A simple question in complex analysis
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Why is the contour integral in upper plane different from the lower plane in this case?
$int_{-infty}^{infty} dkfrac{1}{(k-a)(k+a)(p-k-b)(p-k+b)}$
where $text{Im }a$ and $text{Im }b$ are negative and $p$ is real. Besides, $text{Re }a$, $text{Re }b$, and $p$ are positive.
The the poles in complex plane are shown below:
complex-numbers
asked Nov 21 at 17:20
James Liu
11
migrated from physics.stackexchange.com Nov 22 at 16:49
This question came from our site for active researchers, academics and students of physics.
add a comment |
up vote
-1
down vote
favorite
Why is the contour integral in upper plane different from the lower plane in this case?
$int_{-infty}^{infty} dkfrac{1}{(k-a)(k+a)(p-k-b)(p-k+b)}$
where $text{Im }a$ and $text{Im }b$ are negative and $p$ is real. Besides, $text{Re }a$, $text{Re }b$, and $p$ are positive.
The the poles in complex plane are shown below:
complex-numbers
asked Nov 21 at 17:20
James Liu
11
migrated from physics.stackexchange.com Nov 22 at 16:49
This question came from our site for active researchers, academics and students of physics.
2
Did you calculate the residue at each pole?
– G. Smith
Nov 21 at 17:29
3
Would Mathematics be a better home for this question?
– Qmechanic
Nov 21 at 17:36
Yes, I've calculated the residue at each pole. Since I encounter this question in the loop momentum integration, thus I post here.
– James Liu
Nov 21 at 19:11
If I do the contour integral in the upper plane, I will get $-2pi i left(frac{1}{2a[(p-a)^2-b^2]}-frac{1}{2b[(p+b)^2-a^2]}right)$. But if I integrate in the lower plane, I will get $-2pi i left(frac{1}{2a[(p+a)^2-b^2]}-frac{1}{2b[(p-b)^2-a^2]}right)$
– James Liu
Nov 21 at 19:16
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Why is the contour integral in upper plane different from the lower plane in this case?
$int_{-infty}^{infty} dkfrac{1}{(k-a)(k+a)(p-k-b)(p-k+b)}$
where $text{Im }a$ and $text{Im }b$ are negative and $p$ is real. Besides, $text{Re }a$, $text{Re }b$, and $p$ are positive.
The the poles in complex plane are shown below:
complex-numbers
asked Nov 21 at 17:20
James Liu
11
Why is the contour integral in upper plane different from the lower plane in this case?
$int_{-infty}^{infty} dkfrac{1}{(k-a)(k+a)(p-k-b)(p-k+b)}$
where $text{Im }a$ and $text{Im }b$ are negative and $p$ is real. Besides, $text{Re }a$, $text{Re }b$, and $p$ are positive.
The the poles in complex plane are shown below:
complex-numbers
complex-numbers
asked Nov 21 at 17:20
James Liu
11
asked Nov 21 at 17:20
James Liu
11
asked Nov 21 at 17:20
James Liu
11
asked Nov 21 at 17:20
James Liu
11
asked Nov 21 at 17:20
James Liu
11
11
migrated from physics.stackexchange.com Nov 22 at 16:49
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Nov 22 at 16:49
This question came from our site for active researchers, academics and students of physics.
2
Did you calculate the residue at each pole?
– G. Smith
Nov 21 at 17:29
3
Would Mathematics be a better home for this question?
– Qmechanic
Nov 21 at 17:36
Yes, I've calculated the residue at each pole. Since I encounter this question in the loop momentum integration, thus I post here.
– James Liu
Nov 21 at 19:11
If I do the contour integral in the upper plane, I will get $-2pi i left(frac{1}{2a[(p-a)^2-b^2]}-frac{1}{2b[(p+b)^2-a^2]}right)$. But if I integrate in the lower plane, I will get $-2pi i left(frac{1}{2a[(p+a)^2-b^2]}-frac{1}{2b[(p-b)^2-a^2]}right)$
– James Liu
Nov 21 at 19:16
add a comment |
2
Did you calculate the residue at each pole?
– G. Smith
Nov 21 at 17:29
3
Would Mathematics be a better home for this question?
– Qmechanic
Nov 21 at 17:36
Yes, I've calculated the residue at each pole. Since I encounter this question in the loop momentum integration, thus I post here.
– James Liu
Nov 21 at 19:11
If I do the contour integral in the upper plane, I will get $-2pi i left(frac{1}{2a[(p-a)^2-b^2]}-frac{1}{2b[(p+b)^2-a^2]}right)$. But if I integrate in the lower plane, I will get $-2pi i left(frac{1}{2a[(p+a)^2-b^2]}-frac{1}{2b[(p-b)^2-a^2]}right)$
– James Liu
Nov 21 at 19:16
2
2
Did you calculate the residue at each pole?
