4 ms·

How do we see that f(1) = 0 in the example? That claim is equivalent to: int_0^pi ln[1 - cos(x)] dx = - pi ln[2] Is this easy?

cryvate1284 5y ago
Because the integral is then over ln(1) and the log of 1 is 0.
- prof-dr-ir 5y ago
  No, that is saying that f(0) = 0 which is correct but not what I asked about (nor what was stated in the article). Incidentally, your correct observation is in contradiction with f(a) = 2 pi log(|a|) since the latter only holds for |a| >= 1. This is because f(a) is not a smooth function.
contravariant 5y ago
Yeah I can't immediately spot something that would make it easy. It's much easier to prove that the difference between the integral of log(a^2 - 2 a cos(x) + 1) and 2 pi log(a) goes to 0 when a goes to infinity.
evouga 5y ago
As far as I can tell, it's not trivial, but can be derived by repeatedly exploiting symmetries of the trig functions and properties of logarithms. One observation is that by symmetry int_0^pi ln[1 - cos(x)] dx = int 0^pi ln[1 + cos(x)] dx and so the calculation is equivalent to 1/2 int_0^pi ln[ sin(x)^2 ] dx = 2 int_0^pi/2 ln[ sin(x) ] dx. Now we can use a similar trick again: int_0^pi/2 ln[ sin(x) ] dx = int_0^pi/2 ln[ cos(x) ] dx so 4 int_0^pi/2 ln[ sin(x) ] dx = 2 int_0^pi/2 ln[ sin(x) cos(x) ] = 2 int_0^pi/2 ln[1/2] dx + 2 int_0^pi/2 ln[sin(2x)] dx = - pi ln[2] + int_0^pi ln[sin(u)] du (using substitution u = 2x) = -pi ln[2] + 2 int_0^pi/2 ln[ sin(x) ] dx and so the original integral is -pi ln[2].
kinkrtyavimoodh 5y ago
I think it's not obvious but it reduces to a relatively 'common' integral. 1 - cos(x) can be written as 2*cos^2(x/2) Take the logarithm and you get 2ln (cos(x/2)), which is relatively common and the solution is https://www.quora.com/How-do-I-integrate-log-cos-x-from-0-to-pi-2 https://www.quora.com/How-do-I-integrate-log-cos-x-from-0-to...

cryvate1284 5y ago

Because the integral is then over ln(1) and the log of 1 is 0.

prof-dr-ir 5y ago
No, that is saying that f(0) = 0 which is correct but not what I asked about (nor what was stated in the article). Incidentally, your correct observation is in contradiction with f(a) = 2 pi log(|a|) since the latter only holds for |a| >= 1. This is because f(a) is not a smooth function.

contravariant 5y ago

Yeah I can't immediately spot something that would make it easy. It's much easier to prove that the difference between the integral of log(a^2 - 2 a cos(x) + 1) and 2 pi log(a) goes to 0 when a goes to infinity.

evouga 5y ago

As far as I can tell, it's not trivial, but can be derived by repeatedly exploiting symmetries of the trig functions and properties of logarithms. One observation is that by symmetry int_0^pi ln[1 - cos(x)] dx = int 0^pi ln[1 + cos(x)] dx and so the calculation is equivalent to 1/2 int_0^pi ln[ sin(x)^2 ] dx = 2 int_0^pi/2 ln[ sin(x) ] dx. Now we can use a similar trick again: int_0^pi/2 ln[ sin(x) ] dx = int_0^pi/2 ln[ cos(x) ] dx so 4 int_0^pi/2 ln[ sin(x) ] dx = 2 int_0^pi/2 ln[ sin(x) cos(x) ] = 2 int_0^pi/2 ln[1/2] dx + 2 int_0^pi/2 ln[sin(2x)] dx = - pi ln[2] + int_0^pi ln[sin(u)] du (using substitution u = 2x) = -pi ln[2] + 2 int_0^pi/2 ln[ sin(x) ] dx and so the original integral is -pi ln[2].

kinkrtyavimoodh 5y ago

I think it's not obvious but it reduces to a relatively 'common' integral. 1 - cos(x) can be written as 2*cos^2(x/2) Take the logarithm and you get 2ln (cos(x/2)), which is relatively common and the solution is https://www.quora.com/How-do-I-integrate-log-cos-x-from-0-to-pi-2 https://www.quora.com/How-do-I-integrate-log-cos-x-from-0-to...