Mixing
$ I(r) = int_0^{2pi}frac{cos(t) - r}{1 - 2rcos t + r^2},dt$ is always zero for $rin[0,1)$. Why?
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For $rin [0,1)$ define
$$
I(r) = int_0^{2pi}frac{cos(t) - r}{1 - 2rcos t + r^2},dt.
$$
Numerical experiments hint at $I(r) = 0$ for all $rin [0,1)$ but I can't show this analytically.
This integral appears when you compute the Cauchy transform of $overline z$ over the unit circle. The latter therefore seems to be constantly zero.
real-analysis integration definite-integrals
edited Jan 28 at 7:40
Asaf Karagila♦
307k33438769
asked Jan 26 at 17:16
amsmathamsmath
2,842417
$endgroup$
|
show 3 more comments
$begingroup$
For $rin [0,1)$ define
$$
I(r) = int_0^{2pi}frac{cos(t) - r}{1 - 2rcos t + r^2},dt.
$$
Numerical experiments hint at $I(r) = 0$ for all $rin [0,1)$ but I can't show this analytically.
This integral appears when you compute the Cauchy transform of $overline z$ over the unit circle. The latter therefore seems to be constantly zero.
real-analysis integration definite-integrals
edited Jan 28 at 7:40
Asaf Karagila♦
307k33438769
asked Jan 26 at 17:16
amsmathamsmath
2,842417
$endgroup$
$begingroup$
@Dr.SonnhardGraubner Edited
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– amsmath
Jan 26 at 17:23
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I think your integral isn't zero
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– Dr. Sonnhard Graubner
Jan 26 at 17:31
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@Dr.SonnhardGraubner Well, that was helpful... Why?
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– amsmath
Jan 26 at 17:33
14
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@Dr.SonnhardGraubner: I think the integral is zero.
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– Larry
Jan 26 at 18:01
3
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@Dr.SonnhardGraubner You gave us the proof below. ;-)
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– amsmath
Jan 26 at 18:08
|
show 3 more comments
$begingroup$
For $rin [0,1)$ define
$$
I(r) = int_0^{2pi}frac{cos(t) - r}{1 - 2rcos t + r^2},dt.
$$
Numerical experiments hint at $I(r) = 0$ for all $rin [0,1)$ but I can't show this analytically.
This integral appears when you compute the Cauchy transform of $overline z$ over the unit circle. The latter therefore seems to be constantly zero.
real-analysis integration definite-integrals
edited Jan 28 at 7:40
Asaf Karagila♦
307k33438769
asked Jan 26 at 17:16
amsmathamsmath
2,842417
$endgroup$
For $rin [0,1)$ define
$$
I(r) = int_0^{2pi}frac{cos(t) - r}{1 - 2rcos t + r^2},dt.
$$
Numerical experiments hint at $I(r) = 0$ for all $rin [0,1)$ but I can't show this analytically.
This integral appears when you compute the Cauchy transform of $overline z$ over the unit circle. The latter therefore seems to be constantly zero.
real-analysis integration definite-integrals
real-analysis integration definite-integrals
edited Jan 28 at 7:40
Asaf Karagila♦
307k33438769
asked Jan 26 at 17:16
amsmathamsmath
2,842417
edited Jan 28 at 7:40
Asaf Karagila♦
307k33438769
asked Jan 26 at 17:16
amsmathamsmath
2,842417
edited Jan 28 at 7:40
Asaf Karagila♦
307k33438769
edited Jan 28 at 7:40
Asaf Karagila♦
307k33438769
edited Jan 28 at 7:40
Asaf Karagila♦
307k33438769
307k33438769
asked Jan 26 at 17:16
amsmathamsmath
2,842417
asked Jan 26 at 17:16
amsmathamsmath
2,842417
asked Jan 26 at 17:16
amsmathamsmath
2,842417
2,842417
$begingroup$
@Dr.SonnhardGraubner Edited
$endgroup$
– amsmath
Jan 26 at 17:23
$begingroup$
I think your integral isn't zero
$endgroup$
– Dr. Sonnhard Graubner
Jan 26 at 17:31
$begingroup$
@Dr.SonnhardGraubner Well, that was helpful... Why?
$endgroup$
– amsmath
Jan 26 at 17:33
14
$begingroup$
@Dr.SonnhardGraubner: I think the integral is zero.
$endgroup$
– Larry
Jan 26 at 18:01
3
$begingroup$
@Dr.SonnhardGraubner You gave us the proof below. ;-)
$endgroup$
– amsmath
Jan 26 at 18:08
|
show 3 more comments
$begingroup$
@Dr.SonnhardGraubner Edited
$endgroup$
– amsmath
Jan 26 at 17:23
$begingroup$
I think your integral isn't zero
$endgroup$
– Dr. Sonnhard Graubner
Jan 26 at 17:31
$begingroup$
@Dr.SonnhardGraubner Well, that was helpful... Why?
$endgroup$
– amsmath
Jan 26 at 17:33
14
$begingroup$
@Dr.SonnhardGraubner: I think the integral is zero.
$endgroup$
– Larry
Jan 26 at 18:01
3
$begingroup$
@Dr.SonnhardGraubner You gave us the proof below. ;-)
$endgroup$
– amsmath
Jan 26 at 18:08
$begingroup$
@Dr.SonnhardGraubner Edited
$endgroup$
– amsmath
Jan 26 at 17:23
$begingroup$
@Dr.SonnhardGraubner Edited
$endgroup$
– amsmath
Jan 26 at 17:23
$begingroup$
I think your integral isn't zero
$endgroup$
– Dr. Sonnhard Graubner
Jan 26 at 17:31
$begingroup$
I think your integral isn't zero
$endgroup$
– Dr. Sonnhard Graubner
Jan 26 at 17:31
$begingroup$
@Dr.SonnhardGraubner Well, that was helpful... Why?
$endgroup$
– amsmath
Jan 26 at 17:33
$begingroup$
@Dr.SonnhardGraubner Well, that was helpful... Why?
$endgroup$
– amsmath
Jan 26 at 17:33
14
14
$begingroup$
@Dr.SonnhardGraubner: I think the integral is zero.
$endgroup$
– Larry
Jan 26 at 18:01
$begingroup$
@Dr.SonnhardGraubner: I think the integral is zero.
$endgroup$
– Larry
Jan 26 at 18:01
3
3
$begingroup$
@Dr.SonnhardGraubner You gave us the proof below. ;-)
$endgroup$
– amsmath
Jan 26 at 18:08
$begingroup$
@Dr.SonnhardGraubner You gave us the proof below. ;-)
$endgroup$
– amsmath
Jan 26 at 18:08
|
show 3 more comments
8 Answers
8
active
oldest
votes
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An alternative proof using complex methods:
For $0< r < 1$ let
$$ f_r colon B_frac{1}{r} (0) to mathbb{C} , , , f_r(z) = frac{- ln(1-rz)}{z} , , $$
where $f_r(0) = r $ . Then $f_r$ is holomorphic, so
$$ I(r) equiv - int limits_0^{2pi} ln(1-r mathrm{e}^{mathrm{i}t}) , mathrm{d} t = - mathrm{i} int limits_{S^1} f_r(z) , mathrm{d} z = 0 $$
holds by Cauchy's theorem. If you are not familiar with complex analysis, you can also show this using the Taylor series of the logarithm:
$$ I(r) = sum limits_{n=1}^infty frac{r^n}{n} int limits_0^{2pi} mathrm{e}^{mathrm{i} n t} , mathrm{d} t = 0 , . $$
This implies
begin{align}
int limits_0^{2pi} frac{cos(t) - r}{1 - 2 r cos(t) + r^2} , mathrm{d} t &= - frac{1}{2} frac{mathrm{d}}{mathrm{d} r} int limits_0^{2pi} ln(1 - 2 r cos(t) + r^2) , mathrm{d} t \
&= - frac{1}{2} frac{mathrm{d}}{mathrm{d} r} int limits_0^{2pi} ln[(1 -r mathrm{e}^{mathrm{i} t})(1 -r mathrm{e}^{-mathrm{i} t})] , mathrm{d} t \
&= frac{mathrm{d}}{mathrm{d} r} I(r) = frac{mathrm{d}}{mathrm{d} r} 0 = 0
end{align}
as desired.
answered Jan 26 at 18:32
ComplexYetTrivialComplexYetTrivial
4,9682631
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add a comment |
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(Since it is not mentioned yet) The function
$$
P_r(t) = sum_{n=-infty}^infty r^{|n|} e^{int} =frac{1-r^2}{1-2rcos t +t^2},quad0le r<1, 0le tle 2pi
$$ is called the Poisson kernel. For any continuous function $f:[0,2pi]to Bbb C$ the Poisson integral
$$
u(r,theta) = frac{1}{2pi i}int_0^{2pi} f(t)P_r(theta-t)mathrm{d}ttag{*}
$$ gives the unique solution of the Dirichlet problem $triangle u =0, limlimits_{rto 1^-}u(r,theta)=f(theta)$.
