From: Robin Chapman Subject: Re: Integral of ln( 5/4 + sin(x) ) Date: Thu, 18 Mar 1999 13:16:30 +1000 Newsgroups: sci.math To: Richard Thordsen Keywords: Example of contour integration Richard Thordsen wrote: > > Hi, the integral > > int( ln( 5/4 + sin(x) ), x=0..2*Pi ) > > seems to be 0 . > Is this correct, and why (not)? Yes. More generally one has for any a > 1, int_0^{2 pi} log(a + sin x) dx = 2pi(cosh^{-1}(a) - log 2). As cosh(log 2) = (2 + 1/2)/ = 5/4 this vanishes for a = 5/4. Let I(a) denote the above integral. Then its derivative I'(a) equals int_0^{2pi} dx/(a + sin x). We'll evaluate this. It's convenient to replace x by x + pi/2 so we get I'(a) = J(a) = int_0^{2pi} dx/(a + cos(x)). We now use contour integration. Putting z = e^{ix} gives J(a) = (2/i) int_C dz/(z^2 + 2az + 1) where C is the unit circle. The integrand has exactly one pole inside C, at z = -a + sqrt(a^2 - 1) and it's simple. The calculus of residues gives J(a) = 2pi/sqrt{a^2-1} after some manipulation. Integrating gives I(a) = 2pi cosh^{-1}(a) + C. We need to identify the constant C. As a -> infinity, I(a) = 2pi log(a) + o(1). But cosh^{-1}(a) = log (a + sqrt(a^2-1)) = log(a) + log 2 + o(1). Hence C = -2pi log 2 as required. All of this is quite standard and the results may be found in tables of integrals. [If one doesn't like contour integrals then one can evaluate J(a) by say the half-angle substitution.] -- Robin Chapman + "Going to the chemist in Department of Mathematics, DICS - Australia can be more Macquarie University + exciting than going to NSW 2109, Australia - a nightclub in Wales." rchapman@mpce.mq.edu.au + Howard Jacobson, http://www.maths.ex.ac.uk/~rjc/rjc.html - In the Land of Oz ============================================================================== From: Robin Chapman Subject: Re: Integral of ln( 5/4 + sin(x) ) Date: Sun, 28 Mar 1999 04:49:52 GMT Newsgroups: sci.math In article , John.Rickard@virata.com wrote: > Robin Chapman wrote: > : Richard Thordsen wrote: > : > int( ln( 5/4 + sin(x) ), x=0..2*Pi ) > : > > : > seems to be 0 . > More generally, the same argument gives, for 0 < b < 1, > > int(ln(1 + b^2 + 2b sin x)) = 0 > It's pretty easy to show this is true by contour integration. We can replace sin by cos and then 1 + b^2 + 2b cos x is the square of the modulus of 1 + be^{ix}. The integral in question is twice the real part of int_0^{2pi} log(1 + be^ix) dx. We show this integral vanishes by setting z = e^{ix} and taking the contour integral round the unit circle C. This integral becomes (1/i) int_C log (1 + bz) dz/z. Now 1 + bz lies in the left half plane, so we can take the principal value of the logarithm and the integral vanishes by Cauchy's theorem since the integrand has a removable singularity at z = 0. Robin Chapman + "Going to the chemist in Department of Mathematics, DICS - Australia can be more Macquarie University + exciting than going to NSW 2109, Australia - a nightclub in Wales." rchapman@mpce.mq.edu.au + Howard Jacobson, http://www.maths.ex.ac.uk/~rjc/rjc.html - In the Land of Oz -----------== Posted via Deja News, The Discussion Network ==---------- http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own ============================================================================== From: John Rickard Newsgroups: sci.math Subject: Re: An Integral of The Eero Type... Date: 02 Nov 2001 12:42:03 +0000 (GMT) Bill Taylor wrote: : |> pi : |> _ : |> / \ : |> | : |> | : |> | ln(sin x) dx : |> | : |> \ / : |> ~ 0 Fairly obvious if you look at it the right way and have the right background knowledge. You know that the net gravitational force from a uniform spherical shell is zero in the interior; one way to show this involves noting that the gravitation potential is spherically symmetric and has zero del-squared. Well, the same holds in d dimensions if you have an inverse (d-1) power force law. In two dimensions, you have a 1/r force law and a ln(r) potential. So the integral is the potential on the circle due to a uniform circle of diameter 1 (and density 1) in the plane, and this is equal to the potential at the centre, which is the integral from 0 to pi of ln(1/2), which is -pi ln(2). This may be a complicated method if you take all the preliminaries into account, but it did let me get the answer fairly rapidly in my head. (A similar question has come up on sci.math before -- see the thread "Integral of ln( 5/4 + sin(x) )" from March 1999.) -- John Rickard ============================================================================== From: default Newsgroups: sci.math Subject: Re: An Integral of The Eero Type... Date: Sat, 03 Nov 2001 21:40:14 -0500 the integral says that the geometic mean of |sin(x)| on the circle is 1/2. m = 2m^2 because sin(x)=2sin(x/2)cos(x/2), and m>0 because the integral converges. so m = 1/2.