Compute complex integral inside an open curve
$begingroup$
I need to compute this complex integral:
$$ int_gamma frac{1}{(z-i)(z-2i)} dz $$
$gamma$ is defined as:
$$ gamma (t) = t + i(3e^tcos^2(t)) $$
The parameter t belongs to the following interval: $[frac{-pi}{2},frac{pi}{2}] $
My approach to this question was to use the Residue's Theorem. However, the theorem requires a closed curve.
The problem above does not portray a closed curve. It is an open one.
I know that open curves can be manipulated as closed ones. You treat it artificially as a closed one and, by the end, you take out (or add) something.
How can I approach this integral?
The partial result of the residues is:
$Res(f,i)= i$ and $Res(f,2i)=-i$
Hence:
$2pi i(+i-i)=0 $
P.S.: I also tried a different approach with the formula:
$$ int f(gamma(t))gamma'(t)dt $$
Nonetheless, this approach results in a massive and ugly integral. I think the correct approach is the manipulation described above.
UPDATE
Professor Robert Israel commented bellow and suggested using a third approach with $F(gamma(pi/2)) - F(gamma(-pi/2))$. In order to do that, I needed to find the primitives. I got them with a little help from Wolfram Alpha. As you can see bellow, the primitive is not really friendly. I am still curious on how to solve this problem treating the curve as a closed one and manipulating it:
integration complex-analysis residue-calculus complex-integration
asked Dec 11 '18 at 20:33
Pedro DelfinoPedro Delfino
857
$endgroup$
add a comment |
$begingroup$
I need to compute this complex integral:
$$ int_gamma frac{1}{(z-i)(z-2i)} dz $$
$gamma$ is defined as:
$$ gamma (t) = t + i(3e^tcos^2(t)) $$
The parameter t belongs to the following interval: $[frac{-pi}{2},frac{pi}{2}] $
My approach to this question was to use the Residue's Theorem. However, the theorem requires a closed curve.
The problem above does not portray a closed curve. It is an open one.
I know that open curves can be manipulated as closed ones. You treat it artificially as a closed one and, by the end, you take out (or add) something.
How can I approach this integral?
The partial result of the residues is:
$Res(f,i)= i$ and $Res(f,2i)=-i$
Hence:
$2pi i(+i-i)=0 $
P.S.: I also tried a different approach with the formula:
$$ int f(gamma(t))gamma'(t)dt $$
Nonetheless, this approach results in a massive and ugly integral. I think the correct approach is the manipulation described above.
UPDATE
Professor Robert Israel commented bellow and suggested using a third approach with $F(gamma(pi/2)) - F(gamma(-pi/2))$. In order to do that, I needed to find the primitives. I got them with a little help from Wolfram Alpha. As you can see bellow, the primitive is not really friendly. I am still curious on how to solve this problem treating the curve as a closed one and manipulating it:
integration complex-analysis residue-calculus complex-integration
asked Dec 11 '18 at 20:33
Pedro DelfinoPedro Delfino
857
$endgroup$
1
$begingroup$
The point is that to do it with primitives, which is really the easiest way, you need to split the integral up as in my answer, and then define an appropriate branch of the complex logarithm, as Dr. Israel says. Then, the result falls out. Wolfram is giving you an antiderivative using the Log function, which is not analytic in any open set containing $gamma$, so it won't work.
$endgroup$
– Matematleta
Dec 11 '18 at 23:30
1
$begingroup$
I don't know why Wolfram alpha's giving you such a complicated-looking primitive, but it really isn't. Use partial fraction decomposition.
$endgroup$
– Dylan
Dec 12 '18 at 6:51
add a comment |
$begingroup$
I need to compute this complex integral:
$$ int_gamma frac{1}{(z-i)(z-2i)} dz $$
$gamma$ is defined as:
$$ gamma (t) = t + i(3e^tcos^2(t)) $$
The parameter t belongs to the following interval: $[frac{-pi}{2},frac{pi}{2}] $
My approach to this question was to use the Residue's Theorem. However, the theorem requires a closed curve.
The problem above does not portray a closed curve. It is an open one.
I know that open curves can be manipulated as closed ones. You treat it artificially as a closed one and, by the end, you take out (or add) something.
How can I approach this integral?
