \bigl(\{bx\} - \tfrac12\bigr) &= \sum_{k=0}^{b-1} \sum_{l=0}^{a-1} \int_0^{1/(ab)} \bigl(at + \tfrac kb -\tfrac12\bigr) \bigl(bt + \tfrac la -\tfrac12\bigr) dt \\ &=\int_0^{1/(ab)} \sum_{k=0}^{b-1}\bigl(at + \tfrac kb -\tfrac12\bigr) \sum_{l=0}^{a-1}\bigl(bt + \tfrac la -\tfrac12\bigr) dt \\ &= \int_0^{1/(ab)} \bigl(abt + \tfrac 12(b-1) -\tfrac12b\bigr) \bigl(abt + \tfrac 12(a-1) -\tfrac12a\bigr)dt \\ &= \int_0^{1/(ab)} \bigl(abt - \tfrac 12\bigr)^2dt \\ &= \Bigl[\tfrac1{3ab}\bigl(abt - \tfrac 12\bigr)^3 \Bigr]_0^{1/(ab)} \\ &= \tfrac1{3ab} \bigl(\tfrac18 + \tfrac18\bigr) = \tfrac1{12ab}. \end{aligned}$$

 

[mathisfun1]

Divide the unit interval into $ab$ subintervals $\bigl[\frac r{ab},\,\frac{r+1}{ab}\bigr]$ of length $\frac1{ab}$, for $0\leqslant r <ab.$ In each subinterval, $x = \frac r{ab} + t$ (where $0\leqslant t\leqslant \frac1{ab}$), and $ax = \frac rb + at$. Thus $\{ax\} = \bigl\{ \frac rb \bigr\} + at = \frac kb + at$ for some $k$ with $0\leqslant k <b.$ Similarly, $\{bx\} = \bigl\{ \frac ra \bigr\} + bt = \frac la + bt$ for some $l$ with $0\leqslant l <a.$

You then need to show that as $r$ goes from $0$ to $ab-1$, each of the possible values of the pair $k,\: l$ (with $0\leqslant k <b,\ 0\leqslant l <a$) occurs exactly once. To prove that, you will need to use the fact that $a$ and $b$ are relatively prime.

It will then follow that $$\begin{aligned} \int_0^1\bigl(\{ax\} - \tfrac12\bigr) \bigl(\{bx\} - \tfrac12\bigr) &= \sum_{k=0}^{b-1} \sum_{l=0}^{a-1} \int_0^{1/(ab)} \bigl(at + \tfrac kb -\tfrac12\bigr) \bigl(bt + \tfrac la -\tfrac12\bigr) dt \\ &=\int_0^{1/(ab)} \sum_{k=0}^{b-1}\bigl(at + \tfrac kb -\tfrac12\bigr) \sum_{l=0}^{a-1}\bigl(bt + \tfrac la -\tfrac12\bigr) dt \\ &= \int_0^{1/(ab)} \bigl(abt + \tfrac 12(b-1) -\tfrac12b\bigr) \bigl(abt + \tfrac 12(a-1) -\tfrac12a\bigr)dt \\ &= \int_0^{1/(ab)} \bigl(abt - \tfrac 12\bigr)^2dt \\ &= \Bigl[\tfrac1{3ab}\bigl(abt - \tfrac 12\bigr)^3 \Bigr]_0^{1/(ab)} \\ &= \tfrac1{3ab} \bigl(\tfrac18 + \tfrac18\bigr) = \tfrac1{12ab}. \end{aligned}$$

I am much thankful for your reply.
However, I am unable to understand how fractional part of ax=fractional part of r/b + at.
Also, although I am studying advanced calculus (Relative to XIIth standard, that is I am in XIIth standard but studying a bt ahead for competitive exams, we have never been taught something like this. Maybe, there's another way to do this? I am sure my maths teacher would never give us a question that he had not taught us.

 

[Opalg]

I am much thankful for your reply.
However, I am unable to understand how fractional part of ax=fractional part of r/b + at.
Also, although I am studying advanced calculus (Relative to XIIth standard, that is I am in XIIth standard but studying a bt ahead for competitive exams, we have never been taught something like this. Maybe, there's another way to do this? I am sure my maths teacher would never give us a question that he had not taught us.

There may be a simpler way to do this, but I don't see how. Maybe someone else here can help?

