Integral Calculation

1. Sep 22, 2007

mathslover

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.

Last edited: Sep 22, 2007
2. Sep 22, 2007

ramsey2879

I think you should post this in the calculus group or try yahoo answers.

Last edited: Sep 22, 2007
3. Sep 22, 2007

HallsofIvy

Staff Emeritus
Hardly a number theory question!

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

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. Sep 22, 2007

mathwonk

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. Sep 22, 2007

arildno

ABSOLUTE VALUE SIGN!!

6. Sep 22, 2007

ice109

sin^2 is always positive?

7. Sep 22, 2007

arildno

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. Sep 22, 2007

Gib Z

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. Sep 22, 2007

mathwonk

see joseph kitchen, calculus.

10. Sep 23, 2007

mathslover

I am sorry , the integrand should be (log(sin(x)))^2 and the interval is (Pi/2, 0) .

-Mathslover

11. Sep 23, 2007

traianus

Use a numerical formula (i.g, Gauss's).

12. Sep 24, 2007

mathslover

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. Oct 1, 2007

mathslover

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. Oct 6, 2007

Semperaugustus

Devilishly Clever!!! Can the argument be extended to deal with higher powers?

15. Oct 6, 2007

mathslover

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. May 16, 2009

larstuff

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]