Mixing
Contour intergal of a rational trigonometric function, can't find my mistake.
This is the integral :
$$I = int_{0}^{2pi} frac{dx}{(5+4cos x)^2} $$
Which, according to wolfram alpha, should evaluate to $frac{10pi}{27} $,
but the value i find is $frac{20pi}{27} $.
These are my calculations :
Using complex form of cosine i get $ cos x = frac{t^2+1}{2t}$
where $ t = e^{ix}$ and $ dx = frac{dt}{it}$.
If $partial{D} $ is the unit circle
$$ I = int_{partial{D}} frac{-i}{t(frac{5t+2t^2+2}{t})^2} dt =
-i int_{partial{D}}frac{t}{(t+2)^2(t+1/2)^2}dt = -i I_a$$
$I_a = pi i $ times the residue in -$frac{1}{2}$
The residue is equal to
begin{align}
lim_{zto-frac{1}{2}} frac{d}{dz} left( frac{t(t+1/2)^2}{(t+2)^2(t+1/2)^2}right)
&= lim_{zto-frac{1}{2}} frac{d}{dz} frac{t}{(t+2)^2}
\&= lim_{zto-frac{1}{2}} frac{(t+2)^2-2t(t+2)}{(t+2)^4}
\&= lim_{zto-frac{1}{2}} frac{2-t}{(t+2)^3}
= frac{2+1/2}{(-1/2+2)^3}
\&= frac{5/2}{27/9} = frac{20}{27}
end{align}
so $ I_a = ipifrac{20}{27}$
So our original integral $$I = -i I_a = -i^2 pifrac{20}{27} = frac{20pi}{27}$$
complex-analysis contour-integration trigonometric-integrals
edited Nov 21 '18 at 21:06
LutzL
56.5k42054
asked Nov 21 '18 at 19:56
Nicola FattorusoNicola Fattoruso
43
|
show 6 more comments
This is the integral :
$$I = int_{0}^{2pi} frac{dx}{(5+4cos x)^2} $$
Which, according to wolfram alpha, should evaluate to $frac{10pi}{27} $,
but the value i find is $frac{20pi}{27} $.
These are my calculations :
Using complex form of cosine i get $ cos x = frac{t^2+1}{2t}$
where $ t = e^{ix}$ and $ dx = frac{dt}{it}$.
If $partial{D} $ is the unit circle
$$ I = int_{partial{D}} frac{-i}{t(frac{5t+2t^2+2}{t})^2} dt =
-i int_{partial{D}}frac{t}{(t+2)^2(t+1/2)^2}dt = -i I_a$$
$I_a = pi i $ times the residue in -$frac{1}{2}$
The residue is equal to
begin{align}
lim_{zto-frac{1}{2}} frac{d}{dz} left( frac{t(t+1/2)^2}{(t+2)^2(t+1/2)^2}right)
&= lim_{zto-frac{1}{2}} frac{d}{dz} frac{t}{(t+2)^2}
\&= lim_{zto-frac{1}{2}} frac{(t+2)^2-2t(t+2)}{(t+2)^4}
\&= lim_{zto-frac{1}{2}} frac{2-t}{(t+2)^3}
= frac{2+1/2}{(-1/2+2)^3}
\&= frac{5/2}{27/9} = frac{20}{27}
end{align}
so $ I_a = ipifrac{20}{27}$
So our original integral $$I = -i I_a = -i^2 pifrac{20}{27} = frac{20pi}{27}$$
complex-analysis contour-integration trigonometric-integrals
edited Nov 21 '18 at 21:06
LutzL
56.5k42054
asked Nov 21 '18 at 19:56
Nicola FattorusoNicola Fattoruso
43
Residue theorem asks for a factor $2pi i$ instead of $pi i$ no?
– Math_QED
Nov 21 '18 at 20:09
I tought since it's on the real axis, it was $ipi$
– Nicola Fattoruso
Nov 21 '18 at 20:13
Did you mean to say $dx = frac{dt}{it}$?
– Acccumulation
Nov 21 '18 at 20:13
Yes, edited sorry.
– Nicola Fattoruso
Nov 21 '18 at 20:15
No, the residue theorem always give $2 pi i$.
