In this question, user Franklin Pezzuti Dyer gives the following surprising integral evaluation:
$$\int_0^{\pi/2}\ln \lvert\sin(mx)\rvert \cdot \ln \lvert\sin(nx)\rvert \, dx = \frac{\pi^3}{24} \frac{\gcd^2(m,n)}{mn}+\frac{\pi\ln^2(2)}{2}$$
I've verified this numerically for small values for $m,n$. Is there a proof? Also, can we generalize it to more factors in the integrand?
Okay, I'll prove it for you.
Start with the following well-known identity: $$\int_0^{\pi}\cos(mx)\cos(nx)dx=\frac{\pi}{2}\delta_{mn}\tag{1}$$ ...where $m,n$ are positive integers. Recall also the well-known Fourier Series $$\sum_{n=1}^\infty \frac{\cos(kx)}{k}=-\frac{\ln(2-2\cos(x))}{2}\tag{2}$$ Now, replace $m$ in $(1)$ with $mk$, where both $m,k$ are integers, and divide both sides by $k$ to get $$\int_0^{\pi}\frac{\cos(kmx)}{k}\cos(nx)dx=\frac{\pi\delta_{(mk)n}}{2k}$$ Then sum both sides from $k=1$ to $\infty$ to get $$-\frac{1}{2}\int_0^{\pi}\ln(2-2\cos(mx))\cos(nx)dx=\frac{\pi m}{2n}[m|n]$$ where the brackets on the $RHS$ are Iverson Brackets. A bit more manipulation yields the equality $$\int_0^{\pi}\ln\bigg(\frac{1-\cos(mx)}{2}\bigg)\cos(nx)dx=-\frac{\pi m}{n}[m|n]$$ Now, this time, replace $n$ with $nk$ and divide both sides by $k$. This yields $$\int_0^{\pi}\ln\bigg(\frac{1-\cos(mx)}{2}\bigg)\frac{\cos(knx)}{k}dx=-\frac{\pi m}{k^2n}[m|kn]$$ Then sum from $k=1$ to $\infty$ to get $$-\frac{1}{2}\int_0^{\pi}\ln\bigg(\frac{1-\cos(mx)}{2}\bigg)\ln(2-2\cos(nx))dx=-\sum_{k=1}^{\infty} \frac{\pi m}{k^2n}[m|kn]$$ Now notice the following about the series on the RHS. Due to the Iverson Bracket, the kth term is zero unless $m|kn$, or unless $k$ is divisible by $m/\gcd(m,n)$. Thus, we let $k=jm/\gcd(m,n)$ for the integers $j=1$ to $\infty$ and reindex the sum: $$\begin{align} -\frac{1}{2}\int_0^{\pi}\ln\bigg(\frac{1-\cos(mx)}{2}\bigg)\ln(2-2\cos(nx))dx &=-\sum_{j=1}^{\infty} \frac{\pi m}{(jm/\gcd(m,n))^2n}\\ &=-\frac{\pi\gcd^2(m,n)}{mn}\sum_{j=1}^{\infty} \frac{1}{j^2}\\ &=-\frac{\pi^3\gcd^2(m,n)}{6mn}\\ \end{align}$$ or $$\int_0^{\pi}\ln\bigg(\frac{1-\cos(mx)}{2}\bigg)\ln(2-2\cos(nx))dx=\frac{\pi^3\gcd^2(m,n)}{3mn}\tag{3}$$ Then, by using the result $$\int_0^{\pi}\ln(1-\cos(ax))=-\pi\ln(2)\tag{4}$$ for all positive integers $a$, and the trigonometric identity $$\sin^2(x/2)=\frac{1-\cos(x)}{2}\tag{5}$$ and finally, a substitution $x\to 2x$, the result easily follows from $(3)$ : $$\bbox[lightgray,5px]{\int_0^{\pi/2}\ln \lvert\sin(mx)\rvert \cdot \ln \lvert\sin(nx)\rvert \, dx = \frac{\pi^3}{24} \frac{\gcd^2(m,n)}{mn}+\frac{\pi\ln^2(2)}{2}}$$
