 Concepcion Hale

2022-01-03

Evaluate integral:
${\int }_{0}^{\frac{\pi }{3}}{\mathrm{ln}}^{2}\left(\frac{\mathrm{sin}x}{\mathrm{sin}\left(x+\frac{\pi }{3}\right)}\right)dx$ turtletalk75

It turns out that this integral takes on a very simple form amenable to analysis via residues. Let $u=\frac{\mathrm{sin}x}{\mathrm{sin}\left(x+\frac{\pi }{3}\right)}$. We may then find that (+)
$\mathrm{tan}x=\frac{\left(\frac{\sqrt{3}}{2}\right)u}{1-\left(\frac{u}{2}\right)}$
A little bit of algebra reveals a very nice form for the differential:
$dx=\frac{\sqrt{3}}{2}\frac{du}{1-u+{u}^{2}}$
so the original integral takes on a much simpler-looking form:
$\frac{\sqrt{3}}{2}{\int }_{0}^{1}du\frac{{\mathrm{log}}^{2}u}{1-u+{u}^{2}}$
This is not ready for contour integration yet. We may transform this into such an integral by substituting $u=\frac{1}{v}$ and observing that
${\int }_{0}^{1}du\frac{{\mathrm{log}}^{2}u}{1-u+{u}^{2}}={\int }_{1}^{\mathrm{\infty }}du\frac{{\mathrm{log}}^{2}u}{1-u+{u}^{2}}=\frac{1}{2}{\int }_{0}^{\mathrm{\infty }}du\frac{{\mathrm{log}}^{2}u}{1-u+{u}^{2}}$
We may now analyze that last integral via the residue theorem. Consider the integral
where C is a keyhole contour that passes up and back along the positive real axis. It may be shown that the integral along the large and small circular arcs vanish as the radii of the arcs goes to $\mathrm{\infty }$ and 0, respectively. We may then write the integral in terms of positive contributions just above the real axis and negative contributions just below. The result is
${\oint }_{C}dz\frac{{\mathrm{log}}^{3}z}{1-z+{z}^{2}}=i\left(-6\pi {\int }_{0}^{\mathrm{\infty }}du\frac{{\mathrm{log}}^{2}u}{1-u+{u}^{2}}+8{\pi }^{3}{\int }_{0}^{\mathrm{\infty }}du\frac{1}{1-u+{u}^{2}}\right)+12{\pi }^{2}{\int }_{0}^{\mathrm{\infty }}du\frac{\mathrm{log}u}{1-u+{u}^{2}}$
We set this equal to $i2\pi$ times the sum of the residues of the poles of the integrand within C. The poles are $z\in \left({e}^{i\frac{\pi }{3}},{e}^{i5\frac{\pi }{3}}\right)$. The residues are
$Re{s}_{z={e}^{i\frac{\pi }{3}}}=-\frac{{\pi }^{3}}{27\sqrt{3}}$
$Re{s}_{z={e}^{i5\frac{\pi }{3}}}=\frac{125{\pi }^{3}}{27\sqrt{3}}$
$i2\pi$ times the sum of these residues is then
$i\frac{248{\pi }^{4}}{27\sqrt{3}}$
Equating imaginary parts of the integral to the above quantity, we see that sirpsta3u

Set $t=\frac{\mathrm{sin}x}{\mathrm{sin}\left(x+\frac{\pi }{3}\right)}$, then
${\int }_{0}^{\frac{\pi }{3}}{\mathrm{ln}}^{2}\left[\frac{\mathrm{sin}x}{\mathrm{sin}\left(x+\frac{\pi }{3}\right)}\right]dx=\frac{\sqrt{3}}{2}{\int }_{0}^{1}\frac{{\mathrm{ln}}^{2}t}{1-t+{t}^{2}}dt$ (1)
Method 1:
${\int }_{0}^{1}\frac{{\mathrm{ln}}^{2}t}{1-t+{t}^{2}}dt={\int }_{0}^{1}\frac{1+t}{1+{t}^{3}}{\mathrm{ln}}^{2}tdt$
$={\int }_{0}^{1}\sum _{k=0}^{\mathrm{\infty }}{\left(-1\right)}^{k}{t}^{3k}\left(1+t\right){\mathrm{ln}}^{2}tdt$
$=\sum _{k=0}^{\mathrm{\infty }}{\left(-1\right)}^{k}\left({\int }_{0}^{1}{t}^{3k}{\mathrm{ln}}^{2}tdt+{\int }_{0}^{1}{t}^{3k+1}{\mathrm{ln}}^{2}tdt\right)$ (2)
Using
${\int }_{0}^{1}{x}^{\alpha }{\mathrm{ln}}^{n}xdx=\frac{{\left(-1\right)}^{n}n!}{{\left(\alpha +1\right)}^{n+1}}$, for $n=0,1,2,\dots$ (3)
then (2) turns out to be
${\int }_{0}^{1}\frac{{\mathrm{ln}}^{2}t}{1-t+{t}^{2}}dt=\sum _{k=0}^{\mathrm{\infty }}{\left(-1\right)}^{k}\left[\frac{2}{{\left(3k+1\right)}^{3}}+\frac{2}{{\left(3k+2\right)}^{3}}\right]$ karton

where $L{i}_{s}\left(z\right)$ is a polylogarithm function.
By using the recursive property of those functions:

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