Evaluate integral: ∫0π3ln2(sin⁡xsin⁡(x+π3))dx

Question

Evaluate integral:  ∫0π3ln2(sin⁡xsin⁡(x+π3))dx

turtletalk75 · Accepted Answer

It turns out that this integral takes on a very simple form amenable to analysis via residues. Let u=sin⁡xsin⁡(x+π3). We may then find that (+)  tan⁡x=(32)u1−(u2)  A little bit of algebra reveals a very nice form for the differential:  dx=32du1−u+u2  so the original integral takes on a much simpler-looking form:  32∫01dulog2u1−u+u2  This is not ready for contour integration yet. We may transform this into such an integral by substituting u=1v and observing that  ∫01dulog2u1−u+u2=∫1∞dulog2u1−u+u2=12∫0∞dulog2u1−u+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  ∮Cdzlog3z1−z+z2=i(−6π∫0∞dulog2u1−u+u2+8π3∫0∞du11−u+u2)+12π2∫0∞dulog⁡u1−u+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 · Accepted Answer

Set t=sin⁡xsin⁡(x+π3), then  ∫0π3ln2[sin⁡xsin⁡(x+π3)]dx=32∫01ln2t1−t+t2dt (1)  Method 1:  ∫01ln2t1−t+t2dt=∫011+t1+t3ln2tdt  =∫01∑k=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  ∫01ln2t1−t+t2dt=∑k=0∞(−1)k[2(3k+1)3+2(3k+2)3]  =233∑k=0∞(−1)k[1(k+

karton · Accepted Answer

∫0π/3ln2⁡(sin⁡(x)sin⁡(x+π/3))dx∫0π/3ln2⁡(sin⁡(x)sin⁡(x+π/3))dx=∫−π/6π/6ln2⁡(sin⁡(x+π/6)cos⁡(x))dx=∫−π/6π/6ln2⁡(32tan⁡(x)+12)dxt≡32tan⁡(x)+12→(x=arctan⁡(2t−13)),(dx=32dtt2−t+1):∫0π/3ln2⁡(sin⁡(x)sin⁡(x+π/3)dx=32∫01ln2⁡(t)t2−t+1dtt2−t+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=32∫01ln2⁡(t)(t−tr)(t−tr)dt=3212iI(tr)∫01ln2⁡(t)(1t−tr−1t−tr∗)dt=I∫01ln2⁡(t)tI−tdt=I∫0trln2⁡(tit)1−tdt=I∫0trln⁡(1−t)[2ln⁡(tit)1t]dt=−2I∫0trLi1(t)tln⁡(tit)dt where Lis(z) is a polylogarithm function. By using the recursive property of those functions:

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

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New Questions in Calculus 1