Let $u(r,theta)=rcos theta$. Since it is a real part of the analytic function $z=r e^{itheta}$, we find that $$triangle u=0, limlimits_{rto 1^-}u(r,theta)=cos theta.$$ By $text{(*)}$ it follows that
$$
rcos theta =frac{1}{2pi}int_0^{2pi} frac{cos t(1-r^2)}{1-2rcos (t-theta)+r^2}mathrm{d}t
$$ for all $0le r<1$ and $0le thetale 2pi$. Using the fact
$$
frac{1}{2pi}int_0^{2pi} frac{1-r^2}{1-2rcos (t-theta)+r^2}mathrm{d}t=frac{1}{2pi}sum_{n=-infty}^infty r^{|n|} int_0^{2pi}e^{in(t-theta)}mathrm{d}t =1,
$$ it follows
$$
frac{1}{2pi}int_0^{2pi} frac{left(cos t-rcos thetaright)(1-r^2)}{1-2rcos (t-theta)+r^2}mathrm{d}t =0.
$$ By letting $theta =0$, we obtain
$$
frac{1-r^2}{2pi }int_0^{2pi} frac{cos t-r}{1-2rcos t+r^2}mathrm{d}t =0,$$
or equivalently
$$
int_0^{2pi} frac{cos t-r}{1-2rcos t+r^2}mathrm{d}t =0.$$
answered Jan 26 at 18:51
SongSong
18.5k21551
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That is also very nice. Thank you. In fact, I tried to fuddle around with the Poisson kernel, but I did not see things as clear as you, obviously. ;-)
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– amsmath
Jan 26 at 19:18
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@amsmath I hope this will help :-)
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– Song
Jan 26 at 21:31
add a comment |
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It seems to me that this result is a particular case (or rather 1D adaptation) of Newton's shell theorem. In 2D, the theorem states that no net gravitational force is perceived by objects located inside a uniform hollow sphere.
Here, consider the integral
$$
vec I(M) = int_{Xin C}frac{vec{MX}}{MX^2}mathrm ds
$$
where $M$ is a point at a distance $r<1$ away from the center $O$ of the unit circle $C$, and $mathrm ds$ is a length element of the circle. What you are asking for is a component of $vec I$, the one directed along $vec{OM}$. In fact, we have $vec I=vec 0$. There is a nice geometric proof of this in 2D due to Newton; see here. You can try to adapt it to 1D.
answered Jan 26 at 19:33
HusseinHussein
43114
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interesting physical interpretation (+1)
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– G Cab
Jan 26 at 19:43
add a comment |
$begingroup$
With $gamma$ being the counter-clockwise unit circle and $bargamma$ being the clockwise unit circle,
$$
begin{align}
int_0^{2pi}frac{r-cos(t)}{1-2rcos(t)+r^2},mathrm{d}t
&=int_0^{2pi}frac12left(frac1{r-e^{it}}+frac1{r-e^{-it}}right)mathrm{d}ttag1\
&=frac1{2i}int_gammafrac{,mathrm{d}z}{z(r-z)}-frac1{2i}int_{bargamma}frac{,mathrm{d}z}{z(r-z)}tag2\
&=frac1iint_gammafrac{,mathrm{d}z}{z(r-z)}tag3\
&=frac1{ir}int_gammaleft(frac1z-frac1{z-r}right)mathrm{d}ztag4\
&=left{begin{array}{}
0&text{if }|r|lt1\
frac{2pi}r&text{if }|r|gt1
end{array}right.tag5
end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: $z=e^{it}$ in the left integral and $z=e^{-it}$ in the right integral
$(3)$: $bargamma$ is in the opposite direction from $gamma$
$(4)$: partial fractions
$(5)$: $2pi i$ times the sum of the residues inside $gamma$
answered Jan 27 at 8:24
robjohn♦robjohn
269k27311638
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add a comment |
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Here is yet another slightly different approach.
Let
$$P_r(t) = frac{cos t - r}{1 - 2r cos t + r^2}, qquad r in [0,1).$$
Using $cos t = (e^{it} + e^{-it})/2$, we can rewrite $P_r(t)$ as
begin{align}
P_r (t) &= frac{1}{2} frac{(e^{it} + e^{-it}) - 2r}{1 - r(e^{it} + e^{-it}) + r^2}\
&= -frac{1}{2} frac{2r - e^{it} - e^{-it}}{(r - e^{it})(r - e^{-it})}\
&= -frac{1}{2} left [frac{1}{r - e^{it}} + frac{1}{r - e^{-it}} right ]\
&= frac{1}{2} left [frac{e^{-it}}{1 - r e^{-it}} + frac{e^{it}}{1 - r e^{it}} right ]\
&= frac{1}{2} left [e^{-it} sum_{n = 0}^infty r^n e^{-int} + e^{it} sum_{n = 0}^infty r^n e^{int} right ]\
&= sum_{n = 0}^infty r^n left (frac{e^{i(n + 1)t} + e^{-i(n + 1)t}}{2} right )\
&= sum_{n = 0}^infty r^n cos (n + 1)t.
end{align}
Now
begin{align}
int^{2pi}_0 P_r (t) , dt = int^{2 pi}_0 sum_{n = 0}^infty r^n cos (n + 1)t , dt = sum_{n = 0}^infty r^n int^{2pi}_0 cos (n + 1)t , dt = 0.
end{align}
answered Jan 27 at 3:30
omegadotomegadot
6,4342829
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add a comment |
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Note that $$I(r) =frac{-1}{2r}int_{0}^{2pi}frac{1-2rcos t+r^2+r^2-1}{1-2rcos t+r^2},dt$$ which is same as $$-frac{pi} {r} +frac{1-r^2}{r}int_{0}^{pi}frac{dt}{1+r^2-2rcos t} $$ The integral above is equal to $$frac{pi} {sqrt{(1+r^2)^2-4r^2}}=frac{pi}{1-r^2} $$ and hence $I(r) =0$.
We have used the formula $$int_{0}^{pi}frac{dt}{a+bcos t} =frac{pi} {sqrt{a^2-b^2}},,a>|b|$$ which can be proved using the substitution $$(a+bcos t) (a-bcos u) =a^2-b^2$$
answered Jan 27 at 8:37
Paramanand SinghParamanand Singh
50.9k557168
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Hint: $$cos(t)=frac{1-a^2}{1+a^2}$$ and $$dt=frac{2da}{1+a^2}$$
Now your indefinite integral is given by $$int-frac{2 left(a^2 r+a^2+r-1right)}{left(a^2+1right)
left(a^2 r^2+2 a^2 r+a^2+r^2-2 r+1right)}da$$ and this is rational
The solution of this integral is given by $$-2 left(frac{tan ^{-1}(a)}{2 r}-frac{tan
^{-1}left(frac{a (r+1)}{1-r}right)}{4
r}+frac{tan ^{-1}left(frac{a
(r+1)}{r-1}right)}{4 r}right)$$
answered Jan 26 at 17:42
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
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5
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Thank you very much. So my integral is indeed zero. You can fuse the second and the third term to one term: $-tfrac 1{2r}arctan(tfrac{1+r}{1-r}a)$. The boundaries after substitution are zero and $infty$.