– Math_QED
Nov 21 '18 at 20:15
|
show 6 more comments
This is the integral :
$$I = int_{0}^{2pi} frac{dx}{(5+4cos x)^2} $$
Which, according to wolfram alpha, should evaluate to $frac{10pi}{27} $,
but the value i find is $frac{20pi}{27} $.
These are my calculations :
Using complex form of cosine i get $ cos x = frac{t^2+1}{2t}$
where $ t = e^{ix}$ and $ dx = frac{dt}{it}$.
If $partial{D} $ is the unit circle
$$ I = int_{partial{D}} frac{-i}{t(frac{5t+2t^2+2}{t})^2} dt =
-i int_{partial{D}}frac{t}{(t+2)^2(t+1/2)^2}dt = -i I_a$$
$I_a = pi i $ times the residue in -$frac{1}{2}$
The residue is equal to
begin{align}
lim_{zto-frac{1}{2}} frac{d}{dz} left( frac{t(t+1/2)^2}{(t+2)^2(t+1/2)^2}right)
&= lim_{zto-frac{1}{2}} frac{d}{dz} frac{t}{(t+2)^2}
\&= lim_{zto-frac{1}{2}} frac{(t+2)^2-2t(t+2)}{(t+2)^4}
\&= lim_{zto-frac{1}{2}} frac{2-t}{(t+2)^3}
= frac{2+1/2}{(-1/2+2)^3}
\&= frac{5/2}{27/9} = frac{20}{27}
end{align}
so $ I_a = ipifrac{20}{27}$
So our original integral $$I = -i I_a = -i^2 pifrac{20}{27} = frac{20pi}{27}$$
complex-analysis contour-integration trigonometric-integrals
edited Nov 21 '18 at 21:06
LutzL
56.5k42054
asked Nov 21 '18 at 19:56
Nicola FattorusoNicola Fattoruso
43
This is the integral :
$$I = int_{0}^{2pi} frac{dx}{(5+4cos x)^2} $$
Which, according to wolfram alpha, should evaluate to $frac{10pi}{27} $,
but the value i find is $frac{20pi}{27} $.
These are my calculations :
Using complex form of cosine i get $ cos x = frac{t^2+1}{2t}$
where $ t = e^{ix}$ and $ dx = frac{dt}{it}$.
If $partial{D} $ is the unit circle
$$ I = int_{partial{D}} frac{-i}{t(frac{5t+2t^2+2}{t})^2} dt =
-i int_{partial{D}}frac{t}{(t+2)^2(t+1/2)^2}dt = -i I_a$$
$I_a = pi i $ times the residue in -$frac{1}{2}$
The residue is equal to
begin{align}
lim_{zto-frac{1}{2}} frac{d}{dz} left( frac{t(t+1/2)^2}{(t+2)^2(t+1/2)^2}right)
&= lim_{zto-frac{1}{2}} frac{d}{dz} frac{t}{(t+2)^2}
\&= lim_{zto-frac{1}{2}} frac{(t+2)^2-2t(t+2)}{(t+2)^4}
\&= lim_{zto-frac{1}{2}} frac{2-t}{(t+2)^3}
= frac{2+1/2}{(-1/2+2)^3}
\&= frac{5/2}{27/9} = frac{20}{27}
end{align}
so $ I_a = ipifrac{20}{27}$
So our original integral $$I = -i I_a = -i^2 pifrac{20}{27} = frac{20pi}{27}$$
complex-analysis contour-integration trigonometric-integrals
complex-analysis contour-integration trigonometric-integrals
edited Nov 21 '18 at 21:06
LutzL
56.5k42054
asked Nov 21 '18 at 19:56
Nicola FattorusoNicola Fattoruso
43
edited Nov 21 '18 at 21:06
LutzL
56.5k42054
asked Nov 21 '18 at 19:56
Nicola FattorusoNicola Fattoruso
43
edited Nov 21 '18 at 21:06
LutzL
56.5k42054
edited Nov 21 '18 at 21:06
LutzL
56.5k42054
edited Nov 21 '18 at 21:06
LutzL
56.5k42054
56.5k42054
asked Nov 21 '18 at 19:56
Nicola FattorusoNicola Fattoruso
43
asked Nov 21 '18 at 19:56
Nicola FattorusoNicola Fattoruso
43
asked Nov 21 '18 at 19:56
Nicola FattorusoNicola Fattoruso
43
43
Residue theorem asks for a factor $2pi i$ instead of $pi i$ no?