The partial result of the residues is:
$Res(f,i)= i$ and $Res(f,2i)=-i$
Hence:
$2pi i(+i-i)=0 $
P.S.: I also tried a different approach with the formula:
$$ int f(gamma(t))gamma'(t)dt $$
Nonetheless, this approach results in a massive and ugly integral. I think the correct approach is the manipulation described above.
UPDATE
Professor Robert Israel commented bellow and suggested using a third approach with $F(gamma(pi/2)) - F(gamma(-pi/2))$. In order to do that, I needed to find the primitives. I got them with a little help from Wolfram Alpha. As you can see bellow, the primitive is not really friendly. I am still curious on how to solve this problem treating the curve as a closed one and manipulating it:
integration complex-analysis residue-calculus complex-integration
asked Dec 11 '18 at 20:33
Pedro DelfinoPedro Delfino
857
$endgroup$
I need to compute this complex integral:
$$ int_gamma frac{1}{(z-i)(z-2i)} dz $$
$gamma$ is defined as:
$$ gamma (t) = t + i(3e^tcos^2(t)) $$
The parameter t belongs to the following interval: $[frac{-pi}{2},frac{pi}{2}] $
My approach to this question was to use the Residue's Theorem. However, the theorem requires a closed curve.
The problem above does not portray a closed curve. It is an open one.
I know that open curves can be manipulated as closed ones. You treat it artificially as a closed one and, by the end, you take out (or add) something.
How can I approach this integral?
The partial result of the residues is:
$Res(f,i)= i$ and $Res(f,2i)=-i$
Hence:
$2pi i(+i-i)=0 $
P.S.: I also tried a different approach with the formula:
$$ int f(gamma(t))gamma'(t)dt $$
Nonetheless, this approach results in a massive and ugly integral. I think the correct approach is the manipulation described above.
UPDATE
Professor Robert Israel commented bellow and suggested using a third approach with $F(gamma(pi/2)) - F(gamma(-pi/2))$. In order to do that, I needed to find the primitives. I got them with a little help from Wolfram Alpha. As you can see bellow, the primitive is not really friendly. I am still curious on how to solve this problem treating the curve as a closed one and manipulating it:
integration complex-analysis residue-calculus complex-integration
integration complex-analysis residue-calculus complex-integration
asked Dec 11 '18 at 20:33
Pedro DelfinoPedro Delfino
857
asked Dec 11 '18 at 20:33
Pedro DelfinoPedro Delfino
857
edited Dec 12 '18 at 12:47
Pedro Delfino
asked Dec 11 '18 at 20:33
Pedro DelfinoPedro Delfino
857
asked Dec 11 '18 at 20:33
Pedro DelfinoPedro Delfino
857
asked Dec 11 '18 at 20:33
Pedro DelfinoPedro Delfino
857
857
1
$begingroup$
The point is that to do it with primitives, which is really the easiest way, you need to split the integral up as in my answer, and then define an appropriate branch of the complex logarithm, as Dr. Israel says. Then, the result falls out. Wolfram is giving you an antiderivative using the Log function, which is not analytic in any open set containing $gamma$, so it won't work.
$endgroup$
– Matematleta
Dec 11 '18 at 23:30
1
$begingroup$
I don't know why Wolfram alpha's giving you such a complicated-looking primitive, but it really isn't. Use partial fraction decomposition.
$endgroup$
– Dylan
Dec 12 '18 at 6:51
add a comment |
1
$begingroup$
The point is that to do it with primitives, which is really the easiest way, you need to split the integral up as in my answer, and then define an appropriate branch of the complex logarithm, as Dr. Israel says. Then, the result falls out. Wolfram is giving you an antiderivative using the Log function, which is not analytic in any open set containing $gamma$, so it won't work.
$endgroup$
– Matematleta
Dec 11 '18 at 23:30
1
$begingroup$
I don't know why Wolfram alpha's giving you such a complicated-looking primitive, but it really isn't. Use partial fraction decomposition.
$endgroup$
– Dylan
Dec 12 '18 at 6:51
1
1
$begingroup$
The point is that to do it with primitives, which is really the easiest way, you need to split the integral up as in my answer, and then define an appropriate branch of the complex logarithm, as Dr. Israel says. Then, the result falls out. Wolfram is giving you an antiderivative using the Log function, which is not analytic in any open set containing $gamma$, so it won't work.