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– amsmath
Jan 26 at 18:04
add a comment |
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Not sure why none of the answers mention this approach, but you can just take the derivative with respect to $r$:
begin{align}
I'(r) &= int_0^{2pi}frac{d}{dr}left(frac{cos t - r}{1 - 2rcos t + r^2}right),dt\
&= int_0^{2pi} frac{(r^2-1)-2rcos t + 2cos^2t}{(1 - 2rcos t + r^2)^2} \
&= frac1{i}int_{|z|=1} frac{(r^2-1)-r(z+z^{-1}) + frac12(z+z^{-1})^2}{(1 - r(z+z^{-1}) + r^2)^2}cdotfrac{dz}{z}\
&= frac1{i}int_{|z|=1} frac{z^4-2r z^3+2r^2z^2-2rz+1}{(r^2+1)z(z-r)^2left(z-frac1rright)^2},dz\
end{align}
Since $r in [0,1rangle$, the relevant residues are $$operatorname{Res}(f,0) = frac1{2r^2}$$
$$operatorname{Res}(f,r) = -frac1{2r^2}$$
so $I'(r) = 0$ for $r in [0,1rangle$.
We conclude that $I$ is constant on $[0,1rangle$. Hence $$I(r)=I(0) = int_0^{2pi}cos t,dt = 0$$
answered Jan 27 at 18:19
mechanodroidmechanodroid
28.9k62548
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8 Answers
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8 Answers
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$begingroup$
An alternative proof using complex methods:
For $0< r < 1$ let
$$ f_r colon B_frac{1}{r} (0) to mathbb{C} , , , f_r(z) = frac{- ln(1-rz)}{z} , , $$
where $f_r(0) = r $ . Then $f_r$ is holomorphic, so
$$ I(r) equiv - int limits_0^{2pi} ln(1-r mathrm{e}^{mathrm{i}t}) , mathrm{d} t = - mathrm{i} int limits_{S^1} f_r(z) , mathrm{d} z = 0 $$
holds by Cauchy's theorem. If you are not familiar with complex analysis, you can also show this using the Taylor series of the logarithm:
$$ I(r) = sum limits_{n=1}^infty frac{r^n}{n} int limits_0^{2pi} mathrm{e}^{mathrm{i} n t} , mathrm{d} t = 0 , . $$
This implies
begin{align}
int limits_0^{2pi} frac{cos(t) - r}{1 - 2 r cos(t) + r^2} , mathrm{d} t &= - frac{1}{2} frac{mathrm{d}}{mathrm{d} r} int limits_0^{2pi} ln(1 - 2 r cos(t) + r^2) , mathrm{d} t \
&= - frac{1}{2} frac{mathrm{d}}{mathrm{d} r} int limits_0^{2pi} ln[(1 -r mathrm{e}^{mathrm{i} t})(1 -r mathrm{e}^{-mathrm{i} t})] , mathrm{d} t \
&= frac{mathrm{d}}{mathrm{d} r} I(r) = frac{mathrm{d}}{mathrm{d} r} 0 = 0
end{align}
as desired.
answered Jan 26 at 18:32
ComplexYetTrivialComplexYetTrivial
4,9682631
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add a comment |
$begingroup$
An alternative proof using complex methods:
For $0< r < 1$ let
$$ f_r colon B_frac{1}{r} (0) to mathbb{C} , , , f_r(z) = frac{- ln(1-rz)}{z} , , $$
where $f_r(0) = r $ . Then $f_r$ is holomorphic, so
$$ I(r) equiv - int limits_0^{2pi} ln(1-r mathrm{e}^{mathrm{i}t}) , mathrm{d} t = - mathrm{i} int limits_{S^1} f_r(z) , mathrm{d} z = 0 $$
holds by Cauchy's theorem. If you are not familiar with complex analysis, you can also show this using the Taylor series of the logarithm:
$$ I(r) = sum limits_{n=1}^infty frac{r^n}{n} int limits_0^{2pi} mathrm{e}^{mathrm{i} n t} , mathrm{d} t = 0 , . $$
This implies
begin{align}
int limits_0^{2pi} frac{cos(t) - r}{1 - 2 r cos(t) + r^2} , mathrm{d} t &= - frac{1}{2} frac{mathrm{d}}{mathrm{d} r} int limits_0^{2pi} ln(1 - 2 r cos(t) + r^2) , mathrm{d} t \
&= - frac{1}{2} frac{mathrm{d}}{mathrm{d} r} int limits_0^{2pi} ln[(1 -r mathrm{e}^{mathrm{i} t})(1 -r mathrm{e}^{-mathrm{i} t})] , mathrm{d} t \
&= frac{mathrm{d}}{mathrm{d} r} I(r) = frac{mathrm{d}}{mathrm{d} r} 0 = 0
end{align}
as desired.
answered Jan 26 at 18:32
ComplexYetTrivialComplexYetTrivial
4,9682631
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add a comment |
$begingroup$
An alternative proof using complex methods:
For $0< r < 1$ let
$$ f_r colon B_frac{1}{r} (0) to mathbb{C} , , , f_r(z) = frac{- ln(1-rz)}{z} , , $$
where $f_r(0) = r $ . Then $f_r$ is holomorphic, so
$$ I(r) equiv - int limits_0^{2pi} ln(1-r mathrm{e}^{mathrm{i}t}) , mathrm{d} t = - mathrm{i} int limits_{S^1} f_r(z) , mathrm{d} z = 0 $$
holds by Cauchy's theorem. If you are not familiar with complex analysis, you can also show this using the Taylor series of the logarithm:
$$ I(r) = sum limits_{n=1}^infty frac{r^n}{n} int limits_0^{2pi} mathrm{e}^{mathrm{i} n t} , mathrm{d} t = 0 , . $$
This implies
begin{align}
int limits_0^{2pi} frac{cos(t) - r}{1 - 2 r cos(t) + r^2} , mathrm{d} t &= - frac{1}{2} frac{mathrm{d}}{mathrm{d} r} int limits_0^{2pi} ln(1 - 2 r cos(t) + r^2) , mathrm{d} t \
&= - frac{1}{2} frac{mathrm{d}}{mathrm{d} r} int limits_0^{2pi} ln[(1 -r mathrm{e}^{mathrm{i} t})(1 -r mathrm{e}^{-mathrm{i} t})] , mathrm{d} t \
&= frac{mathrm{d}}{mathrm{d} r} I(r) = frac{mathrm{d}}{mathrm{d} r} 0 = 0
end{align}
as desired.
answered Jan 26 at 18:32
ComplexYetTrivialComplexYetTrivial
4,9682631
$endgroup$
An alternative proof using complex methods:
For $0< r < 1$ let
$$ f_r colon B_frac{1}{r} (0) to mathbb{C} , , , f_r(z) = frac{- ln(1-rz)}{z} , , $$
where $f_r(0) = r $ . Then $f_r$ is holomorphic, so
$$ I(r) equiv - int limits_0^{2pi} ln(1-r mathrm{e}^{mathrm{i}t}) , mathrm{d} t = - mathrm{i} int limits_{S^1} f_r(z) , mathrm{d} z = 0 $$
holds by Cauchy's theorem. If you are not familiar with complex analysis, you can also show this using the Taylor series of the logarithm:
$$ I(r) = sum limits_{n=1}^infty frac{r^n}{n} int limits_0^{2pi} mathrm{e}^{mathrm{i} n t} , mathrm{d} t = 0 , . $$
This implies
begin{align}
int limits_0^{2pi} frac{cos(t) - r}{1 - 2 r cos(t) + r^2} , mathrm{d} t &= - frac{1}{2} frac{mathrm{d}}{mathrm{d} r} int limits_0^{2pi} ln(1 - 2 r cos(t) + r^2) , mathrm{d} t \
&= - frac{1}{2} frac{mathrm{d}}{mathrm{d} r} int limits_0^{2pi} ln[(1 -r mathrm{e}^{mathrm{i} t})(1 -r mathrm{e}^{-mathrm{i} t})] , mathrm{d} t \
&= frac{mathrm{d}}{mathrm{d} r} I(r) = frac{mathrm{d}}{mathrm{d} r} 0 = 0
end{align}
as desired.
answered Jan 26 at 18:32
ComplexYetTrivialComplexYetTrivial
4,9682631
answered Jan 26 at 18:32
ComplexYetTrivialComplexYetTrivial
4,9682631
answered Jan 26 at 18:32
ComplexYetTrivialComplexYetTrivial
4,9682631
answered Jan 26 at 18:32
ComplexYetTrivialComplexYetTrivial
4,9682631
4,9682631
add a comment |
add a comment |
$begingroup$
(Since it is not mentioned yet) The function
$$
P_r(t) = sum_{n=-infty}^infty r^{|n|} e^{int} =frac{1-r^2}{1-2rcos t +t^2},quad0le r<1, 0le tle 2pi
$$ is called the Poisson kernel. For any continuous function $f:[0,2pi]to Bbb C$ the Poisson integral
$$
u(r,theta) = frac{1}{2pi i}int_0^{2pi} f(t)P_r(theta-t)mathrm{d}ttag{*}
$$ gives the unique solution of the Dirichlet problem $triangle u =0, limlimits_{rto 1^-}u(r,theta)=f(theta)$.
