# Integral $\int_{0}^{\pi}\cot\left(at\right)\cot\left(bt\right)\cot\left(ct\right)\left(\sin\left(abct\right)\right)^{3}dt$

For $$a, b, c \in \text{N}$$: $$I(a,b,c)=\int_{0}^{\pi}\cot\left(at\right)\cot\left(bt\right)\cot\left(ct\right)\left(\sin\left(abct\right)\right)^{3}dt$$

Some Results using numerical Evaluation: $$I(4k-2,1,1)=(2k-1)\pi$$ $$I(a,2,1)=\left(\frac{3a-1}{2}\right)\pi$$ Keeping 2 variables constant and changing one, we can see the pattern for many cases.
And confirming for many cases, it comes out to be a multiple of $$\pi$$.

One observation is that answer will be symmetric in terms of $$a, b, c$$.

Maybe there is an Anti-Derivative too, but I can't seem to find it.

Some Examples:

$$\int_{0}^{\pi}\cot\left(10t\right)\cot\left(11t\right)\cot\left(12t\right)\left(\sin\left(1320t\right)\right)^{3}dt=\frac{1979}{2}\pi$$

EDIT:
Here is a table I made too,
Keeping $$c=1$$:

$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c} a/b & 1 &2&3&4&5&6&7&8&9&10 \\\hline 1 &0&1&0&2&0&3&0&4&0&5 \\\hline 2 &1&2.5&4&5.5&7&8.5&10&11.5&13&14.5\\\hline 3 &0&4&0&9&0&13&0&17&0&22\\\hline 4 &2&5.5&9&11.5&14&17.5&21&23.5&26&29.5\\\hline 5&0&7&0&14&0&23&0&30&0&37\\\hline 6&3&8.5&13&17.5&23&26.5&30&35.5&40&44.5\\\hline 7&0&10&0&21&0&30&0&43&0&52\\\hline 8&4&11.5&17&23.5&30&35.5&43&47.5&52&59.5\\\hline 9&0&13&0&26&0&40&0&52&0&69\\\hline 10&5&14.5&22&29.5&37&44.5&52&59.5&69&74.5\\\hline \end{array}$$

Now obviously the table will be symmetric about the diagonal but I made it fully anyways.
Maybe someone can conjecture a form using these.

EDIT 2:
Using more data, maybe:

For $$a, b, c$$: all odd Natural Numbers. $$\boxed{I(a,b,c)=0}$$ Proof:
$$I(a,b,c)=\int_{0}^{\pi}\cot\left(at\right)\cot\left(bt\right)\cot\left(ct\right)\left(\sin\left(abct\right)\right)^{3}dt$$ $$t\to\pi-t$$ $$I(a,b,c)=\int_{0}^{\pi}\cot\left(a\left(\pi-t\right)\right)\cot\left(b\left(\pi-t\right)\right)\cot\left(c\left(\pi-t\right)\right)\left(\sin\left(abc\left(\pi-t\right)\right)\right)^{3}dt$$ For $$a, b, c$$ odd we have $$abc$$ odd too.
Therefore, $$I(a, b, c)=-I(a, b, c)$$ $$I(a,b,c)=0$$

• About your 2nd edit: Well yes, if $a,b,c$ are all odd, then the integrand satisfies $f(x)=f(\pi-x)$, so the integral vanishes. Commented Aug 29, 2023 at 18:29
• @C-RAM Yup! I just realized that. Commented Aug 29, 2023 at 18:29
• In any case, interesting problem. Have you looked at the degree two analogue of this integral? $$\int_0^\pi \cot(at)\cot(bt)\sin^2(abt)dt$$ If not, that would seem like a natural place to start. Commented Aug 29, 2023 at 18:35
• @C-RAM Well a quick Search on Approach0 reveals its been solved here. $$\int_0^\pi \sin^2(ab x) \cot(ax)\cot(bx)dx=\frac{(2\gcd(a,b)-1)\pi}{2}$$ I will try to understand the solution presented there. Commented Aug 29, 2023 at 18:40