Though, I think the Fourier series approach is the cleaner approach for this integral, I'd like to present a somewhat unique alternative method. This integral may be evaluated by exploiting symmetry and the following multiple-angle identity: \begin{equation} \sin(mx)=2^{m-1}\prod_{k=0}^{m-1}\sin\left(x+\frac{\pi k}{m}\right) \end{equation} Your integral is equal to $\frac{1}{2}I(m,n)$ where \begin{equation} I(m,n)=\int_0^{\pi}\ln|\sin(mx)|\cdot\ln|\sin(nx)|dx \end{equation}
First, suppose than $m,n$ are coprime. We see that \begin{equation} \begin{split} I(1,n)&=\int_0^{\pi}\ln|\sin(x)|\cdot\ln|\sin(nx)|dx\\ &=\int_0^{\pi}\ln|\sin(mx)|\cdot\ln|\sin(mnx)|dx\\ \end{split} \end{equation} since the integrand has period $\pi$. Applying the multiple-angle formula for $\sin$ gives us that \begin{equation} I(1,n)=\sum_{k=0}^{m-1}\int_0^{\pi}\ln|\sin(mx)|\cdot\ln|\sin(nx+\pi k/m)|dx+ (m-1)\ln(2)\int_0^{\pi}\ln|\sin(mx)|dx \end{equation} Since $n,m$ are coprime, then the map $k\mapsto nk$ is an automorphism on $\mathbb{Z}/m$, meaning that \begin{equation} \begin{split} I(1,n)&=\sum_{k=0}^{m-1}\int_0^{\pi}\ln|\sin(mx)|\cdot\ln|\sin(nx+\pi nk/m)|dx \\ &\qquad\qquad + (m-1)\ln(2)\int_0^{\pi}\ln|\sin(mx)|dx\\ &=\sum_{k=0}^{m-1}\int_0^{\pi}\ln|\sin(m[x+\pi k/m])|\cdot\ln|\sin(n[x+\pi k/m])|dx \\ &\qquad\qquad+(m-1)\ln(2)\int_0^{\pi}\ln|\sin(mx)|dx\\ \end{split} \end{equation} Again, since the integrands have period $\pi$ we may simplify \begin{equation} \begin{split} I(1,n)&=\sum_{k=0}^{m-1}\int_0^{\pi}\ln|\sin(mx)|\cdot\ln|\sin(nx)|dx+(m-1)\ln(2)\int_0^{\pi}\ln|\sin(x)|dx\\ &=mI(m,n)+(m-1)\ln(2)J\\ \end{split} \end{equation} where \begin{equation} J=\int_0^{\pi}\ln(\sin(x))dx \end{equation} Noting that $I(1,n)=I(n,1)$ and applying the above identity twice gives us that \begin{equation} I(m,n)=\frac{1}{mn}[I(1,1)+\ln(2)J]-\ln(2)J \end{equation}
Now, in the case where $m,n$ are not coprime, then $\frac{m}{\gcd(m,n)},\frac{n}{\gcd(m,n)}$ are coprime, so by the periodicity of the integrand, \begin{equation} I(m,n)=I\left(\frac{m}{\gcd(m,n)},\frac{n}{\gcd(m,n)}\right)=\frac{\gcd^2(m,n)}{mn}[I(1,1)+\ln(2)J]-\ln(2)J \end{equation}
Finally, $I(1,1)$ and $J$ are somewhat famous integrals which may be evaluated in a few different ways. This answer shows that \begin{equation} J=-\pi\ln(2) \end{equation} using an elementary method, and in this answer I use a complex-analytic method to show that \begin{equation} I(1,1)=\pi\ln^2(2)+\frac{\pi^3}{12} \end{equation} Putting these facts together, we have that \begin{equation} I(m,n)=\frac{\pi^3}{12}\frac{\gcd^2(m,n)}{mn}+\pi\ln^2(2) \end{equation} In other words, the original integral is given by \begin{equation} \boxed{\int_0^{\pi/2}\ln|\sin(mx)|\cdot\ln|\sin(nx)|dx= \frac{\pi^3}{24}\frac{\gcd^2(m,n)}{mn}+\frac{\pi\ln^2(2)}{2}} \end{equation} which I think is a really beautiful result.