$endgroup$
– Matematleta
Dec 11 '18 at 23:30
$begingroup$
The point is that to do it with primitives, which is really the easiest way, you need to split the integral up as in my answer, and then define an appropriate branch of the complex logarithm, as Dr. Israel says. Then, the result falls out. Wolfram is giving you an antiderivative using the Log function, which is not analytic in any open set containing $gamma$, so it won't work.
$endgroup$
– Matematleta
Dec 11 '18 at 23:30
1
1
$begingroup$
I don't know why Wolfram alpha's giving you such a complicated-looking primitive, but it really isn't. Use partial fraction decomposition.
$endgroup$
– Dylan
Dec 12 '18 at 6:51
$begingroup$
I don't know why Wolfram alpha's giving you such a complicated-looking primitive, but it really isn't. Use partial fraction decomposition.
$endgroup$
– Dylan
Dec 12 '18 at 6:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here is a brute force approach. First, decompose the integrand:
$int_gamma frac{1}{(z-i)(z-2i)} dz=iint_gamma frac{1}{z-i} dz-iint_gamma frac{1}{z-2i} dz.$
Now, treat each one separately:
For the first integral, complete $gamma $ to a closed curve $Gamma$ by taking $gamma_1$ to be the circular arc centered at $z=i$ of radius $|i-pi/2|$ going from $x=-pi/2$ to $x=pi/2$ and find that the angle $theta $ subtended by this arc is given by $cos frac{1}{2}theta=frac{1}{|i-pi/2|}.$ Then, $iint_{gamma_1} frac{1}{z-i} dz=-theta.$
Now, the argument principle applies to show that $iint_{Gamma} frac{1}{z-i} dz=i(2pi i)(1)=-2pi.$
Therefore, $iint_{Gamma} frac{1}{z-i} dz=iint_{gamma_1} frac{1}{z-i} dz-iint_{gamma} frac{1}{z-i} dzRightarrow int_{gamma} frac{1}{z-i} dz=itheta-2pi i$.
The second integral is treated in the same way, this time with a circle centered at $2i$.
answered Dec 11 '18 at 21:10
MatematletaMatematleta
12k21020
$endgroup$
add a comment |
$begingroup$
Hint: find an antiderivative $F$ of the integrand that is analytic in a region containing the curve $gamma$. For this it suffices that any branch cuts are on the imaginary axis below $2i$.
Then the integral is $F(gamma(pi/2)) - F(gamma(-pi/2))$.
answered Dec 11 '18 at 20:39
Robert IsraelRobert Israel
330k23218473
$endgroup$
$begingroup$
Thanks, Professor Israel. I had a go on your suggestion with a little help from Wolfram Alpha. The primitive I got is not really nice: wolframalpha.com/input/?i=integral+of+1%2F((z-i)(z-2i))+dz
$endgroup$
– Pedro Delfino
Dec 11 '18 at 22:35
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a brute force approach. First, decompose the integrand:
$int_gamma frac{1}{(z-i)(z-2i)} dz=iint_gamma frac{1}{z-i} dz-iint_gamma frac{1}{z-2i} dz.$
Now, treat each one separately:
For the first integral, complete $gamma $ to a closed curve $Gamma$ by taking $gamma_1$ to be the circular arc centered at $z=i$ of radius $|i-pi/2|$ going from $x=-pi/2$ to $x=pi/2$ and find that the angle $theta $ subtended by this arc is given by $cos frac{1}{2}theta=frac{1}{|i-pi/2|}.$ Then, $iint_{gamma_1} frac{1}{z-i} dz=-theta.$
Now, the argument principle applies to show that $iint_{Gamma} frac{1}{z-i} dz=i(2pi i)(1)=-2pi.$
Therefore, $iint_{Gamma} frac{1}{z-i} dz=iint_{gamma_1} frac{1}{z-i} dz-iint_{gamma} frac{1}{z-i} dzRightarrow int_{gamma} frac{1}{z-i} dz=itheta-2pi i$.
