Integral Calculation

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  • #1
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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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  • #2
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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).
I think you should post this in the calculus group or try yahoo answers.
 
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  • #3
HallsofIvy
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Hardly a number theory question!

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

It might help to reduce it to [tex]2\int_0^{2\pi} log(sin(x))dx[/itex]
If not, I tend to suspect the integral cannot be written in terms of elementary functions.
 
  • #4
mathwonk
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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).
 
  • #5
arildno
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Hardly a number theory question!

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

It might help to reduce it to [tex]2\int_0^{2\pi} log(sin(x))dx[/itex]
If not, I tend to suspect the integral cannot be written in terms of elementary functions.
ABSOLUTE VALUE SIGN!! :mad:
 
  • #6
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ABSOLUTE VALUE SIGN!! :mad:
sin^2 is always positive?
 
  • #7
arildno
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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.
 
  • #8
Gib Z
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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.
 
  • #9
mathwonk
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see joseph kitchen, calculus.
 
  • #10
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I am sorry , the integrand should be (log(sin(x)))^2 and the interval is (Pi/2, 0) .


-Mathslover
 
  • #11
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Use a numerical formula (i.g, Gauss's).
 
  • #12
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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
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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
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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
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Hardly a number theory question!

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

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

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