How can you evaluate \int_0^{\pi/2}\log\cos(x)dx ?

kuhse4461a 2021-12-30 Answered
How can you evaluate
0π2logcos(x)dx ?
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Expert Answer

Kirsten Davis
Answered 2021-12-31 Author has 27 answers
For the sake of simplicity, all the integral variables I use are x even there are a lot of substitutions. Because lots of variables could make one confused.
Let I denote the integral value. By substitute x for π2x, we have:
I=0π2logcos(x)dx=0π2logsin(x)dx
And then, we have:
I=0π2log(2cos(x2)sin(x2))dx
=π2log2+0π2logcos(x2)dx+0π2logsin(x2)dx
=π2+20π4logcos(x)dx+20π4+20π4logsin(x)dx
=π2log2+I1+I2
In the second step from bottom, I use the substitution that x=x2
For I1, use the substitution that x=π2x we obtain
I1=2π4π2logsin(x)dx
It gives that I1+I2=2I. So we have
I=π2log2+2I
I=π2log2
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Shawn Kim
Answered 2022-01-01 Author has 25 answers
0π2lncosxdx=I=0π2lnsinxdx
By symmetry we have lncosx=lnsinx on the interval [0,π.2]. This is true for any even/odd function on this interval, as is an exercise in Demidovich-Problems in Analysis. Thus we have
2I=0π2lncosxdx+0π2lnsinxdx=0π2ln(sinxcosx)dx=0π2ln(12sin(2x))dx
All I used was ln(ab)=ln(a)+ln(b) and 2sinxcosx=sin(2x). Now we split the integral back up to obtain
0π2ln(2)dx+0π2ln(sin(2x))dx=2I
But the integral of lnsinu is 2I, thus we have
πln(2)2+I=2II=πln(2)2
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Vasquez
Answered 2022-01-09 Author has 460 answers

We have a well-known identity:
k=1n1sin(πkn)=2n2n
and since logsinx is an improperly Riemann-integrable function over (0,π), it follows that:
0πlogsinθdθ=limn+πnk=1n1logsin(πkn)=πlog2
so:
0π/2logsinθdθ=0π/2logsinθdθ=π2log2

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