Integrals Involving GCD Function

Question

Recently I got interested in Integrals involving GCD.
I was wondering if I could get more such Integrals,
Here are some examples: $$\int_0^\pi \sin^2(ab x) \cot(ax)\cot(bx)dx=\left(\frac{2\color{red}{\gcd(a,b)}-1}{2}\right)\pi$$ $$\int_0^{\pi/2}\ln{\lvert\sin(mx)\rvert}\cdot \ln{\lvert\sin(nx)\rvert}\, dx=\frac{\pi^3}{24}\frac{\color{red}{\gcd^2(m,n)}}{mn}+\frac{\pi\ln^2(2)}{2}$$ $$\int_{0}^{\pi}\arctan\left(\cot\left(mt\right)\right)\arctan\left(\cot\left(nt\right)\right)dt=\frac{\pi^{3}}{12}\cdot\frac{\color{red}{\gcd^2\left(m,n\right)}}{mn}$$ $$\int_{0}^{1}\lfloor{ax}\rfloor \lfloor{bx}\rfloor dx=\frac{\color{red}{\gcd^2\left(a,b\right)}}{12ab}-\frac{a}{4}-\frac{b}{4}-\frac{b}{12a}-\frac{a}{12b}+\frac{ab}{3}+\frac{1}{4}$$
Here is a Conditional One:
If $c$ is not a multiple of $\gcd(a,b)$ then, $$\int_{0}^{\pi/2}\cos^{2}\left(ct\right)\ln\left(\left|\sin at\right|\right)\ln\left(\left|\sin bt\right|\right)dt= \frac{\pi^{3}}{48}\frac{\color{red}{\gcd^{2}\left(a,b\right)}}{ab}+\frac{\pi\ln^{2}\left(2\right)}{4}$$
This is not a question per say but I am just curious about more of these Integrals.
You can answer with a new Integral and maybe its solution.
I will also keep adding more such Integrals.

Answer 1

UPDATE: I added a second example.

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Let $m$ and $n$ be positive integers.

We can use oO_ƲRF_Oo's approach to your previous question to quickly show that $$\int_{0}^{\pi} \operatorname{Cl}_{2}(m \theta) \operatorname{Cl}_{2} (n \theta) \, \mathrm d \theta = \frac{\pi^{5}}{180} \left( \frac{\gcd(m,n)}{\operatorname{lcm}(m,n)} \right)^{2},$$ where $\operatorname{Cl}_{2}(\phi)$ is the Clausen function $$\operatorname{Cl}_{2}(\phi) = - \int_{0}^{\phi} \log \left|2 \sin \left( \frac{x}{2} \right)\right| \, \mathrm dx = \sum_{k=1}^{\infty} \frac{\sin (k \phi)}{k^{2}}.$$

Specifically, we have

$$ \begin{align} \int_{0}^{\pi} \operatorname{Cl}_{2}(m \theta) \operatorname{Cl}_{2} (n \theta) \, \mathrm d \theta &= \int_{0}^{\pi} \sum_{j=1}^{\infty} \frac{\sin(j m \theta)}{j^{2}} \sum_{k=1}^{\infty} \frac{\sin(k n \theta)}{k^{2}} \, \mathrm d\theta \\ &= \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{j^{2}k^{2}} \int_{0}^{\pi} \sin(j m\theta) \sin(k n \theta) \, \mathrm d \theta \\ &= \frac{\pi}{2}\sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{j^{2}k^{2}} \delta_{jm, kn} \\ &= \frac{\pi}{2} \sum_{p=1}^{\infty} \frac{m^{2} \, n^{2}}{p^{2}\operatorname{lcm}^{2}(m,n) \, p^{2} \operatorname{lcm}^{2}(m,n)} \\ &= \frac{\pi}{2} \frac{\gcd^{2}(m,n)}{\operatorname{lcm}^{2}(m,n)} \sum_{p=1}^{\infty} \frac{1}{p^{4}} \\ &= \frac{\pi}{2} \frac{\gcd^{2}(m,n)}{\operatorname{lcm}^{2}(m,n)} \, \zeta(4) \\ &= \frac{\pi^{5}}{180} \left( \frac{\gcd(m,n)}{\operatorname{lcm}(m,n)} \right)^{2}. \end{align}$$

Changing the order of the integration and the summations is justified by Fubini's theorem since $$\int_{0}^{\pi} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \left|\frac{\sin(jm \theta)\sin(k n \theta)}{j^{2}k^{2}} \right| \, \mathrm d \theta \le \int_{0}^{\pi} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{j^{2} k^{2}} \, \mathrm d \theta= \left(\frac{\pi^{2}}{6} \right)^{2} \pi < \infty.$$

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And assuming that changing the order of the integration and the summations can be justified, we also have $$\int_{0}^{\pi} \ln \left| \cot \left(\frac{m\theta }{2} \right) \right| \, \ln \left| \cot \left(\frac{n\theta }{2} \right) \right| \, \mathrm d \theta = \frac{\pi^{3}}{4} \frac{\gcd(m,n)}{\operatorname{lcm}(m,n)} $$ if $m$ and $n$ have the same number of powers of $2$ in their prime factorizations. The value of the integral is zero otherwise.

