In my post, I had proved $$I(a,b)=\int_{0}^{\infty} \frac{\sin (a x) \sin (b x)}{x^{2}} \, \mathrm d x = \frac{\min (a, b) \pi}{2}, $$ in 2021. Recently, I started to investigate the integral further to: $$I(a_1,a_2,\cdots,a_n)=\int_{0}^{\infty} \frac{\sin (a_1x) \sin (a_2 x)\cdots\sin(a_nx)}{x^{2}} \,\mathrm d x.$$ Trying with small number $n$, I found and conjectured that $$ \int_0^{\infty} \frac{\sin (a_1 x) \sin (a_2 x) \cdots \sin (a_n x)}{x^2} \, \mathrm d x = \begin{cases} \frac{\pi}{2^n} \sum_{k=0}^{2^{n-1}}(-1)^{k+1}\left|A_k\right|\quad \quad \text {if } n \text { is even } \\ \frac{1}{2^{n-1}} \sum_{k=0}^{2^{n-1}}(-1)^k A_k\ln|A_k|\quad \text {if } n \text { is odd }\end{cases} $$ where $A_k$ is a sum of $(n-k)$ $ a_i$’s and $k$ $-a_j$’s.
Starting from $n=3$, $$ \begin{aligned} &\quad \sin(ax)\sin(bx)\sin(cx) \\&= \frac{e^{a x i}-e^{-a x i}}{2 i} \cdot \frac{e^{b x i}-e^{-b x i}}{2 i} \cdot \frac{e^{c x i}-e^{-c x i}}{2 i} \\ &= \frac{1}{8 i^3}\left[\left(e^{(a+b+c) i x}-e^{-(a+b+c) i x}\right) -\left(e^{(a+b-c) i x}-e^{-(a+b-c) i}\right) -\left(e^{(a+c-b) i x}-e^{-(a+c-b) i x}\right) \right.\\ & \quad \left. -\left(e^{(b+c-a) i x}-e^{-(b+c-a) i x}\right)\right] \\ &= -\frac{1}{4} [ {\sin ((a+b+c) x)} - {\sin ((a+b-c)x)} -{\sin ((a+c-b) x)} -{\sin ((b+c-a)x)} ] \end{aligned} $$ Via integration by parts, we have $$ \begin{align} &\quad \int_0^{\infty} \frac{\sin (a x) \sin (b x) \sin (c x)}{x^2} \, \mathrm d x \\ &= -\frac{1}{4} \int_0^{\infty} \frac{\sin (A x)-\sin (B x)-\sin (C x)-\sin (D x)}{x^2} \, \mathrm d x \tag{*} \\ &= \frac{1}{4} \int_0^{\infty} \frac{A \cos (A x)-B \cos (B x)-C \cos (C x)-D \cos (D x)}{x} \, \mathrm dx\\ &= \frac{1}{4} \int_0^{\infty}\left[A \cos (A x)-B \cos (B x)-C \cos (C x)-D \cos (D x)\right]\int_0^{\infty} e^{-yx} \, \mathrm d y \,\mathrm d x \\ &= \frac{1}{4} \int_0^{\infty} \int_0^{\infty}\left[A \cos (A x)-B \cos (B x)-C \cos (C x)-D \cos (D x)\right] e^{-yx} \, \mathrm d x \, \mathrm d y \\ &= \frac{1}{4} \int_0^{\infty}\left[ -\frac{A y}{y^2+A^2}+\frac{B y}{y^2+B^2}+\frac{C y}{y^2+C^2}+\frac{D y}{y^2+D} \right] \, \mathrm d y \\ &= \frac{1}{8}\left[-A \ln \left(y^2+A^2\right)+B \ln \left(y^2+B^2\right)+C \ln \left(y^2+C^2\right)+D \ln \left(y^2+D^2 \right)\right]_0^\infty \\ &= \frac{1}{4} \ln \left(|A|^A|B|^{-B}|C|^{-C}|D|^{-D}\right)\end{align} $$ due to $A-B-C-D=0$, where $A=a+b+c, B=a+b-c,C=a+c-b$ and $D=b+c-a$.
In the step $(*)$, we keep non-zero capital letters only.
