Why is the contour integral in upper plane different from the lower plane in this case?
Why is the contour integral in upper plane different from the lower plane in this case?
$int_{-infty}^{infty}mathrm{d}kfrac{1}{(k+a)(k-a)(p-k-b)(p-k+b)}$
where Im[a] and Im[b] are negative and p is real. Besides, Re[a] and Re[b], and p are positive.
The the poles in complex plane are shown below:
enter image description here
$frac{1}{2a[(p-a)^2-b^2]},frac{1}{2b[(p-b)^2-a^2]},-frac{1}{2a[(p+a)^2-b^2]},-frac{1}{2a[(p+b)^2-a^2]}$
I get these residues by just putting $a$ into the other three denominators $(k+a)$, $(p-k-b)$, $(p-k+b)$. and I also use the same procedure to the other three poles $p+b, -a, p-b$ (since each pole is a simple pole). I think the summation of these four poles should give zero. What's the problem?
complex-analysis contour-integration
asked Nov 22 at 6:21
James Liu
11
|
show 2 more comments
Why is the contour integral in upper plane different from the lower plane in this case?
$int_{-infty}^{infty}mathrm{d}kfrac{1}{(k+a)(k-a)(p-k-b)(p-k+b)}$
where Im[a] and Im[b] are negative and p is real. Besides, Re[a] and Re[b], and p are positive.
The the poles in complex plane are shown below:
enter image description here
$frac{1}{2a[(p-a)^2-b^2]},frac{1}{2b[(p-b)^2-a^2]},-frac{1}{2a[(p+a)^2-b^2]},-frac{1}{2a[(p+b)^2-a^2]}$
I get these residues by just putting $a$ into the other three denominators $(k+a)$, $(p-k-b)$, $(p-k+b)$. and I also use the same procedure to the other three poles $p+b, -a, p-b$ (since each pole is a simple pole). I think the summation of these four poles should give zero. What's the problem?
complex-analysis contour-integration
asked Nov 22 at 6:21
James Liu
11
What are the residues that you obtain. The sum of the residues has to be zero in this case. Otherwise, you have a mistake in the calculation.
– Fabian
Nov 22 at 7:06
1
Oh, I see the residues. I believe the second and fourth residue should be minus from what you write.
– Fabian
Nov 22 at 7:09
I get one of the residue $Res_{k=p+b}=frac{1}{(p+b+a)(p+b-a)(-2b)}$ by put p+b into k+a, k-a, p-k-b
– James Liu
Nov 22 at 7:24
We can also get the other answers by the same procedure.
– James Liu
Nov 22 at 7:31
@JamesLiu That is not how you compute a residue. In order for this heuristic to work, you need to change the initial expression it so all signs of $k$ are positive. (i.e. have the last two factors in the denominator be $(k-(p-b))(k-(p+b)),$ otherwise you will be off by a minus sign for two of them.
– spaceisdarkgreen
Nov 22 at 7:46
|
show 2 more comments
Why is the contour integral in upper plane different from the lower plane in this case?
$int_{-infty}^{infty}mathrm{d}kfrac{1}{(k+a)(k-a)(p-k-b)(p-k+b)}$
where Im[a] and Im[b] are negative and p is real. Besides, Re[a] and Re[b], and p are positive.
The the poles in complex plane are shown below:
enter image description here
$frac{1}{2a[(p-a)^2-b^2]},frac{1}{2b[(p-b)^2-a^2]},-frac{1}{2a[(p+a)^2-b^2]},-frac{1}{2a[(p+b)^2-a^2]}$
I get these residues by just putting $a$ into the other three denominators $(k+a)$, $(p-k-b)$, $(p-k+b)$. and I also use the same procedure to the other three poles $p+b, -a, p-b$ (since each pole is a simple pole). I think the summation of these four poles should give zero. What's the problem?
complex-analysis contour-integration
asked Nov 22 at 6:21
James Liu
11
Why is the contour integral in upper plane different from the lower plane in this case?
$int_{-infty}^{infty}mathrm{d}kfrac{1}{(k+a)(k-a)(p-k-b)(p-k+b)}$
where Im[a] and Im[b] are negative and p is real. Besides, Re[a] and Re[b], and p are positive.
The the poles in complex plane are shown below:
enter image description here
$frac{1}{2a[(p-a)^2-b^2]},frac{1}{2b[(p-b)^2-a^2]},-frac{1}{2a[(p+a)^2-b^2]},-frac{1}{2a[(p+b)^2-a^2]}$
I get these residues by just putting $a$ into the other three denominators $(k+a)$, $(p-k-b)$, $(p-k+b)$. and I also use the same procedure to the other three poles $p+b, -a, p-b$ (since each pole is a simple pole). I think the summation of these four poles should give zero. What's the problem?
complex-analysis contour-integration
complex-analysis contour-integration
asked Nov 22 at 6:21
James Liu
11
asked Nov 22 at 6:21
James Liu
11
asked Nov 22 at 6:21
James Liu
11
asked Nov 22 at 6:21
James Liu
11
asked Nov 22 at 6:21
James Liu
11
11
What are the residues that you obtain. The sum of the residues has to be zero in this case. Otherwise, you have a mistake in the calculation.