Using the Fourier series $$\sum_{k=0}^{\infty} \frac{\cos \left((2k+1) \theta\right)}{2k+1} = \Re \operatorname{artanh}(e^{i \theta}) = \Re \, \frac{1}{2} \, \ln \left(i \cot \left(\frac{\theta}{2} \right) \right) = \frac{1}{2} \, \ln \left|\cot \left(\frac{\theta}{2} \right) \right| , $$ we have

$$ \begin{align} \int_{0}^{\pi} \ln \left| \cot \left(\frac{m \theta}{2} \right) \right| \, \ln \left| \cot \left(\frac{n \theta}{2} \right) \right| \, \mathrm d \theta &= 4 \int_{0}^{\pi} \sum_{j=0}^{\infty} \frac{\cos\left((2j+1)m \theta \right)}{2j+1} \sum_{k=0}^{\infty}\frac{\cos\left((2k+1)n \theta\right)}{2k+1} \, \mathrm d \theta \\ &= 2 \pi \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \frac{1}{(2j+1)(2k+1)} \, \delta_{(2j+1)m, (2k+1)n}. \end{align}$$

If $m$ and $n$ have the same number of powers of $2$ in their prime factorizations, then $(2p+1) \operatorname{lcm}(m,n), p=0,1,2, \ldots$ are the common odd multiples of $m$ and $n$.

If $m$ and $n$ don't have the same number of powers of $2$ in their prime factorization, then $m$ and $n$ don't have any odd multiples in common.

See this question for an explanation.

Therefore, if $m$ and $n$ have the same number of powers of $2$ in their prime factorizations, then $$\int_{0}^{\pi} \ln \left| \cot \left(\frac{m \theta}{2} \right) \right| \, \ln \left| \cot \left(\frac{n \theta}{2} \right) \right| \, \mathrm d \theta = 2 \pi \sum_{p=0}^{\infty} \frac{mn}{(2k+1)^{2} \operatorname{lcm}^{2}(m,n)} = \frac{\pi^{3} }{4} \frac{\gcd(m,n)}{\operatorname{lcm}(m,n)}. $$

And if $m$ and $n$ don't have the same number of powers of $2$ in their prime factorizations, then $$\int_{0}^{\pi} \ln \left| \cot \left(\frac{m \theta}{2} \right) \right| \, \ln \left| \cot \left(\frac{n \theta}{2} \right) \right| \, \mathrm d \theta =0. $$

Numerical approximations of the integral for specific values of $m$ and $n$ using Wolfram Alpha and the midpoint method seem to confirm this is correct.

To make this rigorous, I think you would need to use $\sum_{k=0}^{\infty} \frac{r^{2k+1} \cos \left((2k+1)\theta \right)}{2k+1} = \Re \operatorname{artanh}(re^{i \theta})$ and take the limit as $ r \to 1^{-}$.

Answer 2

I have a third example that I'm going to post as an separate answer: namely, $$I(m,n, \phi) = \int_{0}^{\pi} \frac{\sin(m \theta) \sin(n \theta )}{\left(\cosh(\phi) -\cos(m\theta)\right)\left(\cosh(\phi)- \cos(n \theta)\right)} \, \mathrm d \theta = \frac{2 \pi}{\exp \left( \frac{m+n}{\gcd(m,n)} \, \phi \right)-1}.$$

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Let $m$ and $n$ be positive integers, and assume $\phi >0$.

Using the series identity $$\sum_{k=0}^{\infty} e^{-k \phi } \sin(k \theta) = \sum_{k=\color{red}{1}}^{\infty} e^{-k \phi } \sin(k \theta) = \frac{1}{2} \frac{\sin(\theta)}{\cosh(\phi) - \cos(\theta)}, $$ which can be derived by extracting the imaginary part of the identity $$\sum_{k=0}^{\infty} \left(e^{-\phi} e^{i \theta}\right)^{k} = \frac{1}{1-e^{-\theta}e^{i \theta}}, $$ we have

$$ \begin{align} I(m,n, \phi) &= 4\int_{0}^{\pi} \sum_{j=1}^{\infty} e^{- j \phi} \sin(jm \theta) \sum_{k=1}^{\infty} e^{- k \phi } \sin(kn \theta) \, \mathrm d \theta \\ &= 4 \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} e^{-(j+k)\phi} \int_{0}^{\pi} \sin(jm \theta) \sin(kn \theta) \, \mathrm d \theta \\ &= 2 \pi \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} e^{-(j+k)\phi} \, \delta_{jm, kn} \\ &= 2 \pi \sum_{p=1}^{\infty} \exp \left( - \left(\frac{p \operatorname{lcm}(m,n)}{m} + \frac{p \operatorname{lcm}(m,n)}{n} \right) \phi\right) \\ &=2 \pi \sum_{p=1}^{\infty} \exp \left(-p \, \frac{m+n}{\gcd(m,n)} \, \phi \right) \\ &=2 \pi \, \frac{\exp\left(-\frac{m+n}{\gcd(m,n)} \, \phi \right)}{1- \exp\left(-\frac{m+n}{\gcd(m,n)} \, \phi \right)} \\ &= \frac{2 \pi}{\exp \left( \frac{m+n}{\gcd(m,n)} \, \phi\right)-1}. \end{align}$$

For "large" values of the parameters, Wolfram Alpha sometimes says the value of the integral is zero when it actuality the value is nonzero but very small.

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More generally, $$ \int_{0}^{\pi} \frac{\sin(m \theta) \sin(n \theta )}{\left(\cosh(\phi) -\cos(m\theta)\right)\left(\cosh(\lambda)- \cos(n \theta)\right)} \, \mathrm d \theta = \frac{2 \pi}{\exp \left( \frac{m \lambda+n \phi}{\gcd(m,n)} \right)-1}$$ for $\phi, \lambda >0$.