A generalization of this integral is $$\small \int_{0}^{\pi} \ln \left(1-2 e^{- \phi} \cos (mx) + e^{- 2\phi} \right) \ln \left( 1-2 e^{- \lambda} \cos(nx) +e^{-2 \lambda}\right) \, \mathrm dx= \frac{2\pi \gcd^{2} (m,n) \operatorname{Li}_{2} \left(\exp \left(- \, \frac{m\lambda +n \phi}{\gcd(m,n)} \right) \right)}{mn} , $$ where $\operatorname{Li}_{2}(-)$ is the dillogarithm, $m$ and $n$ are positive integers, and $\phi, \lambda \ge 0$.
To prove this result, we'll use the Fourier series $$\sum_{n=1}^{\infty} \frac{e^{- k \phi} \cos(k \theta)}{k} = - \frac{1}{2} \, \ln \left(1-2 e^{- \phi} \cos(\theta) +e^{- 2 \phi} \right),$$ which can be derived by extracting the real part of the identity $$\sum_{k=1}^{\infty} \frac{\left(e^{-\phi}e^{i \theta}\right)^{k}}{k} = - \ln \left(1-e^{- \phi} e^{i \theta} \right). $$
We get
$$ \begin{align} &\int_{0}^{\pi} \ln \left(1-2 e^{- \phi} \cos (mx) + e^{- 2\phi} \right) \ln \left( 1-2 e^{- \lambda} \cos(nx) +e^{-2 \lambda}\right) \, \mathrm dx \\&= 4 \int_{0}^{\pi} \sum_{j=1}^{\infty}\frac{e^{- j \phi}\cos(jm x)}{j} \sum_{k=1}^{\infty} \frac{e^{-k \lambda }\cos(kn x)}{k} \mathrm d x \\ &= 4 \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{e^{- j \phi} e^{- k \lambda}}{jk} \int_{0}^{\pi} \cos(j m x) \cos(kn x) \, \mathrm d x \\ &= 2 \pi \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{e^{- j \phi} e^{- k \lambda}}{jk} \, \delta_{jm, kn} \\ &= 2 \pi \sum_{p=1}^{\infty} \frac{mn }{p \operatorname{lcm}(m,n) p \operatorname{lcm}(m,n)} \, \exp \left(-\frac{p \operatorname{lcm}(m,n)}{m} \phi - \frac{p\operatorname{lcm}(m,n)}{n} \lambda \right) \\ &= \frac{2 \pi \gcd^{2}(m,n)}{mn} \, \sum_{p=1}^{\infty} \frac{1}{p^{2}} \, \exp \left(- p \, \frac{m\lambda +n \phi }{\gcd(m,n)} \right) \\ &= \frac{2 \pi \gcd^{2}(m,n)}{mn} \, \operatorname{Li}_{2} \left(\exp \left(- \, \frac{m \lambda+n \phi}{\gcd(m,n)} \right) \right). \end{align}$$
When the value of the integral is very small, Wolfram Alpha sometimes incorrectly says that the value of the integral is zero.