For examples,
A. When $n=3$, $$ \int_0^{\infty} \frac{\sin x \sin (2 x) \sin (3 x)}{x^2} \, \mathrm d x = \frac{1}{4} \ln \left(6^6 \cdot 2^{-2} \cdot 4^{-4}\right) =\ln \left(\frac{3 \sqrt{3}}{2}\right) $$
B. When $n=5$,
Following the same steps as $n=3$, we have
$$\sin x \sin (2 x) \sin (3 x)\sin(4x)\sin(5x)=\frac{1}{16} \left(\sin (15 x)-\sin (13 x)-\sin (11 x)+\sin (5 x)+\sin (3 x)+\sin x\right) $$ $$\int_0^{\infty} \frac{\sin x \sin (2 x) \sin (3 x) \sin (4 x) \sin (5 x)}{x^2} \, \mathrm d x = \frac{1}{16} \ln \left(\frac{13^{13} \cdot 11^{11}}{15^{15} \cdot 5^5 \cdot 3^3}\right) =\ln \left(\frac{13^{\frac{13}{16}} \cdot 11^{\frac{11}{16}}}{15 \cdot \sqrt[4]{5} \cdot \sqrt[8]{3}}\right)$$
Now if $n>2$ is even, say $n=4$ then \begin{aligned} &\quad \sin (a x) \sin (b x) \sin (e x) \sin (d x) \\ & = \frac{e^{axi} -e^{-a x i}}{2 i} \cdot \frac{e^{bxi}-e^{-b x i}}{2 i} \frac{e^{cx i} e^{-cxi}}{2 i} \cdot \frac{e^{dxi }-e^{-dxi}}{2 i} \\ & = \frac{1}{16}\left[e^{(a+b+c+d) xi}+e^{-(a+b+c+d) xi}\right. -\left(e^{(a+b+c-d) xi}+e^{-(a+b+c-d) xi}\right) \\ & \quad -\left(e^{(a+c+d-b) xi}+e^{-(a+c+d-b) xi}\right) -\left(e^{(a+b+d-c)xi}+e^{-(a+b+d-c) i x}\right) \\ & \quad -\left(e^{(a+c+d-a) xi}+e^{-(b+c+d-a) xi}\right) +\left(e^{(a+b-c-d) xi}+e^{-(a+b-c-d) xi}\right) \\ & \quad +\left(e^{(a+c-b-d) i x}+e^{-(a+c-b-d) xi}\right) \left.+\left(e^{(a+d-b-c) xi}+e^{-(a+d-b-c) xi}\right)\right] \\ & =\frac{1}{8}[\cos ((a+b+c+d) x)-\cos ((a+b+c-d) x) -\cos ((a+c+d-b) x(-\cos ((a+b+d-c) x )\\ & \quad -\cos ((b+c+d-a) x)+\cos ((a+b-c-d) x) +\cos ((a+c-b-d) x)+\cos ((a+d-b-c) x] \end{aligned}
Plugging back with integration by parts yields \begin{align} &\quad \int_0^{\infty} \frac{\sin (a x) \sin (bx) \sin (c x) \sin (d x) }{x^2} \, \mathrm d x \\&= \small{ -\frac{1}{8} \int_0^{\infty} \frac{A \sin (A x)-B \sin (B x)-C \sin (C x) -D \sin (D x)-E \sin (E x)+F \sin (F x)-G \sin (G x)-H \sin (H x)}{x}} \, \mathrm dx \\ &=\frac{\pi}{16}(-|A|+|B|+|C|+|D|+|E|-|F| -|G|-|H|) \end{align} where \begin{align} A &= a+b+c+d, & B &= a+b+c-d, \\ C &= a+c+d-a, & D &= a+b+d-c \\ E &= b+c+d-a, & F &= a+b-c-d, \\ G &= a+c-b-d, & H &= a+d-b-c. \end{align}
For example,
$$ \boxed{\int_0^{\infty} \frac{\sin x \sin (2 x) \sin (3 x) \sin (4 x)}{x^2} \, \mathrm d x = \frac{\pi}{16}(-10+2+6+4+8-4-2-0) = \frac{\pi}{4}}. $$
Conclusion
When $n$ is even, the product of sines can be expressed as a linear sum of cosine of multiple angles. Via integration by parts, it is changed to a sum sine of multiple angles and hence exact value of integral is $h\pi/2$.
When $n$ is odd, the product of sines can be expressed as a linear sum of sine of multiple angles. Via integration by parts, it is changed to a sum of cosine of multiple angles and hence exact value of integral is $\ln k$.
However, it is very hard to prove my conjecture rigorously.