First consider the fact that by changing $$t\rightarrow -t$$ the integral is invariant and we can simply write

$$I(a,b,c)=\frac{1}{2}\int_{-\pi}^{\pi}\cot\left(at\right)\cot\left(bt\right)\cot\left(ct\right)\left(\sin\left(abct\right)\right)^{3}dt$$

Let's take an unconventional approach! Consider the change of variable $$x = e^{i t}\Rightarrow dt = \frac{dx}{i x}$$. Then we can write

$$\cot\left(at\right)\cot\left(bt\right)\cot\left(ct\right)\left(\sin\left(abct\right)\right)^{3}\\ = \frac{1}{8}\left(\frac{x^a+x^{-a}}{x^a-x^{-a}}\right)\left(\frac{x^b+x^{-b}}{x^b-x^{-b}}\right)\left(\frac{x^c+x^{-c}}{x^c-x^{-c}}\right)\left(x^{abc}-x^{-abc}\right)^3$$

Next notice that since $$a, b, c \in \text{N}$$, we have

$$x^{abc}-x^{-abc} = \left(x^a\right)^{bc}-\left(x^{-a}\right)^{bc} = \left(x^a-x^{-a}\right)\left((x^{a})^{bc-1}+\cdots+(x^{a})^{1-bc}\right)$$ Just consider the identity $$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\cdots+y^{n-2}x+y^{n-1})$$ and replace the following $$n\rightarrow bs, x\rightarrow x^{a}, y\rightarrow x^{-a}$$. The same method can be used for $$x^b$$ and $$x^c$$. We get

$$\frac{1}{16}\int\left(x^a + x^{-a}\right)\left(x^b + x^{-b}\right)\left(x^c + x^{-c}\right)\left(\frac{1}{ix}\right)\left(\sum_{k=0}^{bc-1}\left(x^a\right)^k\left(x^{-a}\right)^{bc-1-k}\right)\\ \times \left(\sum_{k=0}^{ac-1}\left(x^b\right)^k\left(x^{-b}\right)^{ac-1-k}\right)\left(\sum_{k=0}^{ab-1}\left(x^c\right)^k\left(x^{-c}\right)^{ab-1-k}\right)dx$$

Now that we have solved the problem of singularities at the origin, we transform back again using $$x = e^{i t}\Rightarrow dx = ixdt$$, we get

$$\frac{1}{16}\int_{-\pi}^{\pi}\left(e^{iat} + e^{-iat}\right)\left(e^{ibt} + e^{-ibt}\right)\left(e^{ict} + e^{-ict}\right)\left(\sum_{k=0}^{bc-1}\left(e^{iat}\right)^k\left(e^{-iat}\right)^{bc-1-k}\right)\\ \times \left(\sum_{k=0}^{ac-1}\left(e^{ibt}\right)^k\left(e^{-ibt}\right)^{ac-1-k}\right)\left(\sum_{k=0}^{ab-1}\left(e^{ict}\right)^k\left(e^{-ict}\right)^{ab-1-k}\right)dt$$

But notice that

$$\left(e^{iat} + e^{-iat}\right)\left(\sum_{k=0}^{bc-1}\left(e^{iat}\right)^k\left(e^{-iat}\right)^{bc-1-k}\right) \\ = \left( \sum_{k=0}^{bc-1}\left(e^{iat}\right)^{k+1}\left(e^{-iat}\right)^{bc-(k+1)} + \sum_{k=0}^{bc-1}\left(e^{iat}\right)^k\left(e^{-iat}\right)^{bc-k} \right)\\ = \left( \sum_{k=1}^{bc}\left(e^{iat}\right)^{k}\left(e^{-iat}\right)^{bc-k} + \sum_{k=0}^{bc-1}\left(e^{iat}\right)^k\left(e^{-iat}\right)^{bc-k} \right)\\ = \left(2\sum_{k=1}^{bc-1}\left(e^{iat}\right)^k\left(e^{-iat}\right)^{bc-k} + e^{iabct} + e^{-iabct} \right) = \left(2\sum_{k=0}^{bc}\left(e^{iat}\right)^k\left(e^{-iat}\right)^{bc-k}\right) \\ = \left(\sum_{k=0}^{\lfloor \frac{bc}{2} \rfloor}\left(\cos\left(abct - 2akt\right)\right) + \left( 1-bc+2\lfloor \frac{bc}{2} \rfloor \right) \right)$$