For $\phi=0$ and $\lambda =0$, we have $$ \begin{align} &\int_{0}^{\pi} \ln \left(2-2 \cos(mx) \right) \ln \left(2-2 \cos(nx)\right) \, \mathrm dx \\ &= \int_{0}^{\pi} \ln \left(4 \ \frac{1-\cos(mx)}{2} \right) \ln \left(4 \ \frac{1+ \cos(nx)}{2} \right) \, \mathrm dx \\ &= \small \pi \ln^{2}(4) + \ln(4) \left( \int_{0}^{\pi} \ln\left(\frac{1- \cos(mx)}{2} \right) \, \mathrm dx + \int_{0}^{\pi} \ln \left(\frac{1- \cos(nx)}{2} \right) \, \mathrm dx \right) \\ &+ \small \int_{0}^{\pi} \ln \left(\frac{1- \cos(mx)}{2} \right) \ln \left(\frac{1-\cos(nx)}{2} \right) \, \mathrm dx \\ &\overset{(1)}{=} \small \pi \ln^{2}(4) + \ln(4) \left(-2 \pi \ln(2) -2 \pi \ln(2) \right)+ 2 \int_{0}^{\pi/2} \ln \left(\frac{1- \cos(2mt)}{2} \right) \ln \left(\frac{1-\cos(2nt)}{2} \right) \, \mathrm dt \\&= \color{red}{-4 \pi \ln^{2}(2) +8 \int_{0}^{\pi/2} \ln |\sin(mt)| \ln |\sin(nt) | \, \mathrm dt} \\ & = \frac{2 \pi \gcd^{2}(m,n)}{mn} \, \operatorname{Li}_{2} (1) \\ &= \frac{2 \pi \gcd^{2}(m,n)}{mn} \, \zeta(2) \\ &= \frac{\pi^{3}}{3} \frac{\gcd^{2}(m,n)}{mn}. \end{align}$$
$(1)$
For an positive integer $m$, $$ \begin{align} \int_{0}^{\pi} \ln \left(\frac{1-\cos(mx)}{2} \right) \, \mathrm dx &= \frac{1}{m} \int_{0}^{m \pi} \ln \left(\frac{1-\cos(u)}{2} \right) \, \mathrm du \\ &= \frac{1}{m} \sum_{k=0}^{m-1} \int_{k \pi}^{(k+1) \pi} \ln \left(\frac{1-\cos(u)}{2} \right) \, \mathrm du \\ &= \frac{1}{m} \sum_{k=0}^{m-1} \int_{0}^{\pi} \ln \left(\frac{1 -(-1)^{k} \cos(v)}{2} \right) \, \mathrm dv \\ &= \frac{1}{m} \sum_{k=0}^{m-1} \int_{0}^{\pi} \ln \left(\frac{1 - \cos(v)}{2} \right) \, \mathrm dv \\ &= \int_{0}^{\pi} \ln \left(\frac{1 - \cos(v)}{2} \right) \, \mathrm dv \\ &= \int_{0}^{\pi } \ln \left(\sin^{2}\left(\frac{u}{2}\right) \right) \, \mathrm dv \\ &= 4 \int_{0}^{\pi/2} \ln \left(\sin (v) \right) \, \mathrm dv \\ &= -2 \pi \ln(2) . \end{align}$$
I was going to use $|r| \le 1$ and $|s| \le 1$ instead of $e^{- \theta}$ and $e^{- \lambda}$, but I chose the later because it looked cleaner.
So, we're starting at this integral: $$I = \int_0^{\pi/2}\ln\lvert\sin(mx)\rvert\cdot \ln\lvert\sin(nx)\rvert\, dx$$
Now, here is to use a special way of writing $\ln|\sin\theta|$ as an infinite sum: $$\ln|\sin\theta| = -\ln 2 - \sum_{k=1}^{\infty} \frac{\cos(2k\theta)}{k}$$ This neat little formula works as long as $\theta$ isn't a multiple of $\pi$ (like $0, \pi, 2\pi$, etc.), which is fine for our integral since we're going from $0$ to $\pi/2$.
