Integral Calc: Solving Log(sin(x))^2 from 0 to Pi/2

mathslover
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Hi guys,recently I came across the following integral and need assistance in solving the problem.

The crux of the problem is calculating the definite integral of log(sin(x)*sin(x)) over
the interval ( 0, Pi/2).
Sorry, I made a mistake in typing the integrand.
It should be (Log(sin(x)))^2 instead.
 
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mathslover said:
Hi guys,recently I came across the following integral and need assistance in solving the problem.

The crux of the problem is calculating the definite integral of log(sin(x)*sin(x)) over
the interval ( 0, Pi/2).

I think you should post this in the calculus group or try yahoo answers.
 
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Hardly a number theory question!

The problem is \int_0^{2\pi} log(sin^2(x))dx?

It might help to reduce it to 2\int_0^{2\pi} log(sin(x))dx[/itex] <br /> If not, I tend to suspect the integral cannot be written in terms of elementary functions.
 
a little research reveals the related problem, where apparently the definite integral is calculable in spite of the difficulty of the indefinite one.

the integral of ln(cos) from 0 to x, is called Lobachevsky's integral f(x), and it satisfies the functional equation f(x) = 2f(pi/4 + x/2) - 2f(pi/4 -x/2) - xln(2).

hence the definite integral from 0 to pi/2 equals pi/2 ln(2).
 
HallsofIvy said:
Hardly a number theory question!

The problem is \int_0^{2\pi} log(sin^2(x))dx?

It might help to reduce it to 2\int_0^{2\pi} log(sin(x))dx[/itex] <br /> If not, I tend to suspect the integral cannot be written in terms of elementary functions.
<br /> <br /> ABSOLUTE VALUE SIGN! <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f621.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":mad:" title="Mad :mad:" data-smilie="4"data-shortname=":mad:" />
 
arildno said:
ABSOLUTE VALUE SIGN! :mad:

sin^2 is always positive?
 
ice109 said:
sin^2 is always positive?
Sure it is; but sin(x) is not in the interval given.
So, when the "2" is re-placed in front of the logarithm, instead of being an exponent within it, you must put sin(x) within the absolute value sign.
 
Halls typoed the interval, it was meant to be pi/2 not 2pi. So in the correct interval, absolute value signs are not needed, but for what arildno read off, they were, no were all just a little bit confused here =] Mathwonk's got some good books, because I can't find Lobachevsky integral anywhere =[ But if the function equation holds, then mathwonks value is correct.
 
see joseph kitchen, calculus.
 
  • #10
I am sorry , the integrand should be (log(sin(x)))^2 and the interval is (Pi/2, 0) .-Mathslover
 
  • #11
Use a numerical formula (i.g, Gauss's).
 
  • #12
I would like to find the definite integral of (log(sin(x)))^2 under the interval (pi/2,0)
Numerical integration can only give a numerical answer

I would like to find the above integral in terms of well-known constants


-Mathslover
 
  • #13
Reading through "Ramanujan's notebook Part 2" and "Collected Papers of Ramanujan ",
I chanced upon an entry which solved my problem beautifully.
I just wish that much more can be learned from Ramanujan's work.





-mathslover
 

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  • #14
Devilishly Clever! Can the argument be extended to deal with higher powers?
 
  • #15
The beautiful point about this calculation is that it is applicable for all positive integers.


Define I(n)=Int((log(sin(x)))^n, {x=0 to pi/2}) then it can be shown that


I(0) = pi/2
I(1) = -(pi/2)*log(2)
I(2) = (pi^3)/(24) + (pi/2)*(log(2))^2

and I(3) is a function of log(2) , pi and Zeta(3)

-mathslover
 
  • #16
HallsofIvy said:
Hardly a number theory question!

The problem is \int_0^{2\pi} log(sin^2(x))dx?

It might help to reduce it to 2\int_0^{2\pi} log(sin(x))dx[/itex] <br /> If not, I tend to suspect the integral cannot be written in terms of elementary functions.
<br /> <br /> This is not very difficult. As log(sin^2(x))=2 log(sin(x))[/itex]&lt;br /&gt; In the same way as log(x^2)=2 log x[/itex]
 
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