– Math_QED
Nov 21 '18 at 20:09
I tought since it's on the real axis, it was $ipi$
– Nicola Fattoruso
Nov 21 '18 at 20:13
Did you mean to say $dx = frac{dt}{it}$?
– Acccumulation
Nov 21 '18 at 20:13
Yes, edited sorry.
– Nicola Fattoruso
Nov 21 '18 at 20:15
No, the residue theorem always give $2 pi i$.
– Math_QED
Nov 21 '18 at 20:15
|
show 6 more comments
Residue theorem asks for a factor $2pi i$ instead of $pi i$ no?
– Math_QED
Nov 21 '18 at 20:09
I tought since it's on the real axis, it was $ipi$
– Nicola Fattoruso
Nov 21 '18 at 20:13
Did you mean to say $dx = frac{dt}{it}$?
– Acccumulation
Nov 21 '18 at 20:13
Yes, edited sorry.
– Nicola Fattoruso
Nov 21 '18 at 20:15
No, the residue theorem always give $2 pi i$.
– Math_QED
Nov 21 '18 at 20:15
Residue theorem asks for a factor $2pi i$ instead of $pi i$ no?
– Math_QED
Nov 21 '18 at 20:09
Residue theorem asks for a factor $2pi i$ instead of $pi i$ no?
– Math_QED
Nov 21 '18 at 20:09
I tought since it's on the real axis, it was $ipi$
– Nicola Fattoruso
Nov 21 '18 at 20:13
I tought since it's on the real axis, it was $ipi$
– Nicola Fattoruso
Nov 21 '18 at 20:13
Did you mean to say $dx = frac{dt}{it}$?
– Acccumulation
Nov 21 '18 at 20:13
Did you mean to say $dx = frac{dt}{it}$?
– Acccumulation
Nov 21 '18 at 20:13
Yes, edited sorry.
– Nicola Fattoruso
Nov 21 '18 at 20:15
Yes, edited sorry.
– Nicola Fattoruso
Nov 21 '18 at 20:15
No, the residue theorem always give $2 pi i$.
– Math_QED
Nov 21 '18 at 20:15
No, the residue theorem always give $2 pi i$.
– Math_QED
Nov 21 '18 at 20:15
|
show 6 more comments
2 Answers
2
active
oldest
votes
You made an error in calculating the factorization of the denominator. You should get $$5t+2t^2+2=(2+t)(1+2t)=2(t+2)(t+1/2),$$ where you missed to transfer the leading factor $2$ to the factorized expression. The correction leads to $I=-frac i4 I_a$ and the additional division by $4$ corrects the residuum that is computed with the correct factor $2pi i$.
You could also use the geometric/binomial series to get the correct residual coefficient. Then, using $u=t+frac12$,
$$
frac{t}{(t+2)^2(t+frac12)^2}=frac{u-tfrac12}{(u+frac32)^2u^2}
=frac49(u-tfrac12)sum_{k=0}^infty (k+1)left(-frac23right)^ku^{k-2}
$$
so that the coefficient for the power $u^{-1}=(t+frac12)^{-2}$ is $frac49(1+frac12cdot2cdotfrac23)=frac{20}{27}$. The integration over the unit circle adds the factor $2pi i$, so that in the final result $I=frac{10pi}{27}$.
answered Nov 21 '18 at 21:23
LutzLLutzL
56.5k42054
+1 for the second method, although I think taking the limit is more straightforward.
– Math_QED
Nov 21 '18 at 21:25
add a comment |
Your mistake is when using the residue theorem:
$$oint_{C^+} f(z) dz = 2pi isum_{singularities} Res(f,singularity)$$
You forgot the factor $2$ in this formula.
Another mistake: $$5t + 2t^2 + 2 = 2(t+2)(t+1/2)$$
You again forgot a factor 2 (which will yield a factor $4$ eventually)
You also made a mistake in calculating the residue.
Note that $(-1/2 + 2)^3 = (3/2)^3 = 27/8$
Correcting these two mistakes, we easily find the correct answer.
answered Nov 21 '18 at 20:12
Math_QEDMath_QED
7,30131449
add a comment |
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2 Answers
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2 Answers
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You made an error in calculating the factorization of the denominator. You should get $$5t+2t^2+2=(2+t)(1+2t)=2(t+2)(t+1/2),$$ where you missed to transfer the leading factor $2$ to the factorized expression. The correction leads to $I=-frac i4 I_a$ and the additional division by $4$ corrects the residuum that is computed with the correct factor $2pi i$.