Let's plug this into our integral. We'll have: $$\ln|\sin(mx)| = -\ln 2 - \sum_{j=1}^{\infty} \frac{\cos(2jmx)}{j}$$ $$\ln|\sin(nx)| = -\ln 2 - \sum_{k=1}^{\infty} \frac{\cos(2knx)}{k}$$
Substituting these into our integral $I$: $$I = \int_0^{\pi/2} \left(-\ln 2 - \sum_{j=1}^{\infty} \frac{\cos(2jmx)}{j}\right) \left(-\ln 2 - \sum_{k=1}^{\infty} \frac{\cos(2knx)}{k}\right) dx$$
Now, we need to multiply these two big expressions. It's going to give us a few terms: $$I = \int_0^{\pi/2} \left[ (\ln 2)^2 + (\ln 2)\sum_{k=1}^{\infty} \frac{\cos(2knx)}{k} + (\ln 2)\sum_{j=1}^{\infty} \frac{\cos(2jmx)}{j} + \left(\sum_{j=1}^{\infty} \frac{\cos(2jmx)}{j}\right)\left(\sum_{k=1}^{\infty} \frac{\cos(2knx)}{k}\right) \right] dx$$
We can handle this integral piece by piece.
First up, we have $\int_0^{\pi/2} (\ln 2)^2 dx$. Since $(\ln 2)^2$ is just a constant, this integral is easy: $$\int_0^{\pi/2} (\ln 2)^2 dx = (\ln 2)^2 \int_0^{\pi/2} dx = (\ln 2)^2 \left[x\right]_0^{\pi/2} = \frac{\pi}{2}(\ln 2)^2$$
Next, let's look at the second term: $(\ln 2)\sum_{k=1}^{\infty} \frac{1}{k} \int_0^{\pi/2} \cos(2knx) dx$. We need to evaluate $\int_0^{\pi/2} \cos(2knx) dx$. The antiderivative of $\cos(2knx)$ is $\frac{\sin(2knx)}{2kn}$. So, $$\int_0^{\pi/2} \cos(2knx) dx = \left[\frac{\sin(2knx)}{2kn}\right]_0^{\pi/2} = \frac{\sin(kn\pi)}{2kn} - \frac{\sin(0)}{2kn} = \frac{\sin(kn\pi)}{2kn}$$ Since $k$ and $n$ are positive integers, $kn$ is also an integer. And we know that $\sin(\text{integer} \times \pi) = 0$. So, this whole second term becomes zero!
Similarly, the third term $(\ln 2)\sum_{j=1}^{\infty} \frac{1}{j} \int_0^{\pi/2} \cos(2jmx) dx$ will also be zero for the same reason.
Now for the last, and most interesting, part: $$\int_0^{\pi/2} \left(\sum_{j=1}^{\infty} \frac{\cos(2jmx)}{j}\right)\left(\sum_{k=1}^{\infty} \frac{\cos(2knx)}{k}\right) dx = \sum_{j=1}^{\infty}\sum_{k=1}^{\infty} \frac{1}{jk} \int_0^{\pi/2} \cos(2jmx)\cos(2knx) dx$$To handle the product of cosines, we can use the identity $\cos A \cos B = \frac{1}{2}(\cos(A-B) + \cos(A+B))$. So,$$\cos(2jmx)\cos(2knx) = \frac{1}{2}(\cos(2(jm-kn)x) + \cos(2(jm+kn)x))$$Now we need to integrate this from $0$ to $\pi/2$:$$\frac{1}{2} \int_0^{\pi/2} [\cos(2(jm-kn)x) + \cos(2(jm+kn)x)] dx$$
Let's consider two scenarios:
If $jm - kn \neq 0$ and $jm + kn \neq 0$: The integral becomes: $$\frac{1}{2} \left[\frac{\sin(2(jm-kn)x)}{2(jm-kn)} + \frac{\sin(2(jm+kn)x)}{2(jm+kn)}\right]_0^{\pi/2} = \frac{1}{2} \left(\frac{\sin((jm-kn)\pi)}{2(jm-kn)} + \frac{\sin((jm+kn)\pi)}{2(jm+kn)}\right)$$ Since $j, m, k, n$ are all integers, both $jm-kn$ and $jm+kn$ are also integers. Therefore, their sines at $\pi$ will be zero, making the entire integral zero in this case.