– G. Smith
Nov 21 at 17:29
Did you calculate the residue at each pole?
– G. Smith
Nov 21 at 17:29
3
3
Would Mathematics be a better home for this question?
– Qmechanic
Nov 21 at 17:36
Would Mathematics be a better home for this question?
– Qmechanic
Nov 21 at 17:36
Yes, I've calculated the residue at each pole. Since I encounter this question in the loop momentum integration, thus I post here.
– James Liu
Nov 21 at 19:11
Yes, I've calculated the residue at each pole. Since I encounter this question in the loop momentum integration, thus I post here.
– James Liu
Nov 21 at 19:11
If I do the contour integral in the upper plane, I will get $-2pi i left(frac{1}{2a[(p-a)^2-b^2]}-frac{1}{2b[(p+b)^2-a^2]}right)$. But if I integrate in the lower plane, I will get $-2pi i left(frac{1}{2a[(p+a)^2-b^2]}-frac{1}{2b[(p-b)^2-a^2]}right)$
– James Liu
Nov 21 at 19:16
If I do the contour integral in the upper plane, I will get $-2pi i left(frac{1}{2a[(p-a)^2-b^2]}-frac{1}{2b[(p+b)^2-a^2]}right)$. But if I integrate in the lower plane, I will get $-2pi i left(frac{1}{2a[(p+a)^2-b^2]}-frac{1}{2b[(p-b)^2-a^2]}right)$
– James Liu
Nov 21 at 19:16
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The integrand vanishes sufficiently fast at large $|k|$ that you can close the contour in either the upper or the lower half-plane. You will get the same answer in either case. In other words, the sum of the residues at the four poles will be zero. Why don't you check that this is indeed true?
Note added: For a general case
$$
frac{1}{(x-a)(x-b)(x-c)(x-d)}
$$
the residue sum is
$$
frac 1{(a - b) (a - c) (a - d)} + frac{1}{(b - a) (b - c) (b - d)} +
frac{1}{(c - a) (c - b) (c - d)} + frac 1{(d - a) (d - b) (d - c)}
$$
A somewhat tedious calculation shows that this sum is zero. A quicker way to see that this must be true is to consider the partial fraction decomposition
$$
frac{1}{(x-a)(x-b)(x-c)(x-d)}=frac{A}{x-a}+ frac{B}{x-b}+frac{C}{x-c}+frac{D}{x-d}
$$
and take the large $x$ limit. Then the LHS is $1/x^4+ O(x^{-5})$ while
$$
RHS= frac{A+B+C+D} x+ O(x^{-2})
$$
Agreement requires $A+B+C+D=0$ togther with similar, but non-trivial-looking, quadratic and cubic relations on the $A,B,C,D$.
answered Nov 21 at 19:17
mike stone
29817
the four residues are $frac{1}{2a[(p-a)^2-b^2]}$, $-frac{1}{2b[(p+b)^2-a^2]}$, $-frac{1}{2a[(p+a)^2-b^2]}$, and $frac{1}{2b[(p-b)^2-a^2]}$. Is it correct?
– James Liu
Nov 21 at 19:26
I'm afraid this is not correct. I get these residues by just putting a into the other three denominators (k+a), (p-k-b), (p-k+b). and so do the other three poles p+b, -a, p-b (since each pole is a simple pole). What's the problem?
– James Liu
Nov 21 at 19:35
Thanks for everyone's answer. I've found that the residues exactly cancel each others.
– James Liu
Nov 22 at 8:35
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The integrand vanishes sufficiently fast at large $|k|$ that you can close the contour in either the upper or the lower half-plane. You will get the same answer in either case. In other words, the sum of the residues at the four poles will be zero. Why don't you check that this is indeed true?
Note added: For a general case
$$
frac{1}{(x-a)(x-b)(x-c)(x-d)}
$$
the residue sum is
$$
frac 1{(a - b) (a - c) (a - d)} + frac{1}{(b - a) (b - c) (b - d)} +
frac{1}{(c - a) (c - b) (c - d)} + frac 1{(d - a) (d - b) (d - c)}
$$
A somewhat tedious calculation shows that this sum is zero. A quicker way to see that this must be true is to consider the partial fraction decomposition
$$
frac{1}{(x-a)(x-b)(x-c)(x-d)}=frac{A}{x-a}+ frac{B}{x-b}+frac{C}{x-c}+frac{D}{x-d}
$$
and take the large $x$ limit. Then the LHS is $1/x^4+ O(x^{-5})$ while
$$
RHS= frac{A+B+C+D} x+ O(x^{-2})
$$
Agreement requires $A+B+C+D=0$ togther with similar, but non-trivial-looking, quadratic and cubic relations on the $A,B,C,D$.
answered Nov 21 at 19:17
mike stone
29817
the four residues are $frac{1}{2a[(p-a)^2-b^2]}$, $-frac{1}{2b[(p+b)^2-a^2]}$, $-frac{1}{2a[(p+a)^2-b^2]}$, and $frac{1}{2b[(p-b)^2-a^2]}$. Is it correct?