You could also use the geometric/binomial series to get the correct residual coefficient. Then, using $u=t+frac12$,
$$
frac{t}{(t+2)^2(t+frac12)^2}=frac{u-tfrac12}{(u+frac32)^2u^2}
=frac49(u-tfrac12)sum_{k=0}^infty (k+1)left(-frac23right)^ku^{k-2}
$$
so that the coefficient for the power $u^{-1}=(t+frac12)^{-2}$ is $frac49(1+frac12cdot2cdotfrac23)=frac{20}{27}$. The integration over the unit circle adds the factor $2pi i$, so that in the final result $I=frac{10pi}{27}$.
answered Nov 21 '18 at 21:23
LutzLLutzL
56.5k42054
+1 for the second method, although I think taking the limit is more straightforward.
– Math_QED
Nov 21 '18 at 21:25
add a comment |
You made an error in calculating the factorization of the denominator. You should get $$5t+2t^2+2=(2+t)(1+2t)=2(t+2)(t+1/2),$$ where you missed to transfer the leading factor $2$ to the factorized expression. The correction leads to $I=-frac i4 I_a$ and the additional division by $4$ corrects the residuum that is computed with the correct factor $2pi i$.
You could also use the geometric/binomial series to get the correct residual coefficient. Then, using $u=t+frac12$,
$$
frac{t}{(t+2)^2(t+frac12)^2}=frac{u-tfrac12}{(u+frac32)^2u^2}
=frac49(u-tfrac12)sum_{k=0}^infty (k+1)left(-frac23right)^ku^{k-2}
$$
so that the coefficient for the power $u^{-1}=(t+frac12)^{-2}$ is $frac49(1+frac12cdot2cdotfrac23)=frac{20}{27}$. The integration over the unit circle adds the factor $2pi i$, so that in the final result $I=frac{10pi}{27}$.
answered Nov 21 '18 at 21:23
LutzLLutzL
56.5k42054
+1 for the second method, although I think taking the limit is more straightforward.
– Math_QED
Nov 21 '18 at 21:25
add a comment |
You made an error in calculating the factorization of the denominator. You should get $$5t+2t^2+2=(2+t)(1+2t)=2(t+2)(t+1/2),$$ where you missed to transfer the leading factor $2$ to the factorized expression. The correction leads to $I=-frac i4 I_a$ and the additional division by $4$ corrects the residuum that is computed with the correct factor $2pi i$.
You could also use the geometric/binomial series to get the correct residual coefficient. Then, using $u=t+frac12$,
$$
frac{t}{(t+2)^2(t+frac12)^2}=frac{u-tfrac12}{(u+frac32)^2u^2}
=frac49(u-tfrac12)sum_{k=0}^infty (k+1)left(-frac23right)^ku^{k-2}
$$
so that the coefficient for the power $u^{-1}=(t+frac12)^{-2}$ is $frac49(1+frac12cdot2cdotfrac23)=frac{20}{27}$. The integration over the unit circle adds the factor $2pi i$, so that in the final result $I=frac{10pi}{27}$.
answered Nov 21 '18 at 21:23
LutzLLutzL
56.5k42054
You made an error in calculating the factorization of the denominator. You should get $$5t+2t^2+2=(2+t)(1+2t)=2(t+2)(t+1/2),$$ where you missed to transfer the leading factor $2$ to the factorized expression. The correction leads to $I=-frac i4 I_a$ and the additional division by $4$ corrects the residuum that is computed with the correct factor $2pi i$.
You could also use the geometric/binomial series to get the correct residual coefficient. Then, using $u=t+frac12$,
$$
frac{t}{(t+2)^2(t+frac12)^2}=frac{u-tfrac12}{(u+frac32)^2u^2}
=frac49(u-tfrac12)sum_{k=0}^infty (k+1)left(-frac23right)^ku^{k-2}
$$
so that the coefficient for the power $u^{-1}=(t+frac12)^{-2}$ is $frac49(1+frac12cdot2cdotfrac23)=frac{20}{27}$. The integration over the unit circle adds the factor $2pi i$, so that in the final result $I=frac{10pi}{27}$.
answered Nov 21 '18 at 21:23
LutzLLutzL
56.5k42054
edited Nov 21 '18 at 21:29
answered Nov 21 '18 at 21:23
LutzLLutzL
56.5k42054
answered Nov 21 '18 at 21:23
LutzLLutzL
56.5k42054
answered Nov 21 '18 at 21:23
LutzLLutzL
56.5k42054
56.5k42054
+1 for the second method, although I think taking the limit is more straightforward.
