How to calculate contour integration of multivalued functions?
$begingroup$
I have a question about contour integral of the multivalude function. The question is from paper arXiv:1211.6767 (page 6-7).
I want to calculate the Fourier transformation of a muti-valued function $f$
$$G(omega,mathbf{k})=int dt d^{d-1}mathbf{x} f(t,mathbf{x})
e^{iomega t-imathbf{k}cdotmathbf{x}}$$
with condition that
$$omega>0,,,,,,,,omega^2-mathbf{k}^2>0$$
so that we can deform the contour from the green one to the red one.
Function $f$ is given by
$$ f(t,mathbf{x})=
big(frac{-1}{t^2-mathbf{x}^2-iepsilon t}big)^Delta
$$
For $t<0$, $f$ can be written as
$$ f(t,mathbf{x})=
frac{e^{ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
For $t>0$, $f$ can be written as
$$ f(t,mathbf{x})=
frac{e^{-ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
First I need to write down the function on the four legs, $1,2,3,4$
On the leg 1, since $t<0$, we should have
$$ f_1(t,mathbf{x})=
frac{e^{ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
From leg 1 to leg 2, we need
$$(t+|x|)rightarrow(t+|x|)e^{2pi i} $$
$$ f_2(t,mathbf{x})=
frac{e^{-ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
On the leg 4, since $t>0$, we should have
$$ f_4(t,mathbf{x})=
frac{e^{-ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
From leg 4 to leg 3, we need
$$(t-|x|)rightarrow(t+|x|)e^{-2pi i} $$
$$ f_3(t,mathbf{x})=
frac{e^{ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
Combining the above result, we have
$$G(omega,mathbf{k})=(e^{-i pi Delta}-e^{+i pi Delta})
int d^{d-1}mathbf{x} int^infty_{|x|} dt
frac{e^{iomega t-imathbf{k}cdotmathbf{x}} }{(t^2-mathbf{x}^2)^Delta} \
+(e^{+i pi Delta}-e^{-i pi Delta})
int d^{d-1}mathbf{x} int^infty_{|x|} dt
frac{e^{-iomega t-imathbf{k}cdotmathbf{x}} }{(t^2-mathbf{x}^2)^Delta}$$
But the authors' result is quite different
$$G(omega,mathbf{k})=2(e^{i pi Delta}-e^{-i pi Delta})
int d^{d-1}mathbf{x} int^infty_{|x|} dt
frac{e^{iomega t-imathbf{k}cdotmathbf{x}} }{(t^2-mathbf{x}^2)^Delta} $$
complex-analysis
$endgroup$
add a comment |
$begingroup$
I have a question about contour integral of the multivalude function. The question is from paper arXiv:1211.6767 (page 6-7).
I want to calculate the Fourier transformation of a muti-valued function $f$
$$G(omega,mathbf{k})=int dt d^{d-1}mathbf{x} f(t,mathbf{x})
e^{iomega t-imathbf{k}cdotmathbf{x}}$$
with condition that
$$omega>0,,,,,,,,omega^2-mathbf{k}^2>0$$
so that we can deform the contour from the green one to the red one.
Function $f$ is given by
$$ f(t,mathbf{x})=
big(frac{-1}{t^2-mathbf{x}^2-iepsilon t}big)^Delta
$$
For $t<0$, $f$ can be written as
$$ f(t,mathbf{x})=
frac{e^{ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
For $t>0$, $f$ can be written as
$$ f(t,mathbf{x})=
frac{e^{-ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
First I need to write down the function on the four legs, $1,2,3,4$
On the leg 1, since $t<0$, we should have
$$ f_1(t,mathbf{x})=
frac{e^{ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
From leg 1 to leg 2, we need
$$(t+|x|)rightarrow(t+|x|)e^{2pi i} $$
$$ f_2(t,mathbf{x})=
frac{e^{-ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
On the leg 4, since $t>0$, we should have
$$ f_4(t,mathbf{x})=
frac{e^{-ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
From leg 4 to leg 3, we need
$$(t-|x|)rightarrow(t+|x|)e^{-2pi i} $$
$$ f_3(t,mathbf{x})=
frac{e^{ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
Combining the above result, we have
$$G(omega,mathbf{k})=(e^{-i pi Delta}-e^{+i pi Delta})
int d^{d-1}mathbf{x} int^infty_{|x|} dt
frac{e^{iomega t-imathbf{k}cdotmathbf{x}} }{(t^2-mathbf{x}^2)^Delta} \
+(e^{+i pi Delta}-e^{-i pi Delta})
int d^{d-1}mathbf{x} int^infty_{|x|} dt
frac{e^{-iomega t-imathbf{k}cdotmathbf{x}} }{(t^2-mathbf{x}^2)^Delta}$$
But the authors' result is quite different
$$G(omega,mathbf{k})=2(e^{i pi Delta}-e^{-i pi Delta})
int d^{d-1}mathbf{x} int^infty_{|x|} dt
frac{e^{iomega t-imathbf{k}cdotmathbf{x}} }{(t^2-mathbf{x}^2)^Delta} $$
complex-analysis
$endgroup$
add a comment |
$begingroup$
I have a question about contour integral of the multivalude function. The question is from paper arXiv:1211.6767 (page 6-7).
