- #1

Pere Callahan

- 586

- 1

Hi,

I'm wondering if a generalization of the residue theorem/formulae to several complex variables could be just as helpful as in the one-dimensional case.

For example if you were to calculate

[tex]\int_\mathhb{R}{\frac{dk}{2\pi}\frac{e^{-ikx}}{1+k^2}}[/tex]

One way would be to observe that the integrand has simple poles at [tex]k=\pm i[/tex] then close the contour of integration in the upper (lower) halfplane when x is less (greater) then zero, to obtain the result

[tex]e^{-|x|}[/tex].

What however in the slightly more complicated case

[tex]\int_{\mathhb{R}^2}{\frac{d^2k}{(2\pi)^2}\frac{e^{-ik_1x}e^{-ik_2y}}{1+k_1^2+k_2^2}}[/tex]

The integrand has "poles" (are they still called that if they are not isolated?) for [tex]k_1^2+k_2^2 = -1[/tex] which might be thought of as a circle in the "imaginary plane".

Is there a similar way to evaulate this integral by somehow closing the contour and applying some generalization of the residue theorem?

If not, how COULD it be calculated? My thought was to do the, say, k_1 integration first. For fixed k_2, there are poles at

[tex]k_1^\pm = \pm i \sqrt{1+k_2^2}[/tex]

with residues

[tex]Res^\pm = \frac{e^{\pm\sqrt{1+k_2^2} x}e^{-ik_2y}}{\pm 2 i \sqrt{1+k_2^2}}[/tex]

where we have to pick the right sign according to whether x is positive or negative.

Then I would do the k_2 integration. The poles are at

[tex]k_2^\pm = \pm i[/tex]

However

[tex]\frac{e^{\pm\sqrt{1+k_2^2} x}e^{-ik_2y}}{\pm 2 i \sqrt{1+k_2^2}}[/tex]

is not holomorphic in a neighbourhood of [tex]\pm i[/tex] so I cannot apply the Residue theorem can I?

Thanks

-Pere

I'm wondering if a generalization of the residue theorem/formulae to several complex variables could be just as helpful as in the one-dimensional case.

For example if you were to calculate

[tex]\int_\mathhb{R}{\frac{dk}{2\pi}\frac{e^{-ikx}}{1+k^2}}[/tex]

One way would be to observe that the integrand has simple poles at [tex]k=\pm i[/tex] then close the contour of integration in the upper (lower) halfplane when x is less (greater) then zero, to obtain the result

[tex]e^{-|x|}[/tex].

What however in the slightly more complicated case

[tex]\int_{\mathhb{R}^2}{\frac{d^2k}{(2\pi)^2}\frac{e^{-ik_1x}e^{-ik_2y}}{1+k_1^2+k_2^2}}[/tex]

The integrand has "poles" (are they still called that if they are not isolated?) for [tex]k_1^2+k_2^2 = -1[/tex] which might be thought of as a circle in the "imaginary plane".

Is there a similar way to evaulate this integral by somehow closing the contour and applying some generalization of the residue theorem?

If not, how COULD it be calculated? My thought was to do the, say, k_1 integration first. For fixed k_2, there are poles at

[tex]k_1^\pm = \pm i \sqrt{1+k_2^2}[/tex]

with residues

[tex]Res^\pm = \frac{e^{\pm\sqrt{1+k_2^2} x}e^{-ik_2y}}{\pm 2 i \sqrt{1+k_2^2}}[/tex]

where we have to pick the right sign according to whether x is positive or negative.

Then I would do the k_2 integration. The poles are at

[tex]k_2^\pm = \pm i[/tex]

However

[tex]\frac{e^{\pm\sqrt{1+k_2^2} x}e^{-ik_2y}}{\pm 2 i \sqrt{1+k_2^2}}[/tex]

is not holomorphic in a neighbourhood of [tex]\pm i[/tex] so I cannot apply the Residue theorem can I?

Thanks

-Pere

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