An integral representation of the modified Bessel function is $$I_\nu(t) = \frac1{2\pi i} \oint_C \exp(t (z+1/z))\frac{\mathrm dz}{z^{1+\nu}}$$ where the contour $C$ starts in $\infty-i\epsilon$ and follows the negative reals axis (branch cut of $z^{1+\nu}$) to the origin, winds around the origin by crossing the positive real axis at some $x>0$, and then goes back out to $\infty+i\epsilon$. Hankel
This is all fine.
However, when I view it on the Riemann sphere, the branch cut (red dashed line on figure below) becomes a "longitude", say the meridian, and the contour (blue solid line) follows it on both sides from the north pole ($\infty$) and winds around the south pole (0).
But now it seems that I can shrink the contour to zero by deforming it away from the meridian, on the other side of the sphere, up to infinity. Without crossing any singularities. Which would seem to imply that the integral vanishes. This is obviously incorrect, but what am I missing? I suppose the Cauchy theorem looks different on the Riemann sphere, but is there a "simple" explanation? Or is it just that there is an essential singularity ON the path ($z=\infty$)?
1 Answer 1
In order to properly enclose the north pole, you need the contour to go around it instead of doing a U-turn once it gets near it. And the problem with this is that the values of the integrand near $\infty$ get very large when $z$ is on the positive real axis. So that tiny bit you are missing, the integral over a small circle around $\infty$ (or a circle of large radius on the original picture), is enough to exactly compensate the integral on $C$ and make the integral over the whole loop vanish.
-
$\begingroup$ thanks for your reply, but I am not sure I follow. I naively thought the original contour goes onto (starts and ends at) the north pole, not enclosing it as such. Or do you mean that the the path I am describing on the sphere does not correspond to the one I am describing in the plane? That the plane contour is missing a circle near infinity? But that's negative infinity.... sorry, not following exactly $\endgroup$Spinor– Spinor2025年04月04日 16:07:09 +00:00Commented Apr 4 at 16:07
You must log in to answer this question.
Explore related questions
See similar questions with these tags.