– James Liu
Nov 21 at 19:26
I'm afraid this is not correct. I get these residues by just putting a into the other three denominators (k+a), (p-k-b), (p-k+b). and so do the other three poles p+b, -a, p-b (since each pole is a simple pole). What's the problem?
– James Liu
Nov 21 at 19:35
Thanks for everyone's answer. I've found that the residues exactly cancel each others.
– James Liu
Nov 22 at 8:35
add a comment |
up vote
1
down vote
accepted
The integrand vanishes sufficiently fast at large $|k|$ that you can close the contour in either the upper or the lower half-plane. You will get the same answer in either case. In other words, the sum of the residues at the four poles will be zero. Why don't you check that this is indeed true?
Note added: For a general case
$$
frac{1}{(x-a)(x-b)(x-c)(x-d)}
$$
the residue sum is
$$
frac 1{(a - b) (a - c) (a - d)} + frac{1}{(b - a) (b - c) (b - d)} +
frac{1}{(c - a) (c - b) (c - d)} + frac 1{(d - a) (d - b) (d - c)}
$$
A somewhat tedious calculation shows that this sum is zero. A quicker way to see that this must be true is to consider the partial fraction decomposition
$$
frac{1}{(x-a)(x-b)(x-c)(x-d)}=frac{A}{x-a}+ frac{B}{x-b}+frac{C}{x-c}+frac{D}{x-d}
$$
and take the large $x$ limit. Then the LHS is $1/x^4+ O(x^{-5})$ while
$$
RHS= frac{A+B+C+D} x+ O(x^{-2})
$$
Agreement requires $A+B+C+D=0$ togther with similar, but non-trivial-looking, quadratic and cubic relations on the $A,B,C,D$.
answered Nov 21 at 19:17
mike stone
29817
the four residues are $frac{1}{2a[(p-a)^2-b^2]}$, $-frac{1}{2b[(p+b)^2-a^2]}$, $-frac{1}{2a[(p+a)^2-b^2]}$, and $frac{1}{2b[(p-b)^2-a^2]}$. Is it correct?
– James Liu
Nov 21 at 19:26
I'm afraid this is not correct. I get these residues by just putting a into the other three denominators (k+a), (p-k-b), (p-k+b). and so do the other three poles p+b, -a, p-b (since each pole is a simple pole). What's the problem?
– James Liu
Nov 21 at 19:35
Thanks for everyone's answer. I've found that the residues exactly cancel each others.
– James Liu
Nov 22 at 8:35
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The integrand vanishes sufficiently fast at large $|k|$ that you can close the contour in either the upper or the lower half-plane. You will get the same answer in either case. In other words, the sum of the residues at the four poles will be zero. Why don't you check that this is indeed true?
Note added: For a general case
$$
frac{1}{(x-a)(x-b)(x-c)(x-d)}
$$
the residue sum is
$$
frac 1{(a - b) (a - c) (a - d)} + frac{1}{(b - a) (b - c) (b - d)} +
frac{1}{(c - a) (c - b) (c - d)} + frac 1{(d - a) (d - b) (d - c)}
$$
A somewhat tedious calculation shows that this sum is zero. A quicker way to see that this must be true is to consider the partial fraction decomposition
$$
frac{1}{(x-a)(x-b)(x-c)(x-d)}=frac{A}{x-a}+ frac{B}{x-b}+frac{C}{x-c}+frac{D}{x-d}
$$
and take the large $x$ limit. Then the LHS is $1/x^4+ O(x^{-5})$ while
$$
RHS= frac{A+B+C+D} x+ O(x^{-2})
$$
Agreement requires $A+B+C+D=0$ togther with similar, but non-trivial-looking, quadratic and cubic relations on the $A,B,C,D$.
answered Nov 21 at 19:17
mike stone
29817
The integrand vanishes sufficiently fast at large $|k|$ that you can close the contour in either the upper or the lower half-plane. You will get the same answer in either case. In other words, the sum of the residues at the four poles will be zero. Why don't you check that this is indeed true?