Let $u(r,theta)=rcos theta$. Since it is a real part of the analytic function $z=r e^{itheta}$, we find that $$triangle u=0, limlimits_{rto 1^-}u(r,theta)=cos theta.$$ By $text{(*)}$ it follows that
$$
rcos theta =frac{1}{2pi}int_0^{2pi} frac{cos t(1-r^2)}{1-2rcos (t-theta)+r^2}mathrm{d}t
$$ for all $0le r<1$ and $0le thetale 2pi$. Using the fact
$$
frac{1}{2pi}int_0^{2pi} frac{1-r^2}{1-2rcos (t-theta)+r^2}mathrm{d}t=frac{1}{2pi}sum_{n=-infty}^infty r^{|n|} int_0^{2pi}e^{in(t-theta)}mathrm{d}t =1,
$$ it follows
$$
frac{1}{2pi}int_0^{2pi} frac{left(cos t-rcos thetaright)(1-r^2)}{1-2rcos (t-theta)+r^2}mathrm{d}t =0.
$$ By letting $theta =0$, we obtain
$$
frac{1-r^2}{2pi }int_0^{2pi} frac{cos t-r}{1-2rcos t+r^2}mathrm{d}t =0,$$
or equivalently
$$
int_0^{2pi} frac{cos t-r}{1-2rcos t+r^2}mathrm{d}t =0.$$
answered Jan 26 at 18:51
SongSong
18.5k21551
$endgroup$
$begingroup$
That is also very nice. Thank you. In fact, I tried to fuddle around with the Poisson kernel, but I did not see things as clear as you, obviously. ;-)
$endgroup$
– amsmath
Jan 26 at 19:18
$begingroup$
@amsmath I hope this will help :-)
$endgroup$
– Song
Jan 26 at 21:31
add a comment |
$begingroup$
(Since it is not mentioned yet) The function
$$
P_r(t) = sum_{n=-infty}^infty r^{|n|} e^{int} =frac{1-r^2}{1-2rcos t +t^2},quad0le r<1, 0le tle 2pi
$$ is called the Poisson kernel. For any continuous function $f:[0,2pi]to Bbb C$ the Poisson integral
$$
u(r,theta) = frac{1}{2pi i}int_0^{2pi} f(t)P_r(theta-t)mathrm{d}ttag{*}
$$ gives the unique solution of the Dirichlet problem $triangle u =0, limlimits_{rto 1^-}u(r,theta)=f(theta)$.
Let $u(r,theta)=rcos theta$. Since it is a real part of the analytic function $z=r e^{itheta}$, we find that $$triangle u=0, limlimits_{rto 1^-}u(r,theta)=cos theta.$$ By $text{(*)}$ it follows that
$$
rcos theta =frac{1}{2pi}int_0^{2pi} frac{cos t(1-r^2)}{1-2rcos (t-theta)+r^2}mathrm{d}t
$$ for all $0le r<1$ and $0le thetale 2pi$. Using the fact
$$
frac{1}{2pi}int_0^{2pi} frac{1-r^2}{1-2rcos (t-theta)+r^2}mathrm{d}t=frac{1}{2pi}sum_{n=-infty}^infty r^{|n|} int_0^{2pi}e^{in(t-theta)}mathrm{d}t =1,
$$ it follows
$$
frac{1}{2pi}int_0^{2pi} frac{left(cos t-rcos thetaright)(1-r^2)}{1-2rcos (t-theta)+r^2}mathrm{d}t =0.
$$ By letting $theta =0$, we obtain
$$
frac{1-r^2}{2pi }int_0^{2pi} frac{cos t-r}{1-2rcos t+r^2}mathrm{d}t =0,$$
or equivalently
$$
int_0^{2pi} frac{cos t-r}{1-2rcos t+r^2}mathrm{d}t =0.$$
answered Jan 26 at 18:51
SongSong
18.5k21551
$endgroup$
$begingroup$
That is also very nice. Thank you. In fact, I tried to fuddle around with the Poisson kernel, but I did not see things as clear as you, obviously. ;-)
$endgroup$
– amsmath
Jan 26 at 19:18
$begingroup$
@amsmath I hope this will help :-)
$endgroup$
– Song
Jan 26 at 21:31
add a comment |
$begingroup$
(Since it is not mentioned yet) The function
$$
P_r(t) = sum_{n=-infty}^infty r^{|n|} e^{int} =frac{1-r^2}{1-2rcos t +t^2},quad0le r<1, 0le tle 2pi
$$ is called the Poisson kernel. For any continuous function $f:[0,2pi]to Bbb C$ the Poisson integral
$$
u(r,theta) = frac{1}{2pi i}int_0^{2pi} f(t)P_r(theta-t)mathrm{d}ttag{*}
$$ gives the unique solution of the Dirichlet problem $triangle u =0, limlimits_{rto 1^-}u(r,theta)=f(theta)$.
Let $u(r,theta)=rcos theta$. Since it is a real part of the analytic function $z=r e^{itheta}$, we find that $$triangle u=0, limlimits_{rto 1^-}u(r,theta)=cos theta.$$ By $text{(*)}$ it follows that
$$
rcos theta =frac{1}{2pi}int_0^{2pi} frac{cos t(1-r^2)}{1-2rcos (t-theta)+r^2}mathrm{d}t
$$ for all $0le r<1$ and $0le thetale 2pi$. Using the fact
$$
frac{1}{2pi}int_0^{2pi} frac{1-r^2}{1-2rcos (t-theta)+r^2}mathrm{d}t=frac{1}{2pi}sum_{n=-infty}^infty r^{|n|} int_0^{2pi}e^{in(t-theta)}mathrm{d}t =1,
$$ it follows
$$
frac{1}{2pi}int_0^{2pi} frac{left(cos t-rcos thetaright)(1-r^2)}{1-2rcos (t-theta)+r^2}mathrm{d}t =0.
$$ By letting $theta =0$, we obtain
$$
frac{1-r^2}{2pi }int_0^{2pi} frac{cos t-r}{1-2rcos t+r^2}mathrm{d}t =0,$$
or equivalently
$$
int_0^{2pi} frac{cos t-r}{1-2rcos t+r^2}mathrm{d}t =0.$$
answered Jan 26 at 18:51
SongSong
18.5k21551
$endgroup$
(Since it is not mentioned yet) The function
$$
P_r(t) = sum_{n=-infty}^infty r^{|n|} e^{int} =frac{1-r^2}{1-2rcos t +t^2},quad0le r<1, 0le tle 2pi
$$ is called the Poisson kernel. For any continuous function $f:[0,2pi]to Bbb C$ the Poisson integral
$$
u(r,theta) = frac{1}{2pi i}int_0^{2pi} f(t)P_r(theta-t)mathrm{d}ttag{*}
$$ gives the unique solution of the Dirichlet problem $triangle u =0, limlimits_{rto 1^-}u(r,theta)=f(theta)$.
