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Science Forum Index » Mathematics Forum » Exact outcome for this integral ?
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| Han de Bruijn |
 Posted: Tue Apr 29, 2008 2:50 am |
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Quote: integral(t=0..1) ln( (a+cos(t))^2 + (b+sin(t))^2 ) dt = ?
Han de Bruijn |
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| David C. Ullrich |
| Posted: Tue Apr 29, 2008 6:29 am |
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On Tue, 29 Apr 2008 09:50:44 +0200, Han de Bruijn
<Han.deBruijn@DTO.TUDelft.NL> wrote:
Quote: 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...
David C. Ullrich |
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| Han de Bruijn |
| Posted: Tue Apr 29, 2008 6:47 am |
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David C. Ullrich wrote:
Quote: On Tue, 29 Apr 2008 09:50:44 +0200, Han de Bruijn
Han.deBruijn@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.
Han de Bruijn |
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| Robert Israel |
| Posted: Tue Apr 29, 2008 1:21 pm |
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Han de Bruijn <Han.deBruijn@DTO.TUDelft.NL> writes:
Quote: integral(t=0..1) ln( (a+cos(t))^2 + (b+sin(t))^2 ) dt = ?
Maple 11:
Quote: 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 /|
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/
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 israel@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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| Guest |
| Posted: Tue Apr 29, 2008 11:54 pm |
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On 29 apr, 20:21, Robert Israel <isr...@math.MyUniversitysInitials.ca>
wrote:
Quote: Han de Bruijn<Han.deBru...@DTO.TUDelft.NL> writes:
> integral(t=0..1) ln( (a+cos(t))^2 + (b+sin(t))^2 ) dt = ?
Maple11:
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 /|
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/
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
Sorry, Robert, what I meant (typo) was:
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 |
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| David C. Ullrich |
| Posted: Thu May 01, 2008 7:36 am |
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On Tue, 29 Apr 2008 13:47:25 +0200, Han de Bruijn
<Han.deBruijn@DTO.TUDelft.NL> wrote:
Quote: David C. Ullrich wrote:
On Tue, 29 Apr 2008 09:50:44 +0200, Han de Bruijn
Han.deBruijn@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.
David C. Ullrich |
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