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sheaf
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I've started working through Zee's book and have got to question I.3.2 - calculation of D(x) in 1+1 dimensions for t=0. The expression to evaluate becomes (omitting constant multipliers for simplicity)
[tex]\int^{\infty}_{-\infty} dk \frac{e^{ikx}}{\sqrt{k^2+m^2}} [/tex]
This is singular at k=+im and k=-im. I'll take the square root to have a branch cut on the imaginary axis from [itex]k=+im\rightarrow+i\infty[/itex] and [itex]k=-im\rightarrow-i\infty[/itex]. If I substitute z=-ik, I can put the branch cut on the real axis from [itex]z=+m\rightarrow+\infty[/itex] and [itex]z=-m\rightarrow-\infty[/itex] and the integral becomes
[tex]\int^{+i\infty}_{-i\infty} dz \frac{e^{zx}}{\sqrt{m^2-z^2}} [/tex]
For positive x, I close the contour in the half of the z plane with negative real value, so that the contribution of the semicircle at infinity vanishes. The answer I want then becomes the integral inwards and outwards along the branch cut. Since with the square root, the integral along the cut an infinitesimal distance above the cut is the negative of the integral an infinitesimal below it, I'm left with (again ignoring factors of 2 etc)
[tex]\int^{-\infty}_{-m} dz \frac{e^{zx}}{\sqrt{m^2-z^2}} [/tex]
ie
[tex]e^{-mx}\int^{-\infty}_{-m} dz \frac{e^{(z+m)x}}{\sqrt{z-m}\sqrt{z+m}} [/tex]
This is as far as I can get. If I expand the exponential, I can separate the integrand into a part which has a singular contribution at z=-m and the rest which doesn't, but beyond this I'm not sure where to go. Since it's an elementary question at the beginning of an introductory book, I expected it to have some simple closed form solution, so I think I must have made a mistake somewhere... Can anybody help ?
(Question I.3.1 is pretty similar, but I have an integral in spherical polars in k space, which changes things slightly. I thought I'd tackle I.3.2 first since it should be simpler).
[tex]\int^{\infty}_{-\infty} dk \frac{e^{ikx}}{\sqrt{k^2+m^2}} [/tex]
This is singular at k=+im and k=-im. I'll take the square root to have a branch cut on the imaginary axis from [itex]k=+im\rightarrow+i\infty[/itex] and [itex]k=-im\rightarrow-i\infty[/itex]. If I substitute z=-ik, I can put the branch cut on the real axis from [itex]z=+m\rightarrow+\infty[/itex] and [itex]z=-m\rightarrow-\infty[/itex] and the integral becomes
[tex]\int^{+i\infty}_{-i\infty} dz \frac{e^{zx}}{\sqrt{m^2-z^2}} [/tex]
For positive x, I close the contour in the half of the z plane with negative real value, so that the contribution of the semicircle at infinity vanishes. The answer I want then becomes the integral inwards and outwards along the branch cut. Since with the square root, the integral along the cut an infinitesimal distance above the cut is the negative of the integral an infinitesimal below it, I'm left with (again ignoring factors of 2 etc)
[tex]\int^{-\infty}_{-m} dz \frac{e^{zx}}{\sqrt{m^2-z^2}} [/tex]
ie
[tex]e^{-mx}\int^{-\infty}_{-m} dz \frac{e^{(z+m)x}}{\sqrt{z-m}\sqrt{z+m}} [/tex]
This is as far as I can get. If I expand the exponential, I can separate the integrand into a part which has a singular contribution at z=-m and the rest which doesn't, but beyond this I'm not sure where to go. Since it's an elementary question at the beginning of an introductory book, I expected it to have some simple closed form solution, so I think I must have made a mistake somewhere... Can anybody help ?
(Question I.3.1 is pretty similar, but I have an integral in spherical polars in k space, which changes things slightly. I thought I'd tackle I.3.2 first since it should be simpler).