Let $u(r,theta)=rcos theta$. Since it is a real part of the analytic function $z=r e^{itheta}$, we find that $$triangle u=0, limlimits_{rto 1^-}u(r,theta)=cos theta.$$ By $text{(*)}$ it follows that
$$
rcos theta =frac{1}{2pi}int_0^{2pi} frac{cos t(1-r^2)}{1-2rcos (t-theta)+r^2}mathrm{d}t
$$ for all $0le r<1$ and $0le thetale 2pi$. Using the fact
$$
frac{1}{2pi}int_0^{2pi} frac{1-r^2}{1-2rcos (t-theta)+r^2}mathrm{d}t=frac{1}{2pi}sum_{n=-infty}^infty r^{|n|} int_0^{2pi}e^{in(t-theta)}mathrm{d}t =1,
$$ it follows
$$
frac{1}{2pi}int_0^{2pi} frac{left(cos t-rcos thetaright)(1-r^2)}{1-2rcos (t-theta)+r^2}mathrm{d}t =0.
$$ By letting $theta =0$, we obtain
$$
frac{1-r^2}{2pi }int_0^{2pi} frac{cos t-r}{1-2rcos t+r^2}mathrm{d}t =0,$$
or equivalently
$$
int_0^{2pi} frac{cos t-r}{1-2rcos t+r^2}mathrm{d}t =0.$$
answered Jan 26 at 18:51
SongSong
18.5k21551
edited Jan 26 at 21:31
answered Jan 26 at 18:51
SongSong
18.5k21551
answered Jan 26 at 18:51
SongSong
18.5k21551
answered Jan 26 at 18:51
SongSong
18.5k21551
18.5k21551
$begingroup$
That is also very nice. Thank you. In fact, I tried to fuddle around with the Poisson kernel, but I did not see things as clear as you, obviously. ;-)
$endgroup$
– amsmath
Jan 26 at 19:18
$begingroup$
@amsmath I hope this will help :-)
$endgroup$
– Song
Jan 26 at 21:31
add a comment |
$begingroup$
That is also very nice. Thank you. In fact, I tried to fuddle around with the Poisson kernel, but I did not see things as clear as you, obviously. ;-)
$endgroup$
– amsmath
Jan 26 at 19:18
$begingroup$
@amsmath I hope this will help :-)
$endgroup$
– Song
Jan 26 at 21:31
$begingroup$
That is also very nice. Thank you. In fact, I tried to fuddle around with the Poisson kernel, but I did not see things as clear as you, obviously. ;-)
$endgroup$
– amsmath
Jan 26 at 19:18
$begingroup$
That is also very nice. Thank you. In fact, I tried to fuddle around with the Poisson kernel, but I did not see things as clear as you, obviously. ;-)
$endgroup$
– amsmath
Jan 26 at 19:18
$begingroup$
@amsmath I hope this will help :-)
$endgroup$
– Song
Jan 26 at 21:31
$begingroup$
@amsmath I hope this will help :-)
$endgroup$
– Song
Jan 26 at 21:31
add a comment |
$begingroup$
It seems to me that this result is a particular case (or rather 1D adaptation) of Newton's shell theorem. In 2D, the theorem states that no net gravitational force is perceived by objects located inside a uniform hollow sphere.
Here, consider the integral
$$
vec I(M) = int_{Xin C}frac{vec{MX}}{MX^2}mathrm ds
$$
where $M$ is a point at a distance $r<1$ away from the center $O$ of the unit circle $C$, and $mathrm ds$ is a length element of the circle. What you are asking for is a component of $vec I$, the one directed along $vec{OM}$. In fact, we have $vec I=vec 0$. There is a nice geometric proof of this in 2D due to Newton; see here. You can try to adapt it to 1D.
answered Jan 26 at 19:33
HusseinHussein
43114
$endgroup$
$begingroup$
interesting physical interpretation (+1)
$endgroup$
– G Cab
Jan 26 at 19:43
add a comment |
$begingroup$
It seems to me that this result is a particular case (or rather 1D adaptation) of Newton's shell theorem. In 2D, the theorem states that no net gravitational force is perceived by objects located inside a uniform hollow sphere.
Here, consider the integral
$$
vec I(M) = int_{Xin C}frac{vec{MX}}{MX^2}mathrm ds
$$
where $M$ is a point at a distance $r<1$ away from the center $O$ of the unit circle $C$, and $mathrm ds$ is a length element of the circle. What you are asking for is a component of $vec I$, the one directed along $vec{OM}$. In fact, we have $vec I=vec 0$. There is a nice geometric proof of this in 2D due to Newton; see here. You can try to adapt it to 1D.
answered Jan 26 at 19:33
HusseinHussein
43114
$endgroup$
$begingroup$
interesting physical interpretation (+1)
$endgroup$
– G Cab
Jan 26 at 19:43
add a comment |
$begingroup$
It seems to me that this result is a particular case (or rather 1D adaptation) of Newton's shell theorem. In 2D, the theorem states that no net gravitational force is perceived by objects located inside a uniform hollow sphere.
Here, consider the integral
$$
vec I(M) = int_{Xin C}frac{vec{MX}}{MX^2}mathrm ds
$$
where $M$ is a point at a distance $r<1$ away from the center $O$ of the unit circle $C$, and $mathrm ds$ is a length element of the circle. What you are asking for is a component of $vec I$, the one directed along $vec{OM}$. In fact, we have $vec I=vec 0$. There is a nice geometric proof of this in 2D due to Newton; see here. You can try to adapt it to 1D.
answered Jan 26 at 19:33
HusseinHussein
43114
$endgroup$
It seems to me that this result is a particular case (or rather 1D adaptation) of Newton's shell theorem. In 2D, the theorem states that no net gravitational force is perceived by objects located inside a uniform hollow sphere.
Here, consider the integral
$$
vec I(M) = int_{Xin C}frac{vec{MX}}{MX^2}mathrm ds
$$
where $M$ is a point at a distance $r<1$ away from the center $O$ of the unit circle $C$, and $mathrm ds$ is a length element of the circle. What you are asking for is a component of $vec I$, the one directed along $vec{OM}$. In fact, we have $vec I=vec 0$. There is a nice geometric proof of this in 2D due to Newton; see here. You can try to adapt it to 1D.
answered Jan 26 at 19:33
HusseinHussein
43114
answered Jan 26 at 19:33
HusseinHussein
43114
answered Jan 26 at 19:33
HusseinHussein
43114
answered Jan 26 at 19:33
HusseinHussein
43114
43114
$begingroup$
interesting physical interpretation (+1)
$endgroup$
– G Cab
Jan 26 at 19:43
add a comment |
$begingroup$
interesting physical interpretation (+1)
$endgroup$
– G Cab
Jan 26 at 19:43
$begingroup$
interesting physical interpretation (+1)
$endgroup$
– G Cab
Jan 26 at 19:43
$begingroup$
interesting physical interpretation (+1)
$endgroup$
– G Cab
Jan 26 at 19:43
add a comment |
$begingroup$
With $gamma$ being the counter-clockwise unit circle and $bargamma$ being the clockwise unit circle,
$$
begin{align}
int_0^{2pi}frac{r-cos(t)}{1-2rcos(t)+r^2},mathrm{d}t
&=int_0^{2pi}frac12left(frac1{r-e^{it}}+frac1{r-e^{-it}}right)mathrm{d}ttag1\
&=frac1{2i}int_gammafrac{,mathrm{d}z}{z(r-z)}-frac1{2i}int_{bargamma}frac{,mathrm{d}z}{z(r-z)}tag2\
&=frac1iint_gammafrac{,mathrm{d}z}{z(r-z)}tag3\
&=frac1{ir}int_gammaleft(frac1z-frac1{z-r}right)mathrm{d}ztag4\
&=left{begin{array}{}
0&text{if }|r|lt1\