To get an idea of what is involved in this problem, I found it helpful to look at a simple particular case. Suppose for example that $a=3$ and $b=5$. Then to calculate the integral \(\displaystyle \int_0^1\bigl(\{3x\} - \tfrac12\bigr) \bigl(\{5x\} - \tfrac12\bigr)dx\) I divided the unit interval into subintervals of length $\frac1{15}$. If you put $x = \frac r{15} + t$ and take $r=7$, say, then you are looking at the subinterval $\bigl[\frac7{15},\frac8{15}\bigr]$. In that interval, $$3x = 3\bigl(\tfrac7{15} + t\bigr) = \tfrac{21}{15} + 3t = \tfrac75 + 3t = 1 + \tfrac25 + 3t.$$ The fractional part of that is $\frac25 + 3t$ (because $0\leqslant t \leqslant \frac1{15}$ and therefore $\frac25\leqslant \frac25 + 3t \leqslant \frac35$). That illustrates why the fractional part of $ax$ is the fractional part of $\frac rb + at$.

 

[mathisfun1]

There may be a simpler way to do this, but I don't see how. Maybe someone else here can help?

To get an idea of what is involved in this problem, I found it helpful to look at a simple particular case. Suppose for example that $a=3$ and $b=5$. Then to calculate the integral \(\displaystyle \int_0^1\bigl(\{3x\} - \tfrac12\bigr) \bigl(\{5x\} - \tfrac12\bigr)dx\) I divided the unit interval into subintervals of length $\frac1{15}$. If you put $x = \frac r{15} + t$ and take $r=7$, say, then you are looking at the subinterval $\bigl[\frac7{15},\frac8{15}\bigr]$. In that interval, $$3x = 3\bigl(\tfrac7{15} + t\bigr) = \tfrac{21}{15} + 3t = \tfrac75 + 3t = 1 + \tfrac25 + 3t.$$ The fractional part of that is $\frac25 + 3t$ (because $0\leqslant t \leqslant \frac1{15}$ and therefore $\frac25\leqslant \frac25 + 3t \leqslant \frac35$). That illustrates why the fractional part of $ax$ is the fractional part of $\frac rb + at$.

Sorry for getting back to you so late, got caught up in school work.
Let's just say for some case, r/b=1.7 and at=0.4
then frac (ax) will be 0.7+0.4=1.1
which is not possible.
I tried a few cases taking different values of a and b and concluded that this will never happen the sum will always be less than one and thus fac(ax) can be correctly written as frac(r/b)+at.
But is there any mathematical way to say this?

 

[Opalg]

Sorry for getting back to you so late, got caught up in school work.
Let's just say for some case, r/b=1.7 and at=0.4
then frac (ax) will be 0.7+0.4=1.1
which is not possible.
I tried a few cases taking different values of a and b and concluded that this will never happen the sum will always be less than one and thus fac(ax) can be correctly written as frac(r/b)+at.
But is there any mathematical way to say this?

Going back to my first comment above, the situation is that for each integer $r$ with $0\leqslant r<ab$, we define $\bigl\{\frac rb \bigr\} = \frac kb$ and $\bigl\{\frac ra \bigr\} = \frac la$. It then follows that $0\leqslant k<b$ and $0\leqslant l<a$. Notice that there are $ab$ possible values of $r$, and there are also $ab$ possible values for the pair $(k,l).$

Now suppose that there are two values of $r$, call them $r$ and $s$, giving rise to the same pair $(k,l)$. In other words $\bigl\{\frac rb \bigr\} =\bigl\{\frac sb \bigr\} =\frac kb$ and $\bigl\{\frac ra \bigr\} =\bigl\{\frac sa \bigr\} =\frac la$. Then the fractional part of $\frac{s-r}b$ is $0$. Therefore $s-r$ is a multiple of $b$. Similarly, $s-r$ is a multiple of $a$. But because $a$ and $b$ are coprime, that implies that $s-r$ is a multiple of $ab$. But both $r$ and $s$ are less than $ab$, so it follows that $s-r=0$, in other words $s=r$.

That says that each value of $r$ corresponds to a different value of the pair $(k,l)$. It then follows from the pigeonhole principle that each possible value of $(k,l)$ must correspond to some value of $r$.

If you know a bit of number theory, there is an alternative way to phrase the same argument. The condition $\bigl\{\frac rb \bigr\} = \frac kb$ is equivalent to $r\equiv k\pmod b$, and similarly $\bigl\{\frac ra \bigr\} = \frac la$ is equivalent to $r\equiv l\pmod a$. Because $a$ and $b$ are coprime, it then follows from the Chinese remainder theorem that $r$ is uniquely determined modulo $ab$.