Note added: For a general case
$$
frac{1}{(x-a)(x-b)(x-c)(x-d)}
$$
the residue sum is
$$
frac 1{(a - b) (a - c) (a - d)} + frac{1}{(b - a) (b - c) (b - d)} +
frac{1}{(c - a) (c - b) (c - d)} + frac 1{(d - a) (d - b) (d - c)}
$$
A somewhat tedious calculation shows that this sum is zero. A quicker way to see that this must be true is to consider the partial fraction decomposition
$$
frac{1}{(x-a)(x-b)(x-c)(x-d)}=frac{A}{x-a}+ frac{B}{x-b}+frac{C}{x-c}+frac{D}{x-d}
$$
and take the large $x$ limit. Then the LHS is $1/x^4+ O(x^{-5})$ while
$$
RHS= frac{A+B+C+D} x+ O(x^{-2})
$$
Agreement requires $A+B+C+D=0$ togther with similar, but non-trivial-looking, quadratic and cubic relations on the $A,B,C,D$.
answered Nov 21 at 19:17
mike stone
29817
answered Nov 21 at 19:17
mike stone
29817
answered Nov 21 at 19:17
mike stone
29817
answered Nov 21 at 19:17
mike stone
29817
29817
the four residues are $frac{1}{2a[(p-a)^2-b^2]}$, $-frac{1}{2b[(p+b)^2-a^2]}$, $-frac{1}{2a[(p+a)^2-b^2]}$, and $frac{1}{2b[(p-b)^2-a^2]}$. Is it correct?
– James Liu
Nov 21 at 19:26
I'm afraid this is not correct. I get these residues by just putting a into the other three denominators (k+a), (p-k-b), (p-k+b). and so do the other three poles p+b, -a, p-b (since each pole is a simple pole). What's the problem?
– James Liu
Nov 21 at 19:35
Thanks for everyone's answer. I've found that the residues exactly cancel each others.
– James Liu
Nov 22 at 8:35
add a comment |
the four residues are $frac{1}{2a[(p-a)^2-b^2]}$, $-frac{1}{2b[(p+b)^2-a^2]}$, $-frac{1}{2a[(p+a)^2-b^2]}$, and $frac{1}{2b[(p-b)^2-a^2]}$. Is it correct?
– James Liu
Nov 21 at 19:26
I'm afraid this is not correct. I get these residues by just putting a into the other three denominators (k+a), (p-k-b), (p-k+b). and so do the other three poles p+b, -a, p-b (since each pole is a simple pole). What's the problem?
– James Liu
Nov 21 at 19:35
Thanks for everyone's answer. I've found that the residues exactly cancel each others.
– James Liu
Nov 22 at 8:35
the four residues are $frac{1}{2a[(p-a)^2-b^2]}$, $-frac{1}{2b[(p+b)^2-a^2]}$, $-frac{1}{2a[(p+a)^2-b^2]}$, and $frac{1}{2b[(p-b)^2-a^2]}$. Is it correct?
– James Liu
Nov 21 at 19:26
the four residues are $frac{1}{2a[(p-a)^2-b^2]}$, $-frac{1}{2b[(p+b)^2-a^2]}$, $-frac{1}{2a[(p+a)^2-b^2]}$, and $frac{1}{2b[(p-b)^2-a^2]}$. Is it correct?
– James Liu
Nov 21 at 19:26
I'm afraid this is not correct. I get these residues by just putting a into the other three denominators (k+a), (p-k-b), (p-k+b). and so do the other three poles p+b, -a, p-b (since each pole is a simple pole). What's the problem?
– James Liu
Nov 21 at 19:35
I'm afraid this is not correct. I get these residues by just putting a into the other three denominators (k+a), (p-k-b), (p-k+b). and so do the other three poles p+b, -a, p-b (since each pole is a simple pole). What's the problem?
– James Liu
Nov 21 at 19:35
Thanks for everyone's answer. I've found that the residues exactly cancel each others.
– James Liu
Nov 22 at 8:35
Thanks for everyone's answer. I've found that the residues exactly cancel each others.
– James Liu
Nov 22 at 8:35
add a comment |
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2
Did you calculate the residue at each pole?
– G. Smith
Nov 21 at 17:29
3
Would Mathematics be a better home for this question?
– Qmechanic
Nov 21 at 17:36
Yes, I've calculated the residue at each pole. Since I encounter this question in the loop momentum integration, thus I post here.
– James Liu
Nov 21 at 19:11
If I do the contour integral in the upper plane, I will get $-2pi i left(frac{1}{2a[(p-a)^2-b^2]}-frac{1}{2b[(p+b)^2-a^2]}right)$. But if I integrate in the lower plane, I will get $-2pi i left(frac{1}{2a[(p+a)^2-b^2]}-frac{1}{2b[(p-b)^2-a^2]}right)$
– James Liu
Nov 21 at 19:16