– Fabian
Nov 22 at 7:06
1
Oh, I see the residues. I believe the second and fourth residue should be minus from what you write.
– Fabian
Nov 22 at 7:09
I get one of the residue $Res_{k=p+b}=frac{1}{(p+b+a)(p+b-a)(-2b)}$ by put p+b into k+a, k-a, p-k-b
– James Liu
Nov 22 at 7:24
We can also get the other answers by the same procedure.
– James Liu
Nov 22 at 7:31
@JamesLiu That is not how you compute a residue. In order for this heuristic to work, you need to change the initial expression it so all signs of $k$ are positive. (i.e. have the last two factors in the denominator be $(k-(p-b))(k-(p+b)),$ otherwise you will be off by a minus sign for two of them.
– spaceisdarkgreen
Nov 22 at 7:46
|
show 2 more comments
What are the residues that you obtain. The sum of the residues has to be zero in this case. Otherwise, you have a mistake in the calculation.
– Fabian
Nov 22 at 7:06
1
Oh, I see the residues. I believe the second and fourth residue should be minus from what you write.
– Fabian
Nov 22 at 7:09
I get one of the residue $Res_{k=p+b}=frac{1}{(p+b+a)(p+b-a)(-2b)}$ by put p+b into k+a, k-a, p-k-b
– James Liu
Nov 22 at 7:24
We can also get the other answers by the same procedure.
– James Liu
Nov 22 at 7:31
@JamesLiu That is not how you compute a residue. In order for this heuristic to work, you need to change the initial expression it so all signs of $k$ are positive. (i.e. have the last two factors in the denominator be $(k-(p-b))(k-(p+b)),$ otherwise you will be off by a minus sign for two of them.
– spaceisdarkgreen
Nov 22 at 7:46
What are the residues that you obtain. The sum of the residues has to be zero in this case. Otherwise, you have a mistake in the calculation.
– Fabian
Nov 22 at 7:06
What are the residues that you obtain. The sum of the residues has to be zero in this case. Otherwise, you have a mistake in the calculation.
– Fabian
Nov 22 at 7:06
1
1
Oh, I see the residues. I believe the second and fourth residue should be minus from what you write.
– Fabian
Nov 22 at 7:09
Oh, I see the residues. I believe the second and fourth residue should be minus from what you write.
– Fabian
Nov 22 at 7:09
I get one of the residue $Res_{k=p+b}=frac{1}{(p+b+a)(p+b-a)(-2b)}$ by put p+b into k+a, k-a, p-k-b
– James Liu
Nov 22 at 7:24
I get one of the residue $Res_{k=p+b}=frac{1}{(p+b+a)(p+b-a)(-2b)}$ by put p+b into k+a, k-a, p-k-b
– James Liu
Nov 22 at 7:24
We can also get the other answers by the same procedure.
– James Liu
Nov 22 at 7:31
We can also get the other answers by the same procedure.
– James Liu
Nov 22 at 7:31
@JamesLiu That is not how you compute a residue. In order for this heuristic to work, you need to change the initial expression it so all signs of $k$ are positive. (i.e. have the last two factors in the denominator be $(k-(p-b))(k-(p+b)),$ otherwise you will be off by a minus sign for two of them.
– spaceisdarkgreen
Nov 22 at 7:46
@JamesLiu That is not how you compute a residue. In order for this heuristic to work, you need to change the initial expression it so all signs of $k$ are positive. (i.e. have the last two factors in the denominator be $(k-(p-b))(k-(p+b)),$ otherwise you will be off by a minus sign for two of them.
– spaceisdarkgreen
Nov 22 at 7:46
|
show 2 more comments
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What are the residues that you obtain. The sum of the residues has to be zero in this case. Otherwise, you have a mistake in the calculation.
– Fabian
Nov 22 at 7:06
1
Oh, I see the residues. I believe the second and fourth residue should be minus from what you write.
– Fabian
Nov 22 at 7:09
I get one of the residue $Res_{k=p+b}=frac{1}{(p+b+a)(p+b-a)(-2b)}$ by put p+b into k+a, k-a, p-k-b
– James Liu
Nov 22 at 7:24
We can also get the other answers by the same procedure.
– James Liu
Nov 22 at 7:31
@JamesLiu That is not how you compute a residue. In order for this heuristic to work, you need to change the initial expression it so all signs of $k$ are positive. (i.e. have the last two factors in the denominator be $(k-(p-b))(k-(p+b)),$ otherwise you will be off by a minus sign for two of them.
– spaceisdarkgreen
Nov 22 at 7:46