I want to calculate the Fourier transformation of a muti-valued function $f$
$$G(omega,mathbf{k})=int dt d^{d-1}mathbf{x} f(t,mathbf{x})
e^{iomega t-imathbf{k}cdotmathbf{x}}$$
with condition that
$$omega>0,,,,,,,,omega^2-mathbf{k}^2>0$$
so that we can deform the contour from the green one to the red one.
Function $f$ is given by
$$ f(t,mathbf{x})=
big(frac{-1}{t^2-mathbf{x}^2-iepsilon t}big)^Delta
$$
For $t<0$, $f$ can be written as
$$ f(t,mathbf{x})=
frac{e^{ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
For $t>0$, $f$ can be written as
$$ f(t,mathbf{x})=
frac{e^{-ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
First I need to write down the function on the four legs, $1,2,3,4$
On the leg 1, since $t<0$, we should have
$$ f_1(t,mathbf{x})=
frac{e^{ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
From leg 1 to leg 2, we need
$$(t+|x|)rightarrow(t+|x|)e^{2pi i} $$
$$ f_2(t,mathbf{x})=
frac{e^{-ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
On the leg 4, since $t>0$, we should have
$$ f_4(t,mathbf{x})=
frac{e^{-ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
From leg 4 to leg 3, we need
$$(t-|x|)rightarrow(t+|x|)e^{-2pi i} $$
$$ f_3(t,mathbf{x})=
frac{e^{ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
Combining the above result, we have
$$G(omega,mathbf{k})=(e^{-i pi Delta}-e^{+i pi Delta})
int d^{d-1}mathbf{x} int^infty_{|x|} dt
frac{e^{iomega t-imathbf{k}cdotmathbf{x}} }{(t^2-mathbf{x}^2)^Delta} \
+(e^{+i pi Delta}-e^{-i pi Delta})
int d^{d-1}mathbf{x} int^infty_{|x|} dt
frac{e^{-iomega t-imathbf{k}cdotmathbf{x}} }{(t^2-mathbf{x}^2)^Delta}$$
But the authors' result is quite different
$$G(omega,mathbf{k})=2(e^{i pi Delta}-e^{-i pi Delta})
int d^{d-1}mathbf{x} int^infty_{|x|} dt
frac{e^{iomega t-imathbf{k}cdotmathbf{x}} }{(t^2-mathbf{x}^2)^Delta} $$
complex-analysis
$endgroup$
I have a question about contour integral of the multivalude function. The question is from paper arXiv:1211.6767 (page 6-7).
I want to calculate the Fourier transformation of a muti-valued function $f$
$$G(omega,mathbf{k})=int dt d^{d-1}mathbf{x} f(t,mathbf{x})
e^{iomega t-imathbf{k}cdotmathbf{x}}$$
with condition that
$$omega>0,,,,,,,,omega^2-mathbf{k}^2>0$$
so that we can deform the contour from the green one to the red one.
Function $f$ is given by
$$ f(t,mathbf{x})=
big(frac{-1}{t^2-mathbf{x}^2-iepsilon t}big)^Delta
$$
For $t<0$, $f$ can be written as
$$ f(t,mathbf{x})=
frac{e^{ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
For $t>0$, $f$ can be written as
$$ f(t,mathbf{x})=
frac{e^{-ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
First I need to write down the function on the four legs, $1,2,3,4$
On the leg 1, since $t<0$, we should have
$$ f_1(t,mathbf{x})=
frac{e^{ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
From leg 1 to leg 2, we need
$$(t+|x|)rightarrow(t+|x|)e^{2pi i} $$
$$ f_2(t,mathbf{x})=
frac{e^{-ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
On the leg 4, since $t>0$, we should have
$$ f_4(t,mathbf{x})=
frac{e^{-ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
From leg 4 to leg 3, we need
$$(t-|x|)rightarrow(t+|x|)e^{-2pi i} $$
$$ f_3(t,mathbf{x})=
frac{e^{ipi Delta}}{ big(t^2-x^2big)^Delta}
$$
Combining the above result, we have
$$G(omega,mathbf{k})=(e^{-i pi Delta}-e^{+i pi Delta})
int d^{d-1}mathbf{x} int^infty_{|x|} dt
frac{e^{iomega t-imathbf{k}cdotmathbf{x}} }{(t^2-mathbf{x}^2)^Delta} \
+(e^{+i pi Delta}-e^{-i pi Delta})
int d^{d-1}mathbf{x} int^infty_{|x|} dt
frac{e^{-iomega t-imathbf{k}cdotmathbf{x}} }{(t^2-mathbf{x}^2)^Delta}$$
But the authors' result is quite different
$$G(omega,mathbf{k})=2(e^{i pi Delta}-e^{-i pi Delta})
int d^{d-1}mathbf{x} int^infty_{|x|} dt
frac{e^{iomega t-imathbf{k}cdotmathbf{x}} }{(t^2-mathbf{x}^2)^Delta} $$
complex-analysis
complex-analysis
edited Dec 7 '18 at 13:13
Craig Thone
asked Dec 7 '18 at 3:46
Craig ThoneCraig Thone
1727
1727
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