frac{2pi}r&text{if }|r|gt1
end{array}right.tag5
end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: $z=e^{it}$ in the left integral and $z=e^{-it}$ in the right integral
$(3)$: $bargamma$ is in the opposite direction from $gamma$
$(4)$: partial fractions
$(5)$: $2pi i$ times the sum of the residues inside $gamma$
answered Jan 27 at 8:24
robjohn♦robjohn
269k27311638
$endgroup$
add a comment |
$begingroup$
With $gamma$ being the counter-clockwise unit circle and $bargamma$ being the clockwise unit circle,
$$
begin{align}
int_0^{2pi}frac{r-cos(t)}{1-2rcos(t)+r^2},mathrm{d}t
&=int_0^{2pi}frac12left(frac1{r-e^{it}}+frac1{r-e^{-it}}right)mathrm{d}ttag1\
&=frac1{2i}int_gammafrac{,mathrm{d}z}{z(r-z)}-frac1{2i}int_{bargamma}frac{,mathrm{d}z}{z(r-z)}tag2\
&=frac1iint_gammafrac{,mathrm{d}z}{z(r-z)}tag3\
&=frac1{ir}int_gammaleft(frac1z-frac1{z-r}right)mathrm{d}ztag4\
&=left{begin{array}{}
0&text{if }|r|lt1\
frac{2pi}r&text{if }|r|gt1
end{array}right.tag5
end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: $z=e^{it}$ in the left integral and $z=e^{-it}$ in the right integral
$(3)$: $bargamma$ is in the opposite direction from $gamma$
$(4)$: partial fractions
$(5)$: $2pi i$ times the sum of the residues inside $gamma$
answered Jan 27 at 8:24
robjohn♦robjohn
269k27311638
$endgroup$
add a comment |
$begingroup$
With $gamma$ being the counter-clockwise unit circle and $bargamma$ being the clockwise unit circle,
$$
begin{align}
int_0^{2pi}frac{r-cos(t)}{1-2rcos(t)+r^2},mathrm{d}t
&=int_0^{2pi}frac12left(frac1{r-e^{it}}+frac1{r-e^{-it}}right)mathrm{d}ttag1\
&=frac1{2i}int_gammafrac{,mathrm{d}z}{z(r-z)}-frac1{2i}int_{bargamma}frac{,mathrm{d}z}{z(r-z)}tag2\
&=frac1iint_gammafrac{,mathrm{d}z}{z(r-z)}tag3\
&=frac1{ir}int_gammaleft(frac1z-frac1{z-r}right)mathrm{d}ztag4\
&=left{begin{array}{}
0&text{if }|r|lt1\
frac{2pi}r&text{if }|r|gt1
end{array}right.tag5
end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: $z=e^{it}$ in the left integral and $z=e^{-it}$ in the right integral
$(3)$: $bargamma$ is in the opposite direction from $gamma$
$(4)$: partial fractions
$(5)$: $2pi i$ times the sum of the residues inside $gamma$
answered Jan 27 at 8:24
robjohn♦robjohn
269k27311638
$endgroup$
With $gamma$ being the counter-clockwise unit circle and $bargamma$ being the clockwise unit circle,
$$
begin{align}
int_0^{2pi}frac{r-cos(t)}{1-2rcos(t)+r^2},mathrm{d}t
&=int_0^{2pi}frac12left(frac1{r-e^{it}}+frac1{r-e^{-it}}right)mathrm{d}ttag1\
&=frac1{2i}int_gammafrac{,mathrm{d}z}{z(r-z)}-frac1{2i}int_{bargamma}frac{,mathrm{d}z}{z(r-z)}tag2\
&=frac1iint_gammafrac{,mathrm{d}z}{z(r-z)}tag3\
&=frac1{ir}int_gammaleft(frac1z-frac1{z-r}right)mathrm{d}ztag4\
&=left{begin{array}{}
0&text{if }|r|lt1\
frac{2pi}r&text{if }|r|gt1
end{array}right.tag5
end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: $z=e^{it}$ in the left integral and $z=e^{-it}$ in the right integral
$(3)$: $bargamma$ is in the opposite direction from $gamma$
$(4)$: partial fractions
$(5)$: $2pi i$ times the sum of the residues inside $gamma$
answered Jan 27 at 8:24
robjohn♦robjohn
269k27311638
edited Jan 27 at 8:45
answered Jan 27 at 8:24
robjohn♦robjohn
269k27311638
answered Jan 27 at 8:24
robjohn♦robjohn
269k27311638
answered Jan 27 at 8:24
robjohn♦robjohn
269k27311638
269k27311638
add a comment |
add a comment |
$begingroup$
Here is yet another slightly different approach.
Let
$$P_r(t) = frac{cos t - r}{1 - 2r cos t + r^2}, qquad r in [0,1).$$
Using $cos t = (e^{it} + e^{-it})/2$, we can rewrite $P_r(t)$ as
begin{align}
P_r (t) &= frac{1}{2} frac{(e^{it} + e^{-it}) - 2r}{1 - r(e^{it} + e^{-it}) + r^2}\
&= -frac{1}{2} frac{2r - e^{it} - e^{-it}}{(r - e^{it})(r - e^{-it})}\
&= -frac{1}{2} left [frac{1}{r - e^{it}} + frac{1}{r - e^{-it}} right ]\
&= frac{1}{2} left [frac{e^{-it}}{1 - r e^{-it}} + frac{e^{it}}{1 - r e^{it}} right ]\
&= frac{1}{2} left [e^{-it} sum_{n = 0}^infty r^n e^{-int} + e^{it} sum_{n = 0}^infty r^n e^{int} right ]\
&= sum_{n = 0}^infty r^n left (frac{e^{i(n + 1)t} + e^{-i(n + 1)t}}{2} right )\
&= sum_{n = 0}^infty r^n cos (n + 1)t.
end{align}
Now
begin{align}
int^{2pi}_0 P_r (t) , dt = int^{2 pi}_0 sum_{n = 0}^infty r^n cos (n + 1)t , dt = sum_{n = 0}^infty r^n int^{2pi}_0 cos (n + 1)t , dt = 0.
end{align}
answered Jan 27 at 3:30
omegadotomegadot
6,4342829
$endgroup$
add a comment |
$begingroup$
Here is yet another slightly different approach.
Let
$$P_r(t) = frac{cos t - r}{1 - 2r cos t + r^2}, qquad r in [0,1).$$
Using $cos t = (e^{it} + e^{-it})/2$, we can rewrite $P_r(t)$ as
begin{align}
P_r (t) &= frac{1}{2} frac{(e^{it} + e^{-it}) - 2r}{1 - r(e^{it} + e^{-it}) + r^2}\
&= -frac{1}{2} frac{2r - e^{it} - e^{-it}}{(r - e^{it})(r - e^{-it})}\
&= -frac{1}{2} left [frac{1}{r - e^{it}} + frac{1}{r - e^{-it}} right ]\
&= frac{1}{2} left [frac{e^{-it}}{1 - r e^{-it}} + frac{e^{it}}{1 - r e^{it}} right ]\
&= frac{1}{2} left [e^{-it} sum_{n = 0}^infty r^n e^{-int} + e^{it} sum_{n = 0}^infty r^n e^{int} right ]\
&= sum_{n = 0}^infty r^n left (frac{e^{i(n + 1)t} + e^{-i(n + 1)t}}{2} right )\
&= sum_{n = 0}^infty r^n cos (n + 1)t.
end{align}
Now
begin{align}
int^{2pi}_0 P_r (t) , dt = int^{2 pi}_0 sum_{n = 0}^infty r^n cos (n + 1)t , dt = sum_{n = 0}^infty r^n int^{2pi}_0 cos (n + 1)t , dt = 0.
end{align}
answered Jan 27 at 3:30
omegadotomegadot
6,4342829
$endgroup$
add a comment |
$begingroup$
Here is yet another slightly different approach.
Let
$$P_r(t) = frac{cos t - r}{1 - 2r cos t + r^2}, qquad r in [0,1).$$
Using $cos t = (e^{it} + e^{-it})/2$, we can rewrite $P_r(t)$ as
begin{align}
P_r (t) &= frac{1}{2} frac{(e^{it} + e^{-it}) - 2r}{1 - r(e^{it} + e^{-it}) + r^2}\
&= -frac{1}{2} frac{2r - e^{it} - e^{-it}}{(r - e^{it})(r - e^{-it})}\
&= -frac{1}{2} left [frac{1}{r - e^{it}} + frac{1}{r - e^{-it}} right ]\
&= frac{1}{2} left [frac{e^{-it}}{1 - r e^{-it}} + frac{e^{it}}{1 - r e^{it}} right ]\
&= frac{1}{2} left [e^{-it} sum_{n = 0}^infty r^n e^{-int} + e^{it} sum_{n = 0}^infty r^n e^{int} right ]\
&= sum_{n = 0}^infty r^n left (frac{e^{i(n + 1)t} + e^{-i(n + 1)t}}{2} right )\
&= sum_{n = 0}^infty r^n cos (n + 1)t.
end{align}
Now
begin{align}
int^{2pi}_0 P_r (t) , dt = int^{2 pi}_0 sum_{n = 0}^infty r^n cos (n + 1)t , dt = sum_{n = 0}^infty r^n int^{2pi}_0 cos (n + 1)t , dt = 0.
end{align}
answered Jan 27 at 3:30
omegadotomegadot
6,4342829
$endgroup$
Here is yet another slightly different approach.