– Math_QED
Nov 21 '18 at 21:25
add a comment |
+1 for the second method, although I think taking the limit is more straightforward.
– Math_QED
Nov 21 '18 at 21:25
+1 for the second method, although I think taking the limit is more straightforward.
– Math_QED
Nov 21 '18 at 21:25
+1 for the second method, although I think taking the limit is more straightforward.
– Math_QED
Nov 21 '18 at 21:25
add a comment |
Your mistake is when using the residue theorem:
$$oint_{C^+} f(z) dz = 2pi isum_{singularities} Res(f,singularity)$$
You forgot the factor $2$ in this formula.
Another mistake: $$5t + 2t^2 + 2 = 2(t+2)(t+1/2)$$
You again forgot a factor 2 (which will yield a factor $4$ eventually)
You also made a mistake in calculating the residue.
Note that $(-1/2 + 2)^3 = (3/2)^3 = 27/8$
Correcting these two mistakes, we easily find the correct answer.
answered Nov 21 '18 at 20:12
Math_QEDMath_QED
7,30131449
add a comment |
Your mistake is when using the residue theorem:
$$oint_{C^+} f(z) dz = 2pi isum_{singularities} Res(f,singularity)$$
You forgot the factor $2$ in this formula.
Another mistake: $$5t + 2t^2 + 2 = 2(t+2)(t+1/2)$$
You again forgot a factor 2 (which will yield a factor $4$ eventually)
You also made a mistake in calculating the residue.
Note that $(-1/2 + 2)^3 = (3/2)^3 = 27/8$
Correcting these two mistakes, we easily find the correct answer.
answered Nov 21 '18 at 20:12
Math_QEDMath_QED
7,30131449
add a comment |
Your mistake is when using the residue theorem:
$$oint_{C^+} f(z) dz = 2pi isum_{singularities} Res(f,singularity)$$
You forgot the factor $2$ in this formula.
Another mistake: $$5t + 2t^2 + 2 = 2(t+2)(t+1/2)$$
You again forgot a factor 2 (which will yield a factor $4$ eventually)
You also made a mistake in calculating the residue.
Note that $(-1/2 + 2)^3 = (3/2)^3 = 27/8$
Correcting these two mistakes, we easily find the correct answer.
answered Nov 21 '18 at 20:12
Math_QEDMath_QED
7,30131449
Your mistake is when using the residue theorem:
$$oint_{C^+} f(z) dz = 2pi isum_{singularities} Res(f,singularity)$$
You forgot the factor $2$ in this formula.
Another mistake: $$5t + 2t^2 + 2 = 2(t+2)(t+1/2)$$
You again forgot a factor 2 (which will yield a factor $4$ eventually)
You also made a mistake in calculating the residue.
Note that $(-1/2 + 2)^3 = (3/2)^3 = 27/8$
Correcting these two mistakes, we easily find the correct answer.
answered Nov 21 '18 at 20:12
Math_QEDMath_QED
7,30131449
edited Nov 21 '18 at 21:23
answered Nov 21 '18 at 20:12
Math_QEDMath_QED
7,30131449
answered Nov 21 '18 at 20:12
Math_QEDMath_QED
7,30131449
answered Nov 21 '18 at 20:12
Math_QEDMath_QED
7,30131449
7,30131449
add a comment |
add a comment |
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Residue theorem asks for a factor $2pi i$ instead of $pi i$ no?
– Math_QED
Nov 21 '18 at 20:09
I tought since it's on the real axis, it was $ipi$
– Nicola Fattoruso
Nov 21 '18 at 20:13
Did you mean to say $dx = frac{dt}{it}$?
– Acccumulation
Nov 21 '18 at 20:13
Yes, edited sorry.
– Nicola Fattoruso
Nov 21 '18 at 20:15
No, the residue theorem always give $2 pi i$.
– Math_QED
Nov 21 '18 at 20:15