Now, what if $jm - kn = 0$? This means $jm = kn$. Since $j, m, k, n \ge 1$, this can definitely happen. In this case, $jm + kn = 2jm \neq 0$. Our integral simplifies to: $$\frac{1}{2} \int_0^{\pi/2} [\cos(0) + \cos(2(2jm)x)] dx = \frac{1}{2} \int_0^{\pi/2} [1 + \cos(4jmx)] dx$$Integrating this gives:$$\frac{1}{2} \left[x + \frac{\sin(4jmx)}{4jm}\right]_0^{\pi/2} = \frac{1}{2} \left(\frac{\pi}{2} + \frac{\sin(2jm\pi)}{4jm} - (0 + 0)\right)$$ Again, since $jm$ is an integer, $\sin(2jm\pi) = 0$. So, the integral in this special case is $\frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$.
So, the sum becomes $\sum_{j=1}^{\infty}\sum_{k=1}^{\infty} \frac{1}{jk} \cdot \delta_{jm,kn} \cdot \frac{\pi}{4}$, where $\delta_{jm,kn}$ is the Kronecker delta, which is 1 if $jm = kn$ and 0 otherwise. This simplifies to $\frac{\pi}{4} \sum_{j,k \ge 1 \text{ such that } jm=kn} \frac{1}{jk}$.
To figure out when $jm = kn$, let's bring in the greatest common divisor of $m$ and $n$. Let $d = \gcd(m,n)$. Then we can write $m = dm'$ and $n = dn'$, where $m'$ and $n'$ are coprime positive integers (they don't share any common factors other than 1). The condition $jm = kn$ now becomes $j(dm') = k(dn')$, which simplifies to $jm' = kn'$. Since $m'$ and $n'$ have no common factors, for this equality to hold, $j$ must be a multiple of $n'$, say $j = \lambda n'$ for some positive integer $\lambda$, and similarly, $k$ must be a multiple of $m'$, say $k = \lambda m'$ for the same $\lambda$.
Substituting these back into our sum: $$\sum_{\lambda=1}^{\infty} \frac{1}{(\lambda n')(\lambda m')} = \sum_{\lambda=1}^{\infty} \frac{1}{\lambda^2 m'n'} = \frac{1}{m'n'} \sum_{\lambda=1}^{\infty} \frac{1}{\lambda^2}$$ We know a famous result that $\sum_{\lambda=1}^{\infty} \frac{1}{\lambda^2} = \zeta(2) = \frac{\pi^2}{6}$ (this is the Basel problem!).
So our sum becomes $\frac{1}{m'n'} \frac{\pi^2}{6}$. Now, let's put back $m' = m/d$ and $n' = n/d$: $$\frac{1}{(m/d)(n/d)} \frac{\pi^2}{6} = \frac{d^2}{mn} \frac{\pi^2}{6}$$
Finally, the contribution of the fourth term to our integral $I$ is $\frac{\pi}{4} \left(\frac{d^2}{mn} \frac{\pi^2}{6}\right) = \frac{\pi^3 d^2}{24mn}$.
Putting it all together, the value of the integral $I$ is the sum of the first part and the fourth part (since the second and third parts were zero): $$I = \frac{\pi}{2}(\ln 2)^2 + \frac{\pi^3 (\gcd(m,n))^2}{24mn}$$
And there you have it!
Let's quickly check the examples they gave:
If $m=n=1$, then $d=\gcd(1,1)=1$. $$I = \frac{\pi}{2}(\ln 2)^2 + \frac{\pi^3 (1)^2}{24(1)(1)} = \frac{\pi}{2}(\ln 2)^2 + \frac{\pi^3}{24}$$ This matches the known result.
If $m=1, n=2$, then $d=\gcd(1,2)=1$. $$I = \frac{\pi}{2}(\ln 2)^2 + \frac{\pi^3 (1)^2}{24(1)(2)} = \frac{\pi}{2}(\ln 2)^2 + \frac{\pi^3}{48}$$ This also matches the known result.