Let
$$P_r(t) = frac{cos t - r}{1 - 2r cos t + r^2}, qquad r in [0,1).$$
Using $cos t = (e^{it} + e^{-it})/2$, we can rewrite $P_r(t)$ as
begin{align}
P_r (t) &= frac{1}{2} frac{(e^{it} + e^{-it}) - 2r}{1 - r(e^{it} + e^{-it}) + r^2}\
&= -frac{1}{2} frac{2r - e^{it} - e^{-it}}{(r - e^{it})(r - e^{-it})}\
&= -frac{1}{2} left [frac{1}{r - e^{it}} + frac{1}{r - e^{-it}} right ]\
&= frac{1}{2} left [frac{e^{-it}}{1 - r e^{-it}} + frac{e^{it}}{1 - r e^{it}} right ]\
&= frac{1}{2} left [e^{-it} sum_{n = 0}^infty r^n e^{-int} + e^{it} sum_{n = 0}^infty r^n e^{int} right ]\
&= sum_{n = 0}^infty r^n left (frac{e^{i(n + 1)t} + e^{-i(n + 1)t}}{2} right )\
&= sum_{n = 0}^infty r^n cos (n + 1)t.
end{align}
Now
begin{align}
int^{2pi}_0 P_r (t) , dt = int^{2 pi}_0 sum_{n = 0}^infty r^n cos (n + 1)t , dt = sum_{n = 0}^infty r^n int^{2pi}_0 cos (n + 1)t , dt = 0.
end{align}
answered Jan 27 at 3:30
omegadotomegadot
6,4342829
answered Jan 27 at 3:30
omegadotomegadot
6,4342829
answered Jan 27 at 3:30
omegadotomegadot
6,4342829
answered Jan 27 at 3:30
omegadotomegadot
6,4342829
6,4342829
add a comment |
add a comment |
$begingroup$
Note that $$I(r) =frac{-1}{2r}int_{0}^{2pi}frac{1-2rcos t+r^2+r^2-1}{1-2rcos t+r^2},dt$$ which is same as $$-frac{pi} {r} +frac{1-r^2}{r}int_{0}^{pi}frac{dt}{1+r^2-2rcos t} $$ The integral above is equal to $$frac{pi} {sqrt{(1+r^2)^2-4r^2}}=frac{pi}{1-r^2} $$ and hence $I(r) =0$.
We have used the formula $$int_{0}^{pi}frac{dt}{a+bcos t} =frac{pi} {sqrt{a^2-b^2}},,a>|b|$$ which can be proved using the substitution $$(a+bcos t) (a-bcos u) =a^2-b^2$$
answered Jan 27 at 8:37
Paramanand SinghParamanand Singh
50.9k557168
$endgroup$
add a comment |
$begingroup$
Note that $$I(r) =frac{-1}{2r}int_{0}^{2pi}frac{1-2rcos t+r^2+r^2-1}{1-2rcos t+r^2},dt$$ which is same as $$-frac{pi} {r} +frac{1-r^2}{r}int_{0}^{pi}frac{dt}{1+r^2-2rcos t} $$ The integral above is equal to $$frac{pi} {sqrt{(1+r^2)^2-4r^2}}=frac{pi}{1-r^2} $$ and hence $I(r) =0$.
We have used the formula $$int_{0}^{pi}frac{dt}{a+bcos t} =frac{pi} {sqrt{a^2-b^2}},,a>|b|$$ which can be proved using the substitution $$(a+bcos t) (a-bcos u) =a^2-b^2$$
answered Jan 27 at 8:37
Paramanand SinghParamanand Singh
50.9k557168
$endgroup$
add a comment |
$begingroup$
Note that $$I(r) =frac{-1}{2r}int_{0}^{2pi}frac{1-2rcos t+r^2+r^2-1}{1-2rcos t+r^2},dt$$ which is same as $$-frac{pi} {r} +frac{1-r^2}{r}int_{0}^{pi}frac{dt}{1+r^2-2rcos t} $$ The integral above is equal to $$frac{pi} {sqrt{(1+r^2)^2-4r^2}}=frac{pi}{1-r^2} $$ and hence $I(r) =0$.
We have used the formula $$int_{0}^{pi}frac{dt}{a+bcos t} =frac{pi} {sqrt{a^2-b^2}},,a>|b|$$ which can be proved using the substitution $$(a+bcos t) (a-bcos u) =a^2-b^2$$
answered Jan 27 at 8:37
Paramanand SinghParamanand Singh
50.9k557168
$endgroup$
Note that $$I(r) =frac{-1}{2r}int_{0}^{2pi}frac{1-2rcos t+r^2+r^2-1}{1-2rcos t+r^2},dt$$ which is same as $$-frac{pi} {r} +frac{1-r^2}{r}int_{0}^{pi}frac{dt}{1+r^2-2rcos t} $$ The integral above is equal to $$frac{pi} {sqrt{(1+r^2)^2-4r^2}}=frac{pi}{1-r^2} $$ and hence $I(r) =0$.
We have used the formula $$int_{0}^{pi}frac{dt}{a+bcos t} =frac{pi} {sqrt{a^2-b^2}},,a>|b|$$ which can be proved using the substitution $$(a+bcos t) (a-bcos u) =a^2-b^2$$
answered Jan 27 at 8:37
Paramanand SinghParamanand Singh
50.9k557168
answered Jan 27 at 8:37
Paramanand SinghParamanand Singh
50.9k557168
answered Jan 27 at 8:37
Paramanand SinghParamanand Singh
50.9k557168
answered Jan 27 at 8:37
Paramanand SinghParamanand Singh
50.9k557168
50.9k557168
add a comment |
add a comment |
$begingroup$
Hint: $$cos(t)=frac{1-a^2}{1+a^2}$$ and $$dt=frac{2da}{1+a^2}$$
Now your indefinite integral is given by $$int-frac{2 left(a^2 r+a^2+r-1right)}{left(a^2+1right)
left(a^2 r^2+2 a^2 r+a^2+r^2-2 r+1right)}da$$ and this is rational
The solution of this integral is given by $$-2 left(frac{tan ^{-1}(a)}{2 r}-frac{tan
^{-1}left(frac{a (r+1)}{1-r}right)}{4
r}+frac{tan ^{-1}left(frac{a
(r+1)}{r-1}right)}{4 r}right)$$
answered Jan 26 at 17:42
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
5
$begingroup$
Thank you very much. So my integral is indeed zero. You can fuse the second and the third term to one term: $-tfrac 1{2r}arctan(tfrac{1+r}{1-r}a)$. The boundaries after substitution are zero and $infty$.
$endgroup$
– amsmath
Jan 26 at 18:04
add a comment |
$begingroup$
Hint: $$cos(t)=frac{1-a^2}{1+a^2}$$ and $$dt=frac{2da}{1+a^2}$$
Now your indefinite integral is given by $$int-frac{2 left(a^2 r+a^2+r-1right)}{left(a^2+1right)
left(a^2 r^2+2 a^2 r+a^2+r^2-2 r+1right)}da$$ and this is rational
The solution of this integral is given by $$-2 left(frac{tan ^{-1}(a)}{2 r}-frac{tan
^{-1}left(frac{a (r+1)}{1-r}right)}{4
r}+frac{tan ^{-1}left(frac{a
(r+1)}{r-1}right)}{4 r}right)$$
answered Jan 26 at 17:42
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
5
$begingroup$
Thank you very much. So my integral is indeed zero. You can fuse the second and the third term to one term: $-tfrac 1{2r}arctan(tfrac{1+r}{1-r}a)$. The boundaries after substitution are zero and $infty$.
$endgroup$
– amsmath
Jan 26 at 18:04
add a comment |
$begingroup$
Hint: $$cos(t)=frac{1-a^2}{1+a^2}$$ and $$dt=frac{2da}{1+a^2}$$
Now your indefinite integral is given by $$int-frac{2 left(a^2 r+a^2+r-1right)}{left(a^2+1right)
left(a^2 r^2+2 a^2 r+a^2+r^2-2 r+1right)}da$$ and this is rational
The solution of this integral is given by $$-2 left(frac{tan ^{-1}(a)}{2 r}-frac{tan
^{-1}left(frac{a (r+1)}{1-r}right)}{4
r}+frac{tan ^{-1}left(frac{a
(r+1)}{r-1}right)}{4 r}right)$$
answered Jan 26 at 17:42
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
Hint: $$cos(t)=frac{1-a^2}{1+a^2}$$ and $$dt=frac{2da}{1+a^2}$$
Now your indefinite integral is given by $$int-frac{2 left(a^2 r+a^2+r-1right)}{left(a^2+1right)
left(a^2 r^2+2 a^2 r+a^2+r^2-2 r+1right)}da$$ and this is rational
The solution of this integral is given by $$-2 left(frac{tan ^{-1}(a)}{2 r}-frac{tan
^{-1}left(frac{a (r+1)}{1-r}right)}{4
r}+frac{tan ^{-1}left(frac{a
(r+1)}{r-1}right)}{4 r}right)$$
answered Jan 26 at 17:42
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
edited Jan 26 at 17:48
answered Jan 26 at 17:42
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered Jan 26 at 17:42
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered Jan 26 at 17:42
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
78k42866
5
$begingroup$
Thank you very much. So my integral is indeed zero. You can fuse the second and the third term to one term: $-tfrac 1{2r}arctan(tfrac{1+r}{1-r}a)$. The boundaries after substitution are zero and $infty$.
