Exact outcome for this integral ?
Han de Bruijn
Han de Bruijn
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
> > integral(t=0..1) ln( (a+cos(t))^2 + (b+sin(t))^2 ) dt = ?
Did you really mean that, or did you mean
integral(t=0..2pi) ln( (a+cos(t))^2 + (b+sin(t))^2 ) dt
or equivalently
integral(t=0..1) ln( (a+cos(2pi*t))^2 + (b+sin(2pi*t))^2 ) dt
? I can tell you what _those_ integrals are...
>Han de Bruijn
David C. Ullrich
Han de Bruijn
Yes, I've made a typo. Sorry. And yes, I meant one of _those_ integrals.
If you tell me what they are, can you please also give me a hint of how
you've arrived at the results. I can understand some (complex) calculus.
Han de Bruijn
Robert Israel
> > integral(t=0..1) ln( (a+cos(t))^2 + (b+sin(t))^2 ) dt = ?
Maple 11:
> int( ln((a+cos(t))^2 + (b+sin(t))^2), t) ;
2 2
ln(2) ln(exp(t I)) I - ln(exp(t I)) ln((2 a exp(t I)
3 4 2
+ 2 a exp(t I) + 2 a exp(t I) + exp(t I) + 2 exp(t I) + 1
2 2 3 / 2
+ 2 b exp(t I) + 2 b exp(t I) + 2 b exp(t I)) / exp(t I) )
/
/ -----
| \
I + | )
| /
| -----
\_R1 = %1
\
/ _R1 - exp(t I) _R1 - exp(t I) \|
|ln(exp(t I)) ln(--------------) + dilog(--------------)|| I
\ _R1 _R1 /|
|
/
2
- ln(exp(t I)) I
4 3 2 2 2
%1 := RootOf(1 + _Z + (2 b + 2 a) _Z + (2 a + 2 b + 2) _Z
+ (2 b + 2 a) _Z)
Now use FTC (being careful to watch out for branch cuts in the antiderivative).
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
umu...@gmail.com
wrote:
integral(t=0..(2.pi)) ln( (a+cos(t))^2 + (b+sin(t))^2 ) dt
And I'm expecting a much simpler outcome with this.
Han de Bruijn
umu...@gmail.com
The answer has been given a while ago by The World Wide Wade. And I've
expressed my enthousiasm for this in another thread, namely:
http://groups.google.nl/group/sci.math/browse_thread/thread/b1634000e5babd75/
It just took a while before it sunk into my brain. The exponent of the
above integral divided by (2*pi), is incredibly simple, namely:
CGM_O(r) = r^2 for r >= 1
= 1 for r <= 1
Here CGM_O = Continuous Geometric Mean of Circle; r = radius of point
in the plane with respect to circle, divided by radius of the circle.
The result is "easily" (uhm ..) obtained with complex function
theory.
Thanks to the other debaters as well.
Han de Bruijn
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
>David C. Ullrich wrote:
>
>> On Tue, 29 Apr 2008 09:50:44 +0200, Han de Bruijn
>> <Han.de...@DTO.TUDelft.NL> wrote:
>>
>>>>integral(t=0..1) ln( (a+cos(t))^2 + (b+sin(t))^2 ) dt = ?
>>
>> Did you really mean that, or did you mean
>>
>> integral(t=0..2pi) ln( (a+cos(t))^2 + (b+sin(t))^2 ) dt
>>
>> or equivalently
>>
>> integral(t=0..1) ln( (a+cos(2pi*t))^2 + (b+sin(2pi*t))^2 ) dt
>>
>> ? I can tell you what _those_ integrals are...
>
>Yes, I've made a typo. Sorry. And yes, I meant one of _those_ integrals.
>If you tell me what they are, can you please also give me a hint of how
>you've arrived at the results. I can understand some (complex) calculus.
There are two cases (really three, I suppose).
Assume first that a^2 + b^2 > 1. Let's define
I = integral(t=0..2pi) ln( (a+cos(t))^2 + (b+sin(t))^2 ) dt .
If a^2 + b^2 > 1 then I = (2pi) log(sqrt(a^2+b^2)).
The reason is this: The function u(z) = log|z| is harmonic
in the region C' = the complex plane minus the origin.
So if D is a disk contained in that region then the value
of u at the center is equal to the average of the values
of u on the boundary. _Note_ that the disk needs to
be contained in C' (!).
This says that if z <> 0 and 0 < r < |z| then
(*) u(z) = 1/(2 pi) int_0^(2 pi) u(z + r e^{it}) dt.
Now let z = a + bi and let r = 1. Our assumption
that a^2 + b^2 > 1 shows that r < |z|. so you can
apply (*) to find the value of I in this case.
Now if r > |z| then the disk with center z and radius r
is not contained in the region where u is harmonic,
so we can't apply (*). In that case we use a little
trick: Since u(z) depends on on the value of |z|
we see that
1/(2 pi) int_0^(2 pi) u(z + r e^{it}) dt
= 1/(2 pi) int_0^(2 pi) u(r + z e^{-it}) dt.
The last integral is the average of u on the disk
with center r and radius |z|: now if |z| < r then
_that_ disk is contained in the region where
u is harmonic, so we see that _if_ |z| < r then
1/(2 pi) int_0^(2 pi) u(z + r e^{it}) dt = u(r).
If a^2 + b^2 < 1 then apply this with z = a + ib
and r = 1 to show that I = 0.
We haven't covered the case a^2 + b^2 = 1. By
some sort of limiting argument it follows that
I = 0 in that case as well.