Where $$\lfloor \rfloor$$ represents floor function. The last term is aded since if $$\left(bs\right)$$ is an even number, then one term remains (the middle term) which is equal to one. Also since $$\int_{-\pi}^{\pi}e^{\pm int}dt = \begin{cases} 0, \quad &\text{if} \quad n\neq 0 \\ 2\pi, \quad &\text{if} \quad n=0 \end{cases}$$, one can solve the integral without converting to cosine and by finding the constant terms in exponential form! For convenience we name $$m_a = \left( 1-bc+2\lfloor \frac{bc}{2} \rfloor \right), m_b = \left( 1-ac+2\lfloor \frac{ac}{2} \rfloor \right), m_c = \left( 1-ab+2\lfloor \frac{ab}{2} \rfloor \right)$$

The problem transforms into

$$I(a,b,c) = \frac{1}{16}\int_{-\pi}^{\pi}\left(\sum_{k=0}^{\lfloor \frac{bc}{2} \rfloor}\left(\cos\left(abct - 2akt\right)\right) + m_a \right)\\ \times \left(\sum_{k=0}^{\lfloor \frac{ac}{2} \rfloor}\left(\cos\left(abct - 2bkt\right)\right) + m_b\right)\left(\sum_{k=0}^{\lfloor \frac{ab}{2} \rfloor}\left(\cos\left(abct - 2ckt\right)\right) + m_c \right)$$

Now we multiply the parentheses to get 8 different terms from

$$\left(\sum_{k=0}^{\lfloor \frac{bc}{2} \rfloor}\left(\cos\left(abct - 2akt\right)\right) + m_a \right)\left(\sum_{k=0}^{\lfloor \frac{ac}{2} \rfloor}\left(\cos\left(abct - 2bkt\right)\right) + m_b\right) \\ \times\left(\sum_{k=0}^{\lfloor \frac{ab}{2} \rfloor}\left(\cos\left(abct - 2ckt\right)\right) + m_c \right)$$ for example the multiplication of the three cosine terms results in

$$\sum_{i=0}^{\lfloor \frac{bc}{2} \rfloor}\sum_{j=0}^{\lfloor \frac{ac}{2} \rfloor}\sum_{k=0}^{\lfloor \frac{ab}{2} \rfloor}\cos\left(abct - 2ait\right)\cos\left(abct - 2bjt\right)\cos\left(abct - 2ckt\right) = \\ \frac{1}{4}\sum_{i=0}^{\lfloor \frac{bc}{2} \rfloor}\sum_{j=0}^{\lfloor \frac{ac}{2} \rfloor}\sum_{k=0}^{\lfloor \frac{ab}{2} \rfloor} \left[ \cos\left( 3abct-2ait-2bjt-2ckt \right) + \cos\left( abct+2ckt-2ait-2bjt \right) \\ \cos\left( abct+2ait-2ckt-2bjt \right) + \cos\left( abct+2bjt-2ait-2ckt \right) \right]$$

and now each term can be integrated. The rest of the terms can be integrated in this fashion. However the result is messy, but the closed form solution exists, it just takes an eternity to write it down! Also there are some values of $$t$$ for which the sine terms become zero and the above identity does not hold however these points are measure zero and can be neglected! $$t=0,\pi,-\pi$$ are amongst them!

• Nice! I arrived at the same conclusion and here I was thinking of generalizing this to higher degree variations. Commented Aug 30, 2023 at 7:44
• Yes exactly, it comes down to the oddness or evenness of integers $a,b,c$! Commented Aug 30, 2023 at 7:50

$$I(a,b,c)=\int_{0}^{\pi}\cot\left(at\right)\cot\left(bt\right)\cot\left(ct\right)\left(\sin\left(abct\right)\right)^{3}dt$$