The second integral is treated in the same way, this time with a circle centered at $2i$.
answered Dec 11 '18 at 21:10
MatematletaMatematleta
12k21020
$endgroup$
add a comment |
$begingroup$
Here is a brute force approach. First, decompose the integrand:
$int_gamma frac{1}{(z-i)(z-2i)} dz=iint_gamma frac{1}{z-i} dz-iint_gamma frac{1}{z-2i} dz.$
Now, treat each one separately:
For the first integral, complete $gamma $ to a closed curve $Gamma$ by taking $gamma_1$ to be the circular arc centered at $z=i$ of radius $|i-pi/2|$ going from $x=-pi/2$ to $x=pi/2$ and find that the angle $theta $ subtended by this arc is given by $cos frac{1}{2}theta=frac{1}{|i-pi/2|}.$ Then, $iint_{gamma_1} frac{1}{z-i} dz=-theta.$
Now, the argument principle applies to show that $iint_{Gamma} frac{1}{z-i} dz=i(2pi i)(1)=-2pi.$
Therefore, $iint_{Gamma} frac{1}{z-i} dz=iint_{gamma_1} frac{1}{z-i} dz-iint_{gamma} frac{1}{z-i} dzRightarrow int_{gamma} frac{1}{z-i} dz=itheta-2pi i$.
The second integral is treated in the same way, this time with a circle centered at $2i$.
answered Dec 11 '18 at 21:10
MatematletaMatematleta
12k21020
$endgroup$
add a comment |
$begingroup$
Here is a brute force approach. First, decompose the integrand:
$int_gamma frac{1}{(z-i)(z-2i)} dz=iint_gamma frac{1}{z-i} dz-iint_gamma frac{1}{z-2i} dz.$
Now, treat each one separately:
For the first integral, complete $gamma $ to a closed curve $Gamma$ by taking $gamma_1$ to be the circular arc centered at $z=i$ of radius $|i-pi/2|$ going from $x=-pi/2$ to $x=pi/2$ and find that the angle $theta $ subtended by this arc is given by $cos frac{1}{2}theta=frac{1}{|i-pi/2|}.$ Then, $iint_{gamma_1} frac{1}{z-i} dz=-theta.$
Now, the argument principle applies to show that $iint_{Gamma} frac{1}{z-i} dz=i(2pi i)(1)=-2pi.$
Therefore, $iint_{Gamma} frac{1}{z-i} dz=iint_{gamma_1} frac{1}{z-i} dz-iint_{gamma} frac{1}{z-i} dzRightarrow int_{gamma} frac{1}{z-i} dz=itheta-2pi i$.
The second integral is treated in the same way, this time with a circle centered at $2i$.
answered Dec 11 '18 at 21:10
MatematletaMatematleta
12k21020
$endgroup$
Here is a brute force approach. First, decompose the integrand:
$int_gamma frac{1}{(z-i)(z-2i)} dz=iint_gamma frac{1}{z-i} dz-iint_gamma frac{1}{z-2i} dz.$
Now, treat each one separately:
For the first integral, complete $gamma $ to a closed curve $Gamma$ by taking $gamma_1$ to be the circular arc centered at $z=i$ of radius $|i-pi/2|$ going from $x=-pi/2$ to $x=pi/2$ and find that the angle $theta $ subtended by this arc is given by $cos frac{1}{2}theta=frac{1}{|i-pi/2|}.$ Then, $iint_{gamma_1} frac{1}{z-i} dz=-theta.$
Now, the argument principle applies to show that $iint_{Gamma} frac{1}{z-i} dz=i(2pi i)(1)=-2pi.$
Therefore, $iint_{Gamma} frac{1}{z-i} dz=iint_{gamma_1} frac{1}{z-i} dz-iint_{gamma} frac{1}{z-i} dzRightarrow int_{gamma} frac{1}{z-i} dz=itheta-2pi i$.
The second integral is treated in the same way, this time with a circle centered at $2i$.
answered Dec 11 '18 at 21:10
MatematletaMatematleta
12k21020
edited Dec 11 '18 at 21:57
answered Dec 11 '18 at 21:10
MatematletaMatematleta
12k21020
answered Dec 11 '18 at 21:10
MatematletaMatematleta
12k21020
answered Dec 11 '18 at 21:10
MatematletaMatematleta
12k21020
12k21020
add a comment |
add a comment |
$begingroup$
Hint: find an antiderivative $F$ of the integrand that is analytic in a region containing the curve $gamma$. For this it suffices that any branch cuts are on the imaginary axis below $2i$.
