Contour integral along the imaginary axis

[dipole]

I'd like to evaluate the integral,
[tex] \int^{i\infty}_{-i\infty} \frac{e^{iz}}{z^2 + a^2}dz [/tex]

along the imaginary axis. This function has poles at [itex] z = \pm ia [/itex], with corresponding residues [itex] \textrm{res}(\frac{e^{iz}}{z^2 + a^2},\pm ia) = \pm\frac{e^{\mp a}}{2ai} [/itex]

My question is - I'm not sure what contour to use. If I go from a segment [itex] (-iR, iR) [/itex], while skirting around the poles, and close it with a semi-circle in the right-half plane the resulting arc in the lower-right half plane will not vanish according to Jordan's lemma... I can't find any examples in my book about how to do contours along the imaginary axis - they all go along the real axis and mostly make use of Jordan's Lemma to simplify things, which doesn't seem applicable here.

Any suggestions?

 

[Dick]

I'd like to evaluate the integral,
[tex] \int^{i\infty}_{-i\infty} \frac{e^{iz}}{z^2 + a^2}dz [/tex]

along the imaginary axis. This function has poles at [itex] z = \pm ia [/itex], with corresponding residues [itex] \textrm{res}(\frac{e^{iz}}{z^2 + a^2},\pm ia) = \pm\frac{e^{\mp a}}{2ai} [/itex]

My question is - I'm not sure what contour to use. If I go from a segment [itex] (-iR, iR) [/itex], while skirting around the poles, and close it with a semi-circle in the right-half plane the resulting arc in the lower-right half plane will not vanish according to Jordan's lemma... I can't find any examples in my book about how to do contours along the imaginary axis - they all go along the real axis and mostly make use of Jordan's Lemma to simplify things, which doesn't seem applicable here.

Any suggestions?

Why don't you change the variable z to u=iz and write the contour integral in terms of u. That will give you an integral along the real axis, right?

 

[dipole]

Hmm well that seems easy... so making the substitution [itex] u = iz [/itex], I'd have,

[tex] -i\int^{\infty}_{-\infty} \frac{e^{u}}{a^2-u^2}du [/tex]

which clearly won't converge. Seems like a silly integral to ask us to do, but does this look correct?

 

[Dick]

Hmm well that seems easy... so making the substitution [itex] u = iz [/itex], I'd have,

[tex] -i\int^{\infty}_{-\infty} \frac{e^{u}}{a^2-u^2}du [/tex]

which clearly won't converge. Seems like a silly integral to ask us to do, but does this look correct?

Yeah, I didn't check all of the signs, but I think that integral has some problems. The same ones your original integral does.

 

[dipole]

It's definitely the correct integral, we were asked to integrate along the real axis first, which was easy enough, and then along the imaginary axis.

Is it true in general that if an integral converges along the real axis it will diverge along the imaginary axis? I know analytic functions have to be unbounded somewhere - this a manifestation of that?

 

[Dick]

It's definitely the correct integral, we were asked to integrate along the real axis first, which was easy enough, and then along the imaginary axis.

Is it true in general that if an integral converges along the real axis it will diverge along the imaginary axis? I know analytic functions have to be unbounded somewhere - this a manifestation of that?

No, that's not true. It is true analytic functions can't be bounded unless they are constant. But 1/(z^2-i) converges along both the real and imaginary axes.

 

[vanhees71]

If [itex]a[/itex] is real the integral along the imaginary axis doesn't exist. You can however calculate an integral using a contour that circumvents the poles along the imaginary axis somehow. The result will however depend on the way how you avoid the singularities.

Such integrals appear in physics when looking for Green's functions of oscillators or for relativistic wave equations. Then the physical problem tells you, which Green's function (retarded, advanced, time-ordered, anti-timeordered etc.) you want to calculate and this determines, how to avoid the singularities.