How to calculate contour integration of multivalued functions?












1












$begingroup$


enter image description here



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







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    enter image description here



    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







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      enter image description here



      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







      share|cite|improve this question











      $endgroup$




      enter image description here



      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 13:13







      Craig Thone

















      asked Dec 7 '18 at 3:46









      Craig ThoneCraig Thone

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