Then the integral is $F(gamma(pi/2)) - F(gamma(-pi/2))$.
answered Dec 11 '18 at 20:39
Robert IsraelRobert Israel
330k23218473
$endgroup$
$begingroup$
Thanks, Professor Israel. I had a go on your suggestion with a little help from Wolfram Alpha. The primitive I got is not really nice: wolframalpha.com/input/?i=integral+of+1%2F((z-i)(z-2i))+dz
$endgroup$
– Pedro Delfino
Dec 11 '18 at 22:35
add a comment |
$begingroup$
Hint: find an antiderivative $F$ of the integrand that is analytic in a region containing the curve $gamma$. For this it suffices that any branch cuts are on the imaginary axis below $2i$.
Then the integral is $F(gamma(pi/2)) - F(gamma(-pi/2))$.
answered Dec 11 '18 at 20:39
Robert IsraelRobert Israel
330k23218473
$endgroup$
$begingroup$
Thanks, Professor Israel. I had a go on your suggestion with a little help from Wolfram Alpha. The primitive I got is not really nice: wolframalpha.com/input/?i=integral+of+1%2F((z-i)(z-2i))+dz
$endgroup$
– Pedro Delfino
Dec 11 '18 at 22:35
add a comment |
$begingroup$
Hint: find an antiderivative $F$ of the integrand that is analytic in a region containing the curve $gamma$. For this it suffices that any branch cuts are on the imaginary axis below $2i$.
Then the integral is $F(gamma(pi/2)) - F(gamma(-pi/2))$.
answered Dec 11 '18 at 20:39
Robert IsraelRobert Israel
330k23218473
$endgroup$
Hint: find an antiderivative $F$ of the integrand that is analytic in a region containing the curve $gamma$. For this it suffices that any branch cuts are on the imaginary axis below $2i$.
Then the integral is $F(gamma(pi/2)) - F(gamma(-pi/2))$.
answered Dec 11 '18 at 20:39
Robert IsraelRobert Israel
330k23218473
answered Dec 11 '18 at 20:39
Robert IsraelRobert Israel
330k23218473
answered Dec 11 '18 at 20:39
Robert IsraelRobert Israel
330k23218473
answered Dec 11 '18 at 20:39
Robert IsraelRobert Israel
330k23218473
330k23218473
$begingroup$
Thanks, Professor Israel. I had a go on your suggestion with a little help from Wolfram Alpha. The primitive I got is not really nice: wolframalpha.com/input/?i=integral+of+1%2F((z-i)(z-2i))+dz
$endgroup$
– Pedro Delfino
Dec 11 '18 at 22:35
add a comment |
$begingroup$
Thanks, Professor Israel. I had a go on your suggestion with a little help from Wolfram Alpha. The primitive I got is not really nice: wolframalpha.com/input/?i=integral+of+1%2F((z-i)(z-2i))+dz
$endgroup$
– Pedro Delfino
Dec 11 '18 at 22:35
$begingroup$
Thanks, Professor Israel. I had a go on your suggestion with a little help from Wolfram Alpha. The primitive I got is not really nice: wolframalpha.com/input/?i=integral+of+1%2F((z-i)(z-2i))+dz
$endgroup$
– Pedro Delfino
Dec 11 '18 at 22:35
$begingroup$
Thanks, Professor Israel. I had a go on your suggestion with a little help from Wolfram Alpha. The primitive I got is not really nice: wolframalpha.com/input/?i=integral+of+1%2F((z-i)(z-2i))+dz
$endgroup$
– Pedro Delfino
Dec 11 '18 at 22:35
add a comment |
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$begingroup$
The point is that to do it with primitives, which is really the easiest way, you need to split the integral up as in my answer, and then define an appropriate branch of the complex logarithm, as Dr. Israel says. Then, the result falls out. Wolfram is giving you an antiderivative using the Log function, which is not analytic in any open set containing $gamma$, so it won't work.
$endgroup$
– Matematleta
Dec 11 '18 at 23:30
1
$begingroup$
I don't know why Wolfram alpha's giving you such a complicated-looking primitive, but it really isn't. Use partial fraction decomposition.
$endgroup$
– Dylan
Dec 12 '18 at 6:51