Consider the sequence of functions, $$f_{n}\left(x\right)=\frac{\sin\left(nx\right)}{\sin x}$$ Then, $$f_{n+2}(x)=f_n(x)+2\cos((n+1)x)$$ Therefore, \begin{align} f_{2n}(x)&=2\sum_{k=1}^n\cos((2k-1)x)\\ f_{2n+1}(x)&=1+2\sum_{k=1}^n\cos(2kx) \end{align} Next, see that: $$\sin^3(abcx)\cot(ax)\cot(bx)\cot(cx)=\frac{\sin(abcx)\cos(ax)}{\sin(ax)}\cdot\frac{\sin(abcx)\cos(bx)}{\sin(bx)}\cdot\frac{\sin(abcx)\cos(cx)}{\sin(cx)}$$ Hence, $$I(a,b,c)=\frac{1}{8}\int_{0}^{\pi}\bigg(f_{bc+1}\left(ax\right)+f_{bc-1}\left(ax\right)\bigg)\bigg(f_{ca+1}\left(bx\right)+f_{ca-1}\left(bx\right)\bigg)\bigg(f_{ab+1}\left(cx\right)+f_{ab-1}\left(cx\right)\bigg)dx$$ Since, $$f(a,b,c)$$ is a symmetric expression in $$a, b, c$$, we need to consider four cases:

1. All are Odd.
2. 1 Even 2 Odd.
3. 2 Even 1 Odd
4. All are Even

For the First Case: $$f(\text{odd},\text{odd},\text{odd})=0$$ as the Integrand satisfies $$g(x)=g(\pi-x)$$ if all three are odd and hence vanishes.
Considering the Second Case:
Let $$a=2u$$, $$b=2v-1$$, $$c=2w-1$$:

$$I(a,b,c)=\int_{0}^{\pi}\left(\sum_{k=1}^{2vw-v-w+1}\cos\left(\left(2k-1\right)2ux\right)+\sum_{k=1}^{2vw-v-w}\cos\left(\left(2k-1\right)2ux\right)\right)\left(1+\sum_{k=1}^{u\left(2w-1\right)}\cos\left(2k\left(2v-1\right)x\right)+\sum_{k=1}^{u\left(2w-1\right)-1}\cos\left(2k\left(2v-1\right)x\right)\right)\left(1+\sum_{k=1}^{u\left(2v-1\right)}\cos\left(2k\left(2w-1\right)x\right)+\sum_{k=1}^{u\left(2v-1\right)-1}\cos\left(2k\left(2w-1\right)x\right)\right)dx$$ $$=\int_{0}^{\pi}\left(\cos\left(abcx\right)+2\underbrace{\sum_{k=1}^{2vw-v-w}\cos\left(\left(2k-1\right)ax\right)}_{S_1}\right)\left(1+\cos\left(acbx\right)+2\underbrace{\sum_{k=1}^{u\left(2w-1\right)-1}\cos\left(2kbx\right)}_{S_2}\right)\left(1+\cos\left(abcx\right)+2\underbrace{\sum_{k=1}^{u\left(2v-1\right)-1}\cos\left(2kcx\right)}_{S_3}\right)dx$$
Using,
For Positive Integers $$m, n$$: $$\int_0^{\pi}\cos(mx)\cos(nx) \ dx=\frac{\pi}{2}[m=n]$$ $$\int_0^{\pi}\cos(mx)\cos^2(nx) \ dx=\frac{\pi}{4}[m=2n]$$ $$\int_0^{\pi}\cos(mx)\cos(nx)\cos(ox) \ dx=\frac{\pi}{4}\bigg([m=n+o]+[n=m+o]+[o=m+n]\bigg)$$
We arrive at,

$$=\pi+\left(4\int_{0}^{\pi}S_{1}S_{2}dx+4\int_{0}^{\pi}S_{1}S_{2}\cos\left(abcx\right)dx+4\int_{0}^{\pi}S_{1}S_{3}dx+4\int_{0}^{\pi}S_{1}S_{3}\cos\left(abcx\right)dx+4\int_{0}^{\pi}S_{2}S_{3}\cos\left(abcx\right)dx+8\int_{0}^{\pi}S_{1}S_{2}S_{3}dx\right)$$
Although we can find closed form for each of these Integrals, it is not going to be elegant or as simple as $$\gcd$$ in the degree two analogue of this problem.
And then we also have the other three cases to consider.