$endgroup$
– amsmath
Jan 26 at 18:04
add a comment |
5
$begingroup$
Thank you very much. So my integral is indeed zero. You can fuse the second and the third term to one term: $-tfrac 1{2r}arctan(tfrac{1+r}{1-r}a)$. The boundaries after substitution are zero and $infty$.
$endgroup$
– amsmath
Jan 26 at 18:04
5
5
$begingroup$
Thank you very much. So my integral is indeed zero. You can fuse the second and the third term to one term: $-tfrac 1{2r}arctan(tfrac{1+r}{1-r}a)$. The boundaries after substitution are zero and $infty$.
$endgroup$
– amsmath
Jan 26 at 18:04
$begingroup$
Thank you very much. So my integral is indeed zero. You can fuse the second and the third term to one term: $-tfrac 1{2r}arctan(tfrac{1+r}{1-r}a)$. The boundaries after substitution are zero and $infty$.
$endgroup$
– amsmath
Jan 26 at 18:04
add a comment |
$begingroup$
Not sure why none of the answers mention this approach, but you can just take the derivative with respect to $r$:
begin{align}
I'(r) &= int_0^{2pi}frac{d}{dr}left(frac{cos t - r}{1 - 2rcos t + r^2}right),dt\
&= int_0^{2pi} frac{(r^2-1)-2rcos t + 2cos^2t}{(1 - 2rcos t + r^2)^2} \
&= frac1{i}int_{|z|=1} frac{(r^2-1)-r(z+z^{-1}) + frac12(z+z^{-1})^2}{(1 - r(z+z^{-1}) + r^2)^2}cdotfrac{dz}{z}\
&= frac1{i}int_{|z|=1} frac{z^4-2r z^3+2r^2z^2-2rz+1}{(r^2+1)z(z-r)^2left(z-frac1rright)^2},dz\
end{align}
Since $r in [0,1rangle$, the relevant residues are $$operatorname{Res}(f,0) = frac1{2r^2}$$
$$operatorname{Res}(f,r) = -frac1{2r^2}$$
so $I'(r) = 0$ for $r in [0,1rangle$.
We conclude that $I$ is constant on $[0,1rangle$. Hence $$I(r)=I(0) = int_0^{2pi}cos t,dt = 0$$
answered Jan 27 at 18:19
mechanodroidmechanodroid
28.9k62548
$endgroup$
add a comment |
$begingroup$
Not sure why none of the answers mention this approach, but you can just take the derivative with respect to $r$:
begin{align}
I'(r) &= int_0^{2pi}frac{d}{dr}left(frac{cos t - r}{1 - 2rcos t + r^2}right),dt\
&= int_0^{2pi} frac{(r^2-1)-2rcos t + 2cos^2t}{(1 - 2rcos t + r^2)^2} \
&= frac1{i}int_{|z|=1} frac{(r^2-1)-r(z+z^{-1}) + frac12(z+z^{-1})^2}{(1 - r(z+z^{-1}) + r^2)^2}cdotfrac{dz}{z}\
&= frac1{i}int_{|z|=1} frac{z^4-2r z^3+2r^2z^2-2rz+1}{(r^2+1)z(z-r)^2left(z-frac1rright)^2},dz\
end{align}
Since $r in [0,1rangle$, the relevant residues are $$operatorname{Res}(f,0) = frac1{2r^2}$$
$$operatorname{Res}(f,r) = -frac1{2r^2}$$
so $I'(r) = 0$ for $r in [0,1rangle$.
We conclude that $I$ is constant on $[0,1rangle$. Hence $$I(r)=I(0) = int_0^{2pi}cos t,dt = 0$$
answered Jan 27 at 18:19
mechanodroidmechanodroid
28.9k62548
$endgroup$
add a comment |
$begingroup$
Not sure why none of the answers mention this approach, but you can just take the derivative with respect to $r$:
begin{align}
I'(r) &= int_0^{2pi}frac{d}{dr}left(frac{cos t - r}{1 - 2rcos t + r^2}right),dt\
&= int_0^{2pi} frac{(r^2-1)-2rcos t + 2cos^2t}{(1 - 2rcos t + r^2)^2} \
&= frac1{i}int_{|z|=1} frac{(r^2-1)-r(z+z^{-1}) + frac12(z+z^{-1})^2}{(1 - r(z+z^{-1}) + r^2)^2}cdotfrac{dz}{z}\
&= frac1{i}int_{|z|=1} frac{z^4-2r z^3+2r^2z^2-2rz+1}{(r^2+1)z(z-r)^2left(z-frac1rright)^2},dz\
end{align}
Since $r in [0,1rangle$, the relevant residues are $$operatorname{Res}(f,0) = frac1{2r^2}$$
$$operatorname{Res}(f,r) = -frac1{2r^2}$$
so $I'(r) = 0$ for $r in [0,1rangle$.
We conclude that $I$ is constant on $[0,1rangle$. Hence $$I(r)=I(0) = int_0^{2pi}cos t,dt = 0$$
answered Jan 27 at 18:19
mechanodroidmechanodroid
28.9k62548
$endgroup$
Not sure why none of the answers mention this approach, but you can just take the derivative with respect to $r$:
begin{align}
I'(r) &= int_0^{2pi}frac{d}{dr}left(frac{cos t - r}{1 - 2rcos t + r^2}right),dt\
&= int_0^{2pi} frac{(r^2-1)-2rcos t + 2cos^2t}{(1 - 2rcos t + r^2)^2} \
&= frac1{i}int_{|z|=1} frac{(r^2-1)-r(z+z^{-1}) + frac12(z+z^{-1})^2}{(1 - r(z+z^{-1}) + r^2)^2}cdotfrac{dz}{z}\
&= frac1{i}int_{|z|=1} frac{z^4-2r z^3+2r^2z^2-2rz+1}{(r^2+1)z(z-r)^2left(z-frac1rright)^2},dz\
end{align}
Since $r in [0,1rangle$, the relevant residues are $$operatorname{Res}(f,0) = frac1{2r^2}$$
$$operatorname{Res}(f,r) = -frac1{2r^2}$$
so $I'(r) = 0$ for $r in [0,1rangle$.
We conclude that $I$ is constant on $[0,1rangle$. Hence $$I(r)=I(0) = int_0^{2pi}cos t,dt = 0$$
answered Jan 27 at 18:19
mechanodroidmechanodroid
28.9k62548
edited Jan 30 at 8:05
answered Jan 27 at 18:19
mechanodroidmechanodroid
28.9k62548
answered Jan 27 at 18:19
mechanodroidmechanodroid
28.9k62548
answered Jan 27 at 18:19
mechanodroidmechanodroid
28.9k62548
28.9k62548
add a comment |
add a comment |
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$begingroup$
@Dr.SonnhardGraubner Edited
$endgroup$
– amsmath
Jan 26 at 17:23
$begingroup$
I think your integral isn't zero
$endgroup$
– Dr. Sonnhard Graubner
Jan 26 at 17:31
$begingroup$
@Dr.SonnhardGraubner Well, that was helpful... Why?
$endgroup$
– amsmath
Jan 26 at 17:33
14
$begingroup$
@Dr.SonnhardGraubner: I think the integral is zero.
$endgroup$
– Larry
Jan 26 at 18:01
3
$begingroup$
@Dr.SonnhardGraubner You gave us the proof below. ;-)
$endgroup$
– amsmath
Jan 26 at 18:08