Evaluate integral: \int_0^{\pi/3}\ln^2(\frac{\sin x}{\sin(x+\pi/3)})dx

Concepcion Hale

Concepcion Hale

Answered question

2022-01-03

Evaluate integral:
0π3ln2(sinxsin(x+π3))dx

Answer & Explanation

turtletalk75

turtletalk75

Beginner2022-01-04Added 29 answers

It turns out that this integral takes on a very simple form amenable to analysis via residues. Let u=sinxsin(x+π3). We may then find that (+)
tanx=(32)u1(u2)
A little bit of algebra reveals a very nice form for the differential:
dx=32du1u+u2
so the original integral takes on a much simpler-looking form:
3201dulog2u1u+u2
This is not ready for contour integration yet. We may transform this into such an integral by substituting u=1v and observing that
01dulog2u1u+u2=1dulog2u1u+u2=120dulog2u1u+u2
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 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
Cdzlog3z1z+z2=i(6π0dulog2u1u+u2+8π30du11u+u2)+12π20dulogu1u+u2
We set this equal to i2π times the sum of the residues of the poles of the integrand within C. The poles are z(eiπ3,ei5π3). The residues are
Resz=eiπ3=π3273
Resz=ei5π3=125π3273
i2π times the sum of these residues is then
i248π4273
Equating imaginary parts of the integral to the above quantity, we see that
sirpsta3u

sirpsta3u

Beginner2022-01-05Added 42 answers

Set t=sinxsin(x+π3), then
0π3ln2[sinxsin(x+π3)]dx=3201ln2t1t+t2dt (1)
Method 1:
01ln2t1t+t2dt=011+t1+t3ln2tdt
=01k=0(1)kt3k(1+t)ln2tdt
=k=0(1)k(01t3kln2tdt+01t3k+1ln2tdt) (2)
Using
01xαlnnxdx=(1)nn!(α+1)n+1, for n=0,1,2, (3)
then (2) turns out to be
01ln2t1t+t2dt=k=0(1)k[2(3k+1)3+2(3k+2)3]
=233k=0(1)k[1(k+
karton

karton

Expert2022-01-11Added 613 answers

0π/3ln2(sin(x)sin(x+π/3))dx0π/3ln2(sin(x)sin(x+π/3))dx=π/6π/6ln2(sin(x+π/6)cos(x))dx=π/6π/6ln2(32tan(x)+12)dxt32tan(x)+12(x=arctan(2t13)),(dx=32dtt2t+1):0π/3ln2(sin(x)sin(x+π/3)dx=3201ln2(t)t2t+1dtt2t+1=0 roots are given by tr=1+3i2=eiπ/3 and tI=1tr such that:0π/3ln2(sin(x)sin(x+π/3))dx=3201ln2(t)(ttr)(ttr)dt=3212iI(tr)01ln2(t)(1ttr1ttr)dt=I01ln2(t)tItdt=I0trln2(tit)1tdt=I0trln(1t)[2ln(tit)1t]dt=2I0trLi1(t)tln(tit)dt
where Lis(z) is a polylogarithm function.
By using the recursive property of those functions:

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