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 ($z=\infty$)?
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 becomes a "longitude", say the meridian, and the contour 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 ($z=\infty$)?
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 ($z=\infty$)?
An integral representation of the modified Bessel function is $$I_\nu(t) = \frac1{2\pi i} \oint_C \exp(t (z+1/z))\frac{dz}{z^{1+\nu}}$$$$I_\nu(t) = \frac1{2\pi i} \oint_C \exp(t (z+1/z))\frac{\mathrm dz}{z^{1+\nu}}$$ where the contour C$C$ 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 becomes a "longitude""longitude", say the meridian, and the contour 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""simple" explanation? Or is it just that there is an essential singularity ON the path ($z=\infty$)?
An integral representation of the modified Bessel function is $$I_\nu(t) = \frac1{2\pi i} \oint_C \exp(t (z+1/z))\frac{dz}{z^{1+\nu}}$$ 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 becomes a "longitude", say the meridian, and the contour 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 ($z=\infty$)?
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 becomes a "longitude", say the meridian, and the contour 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 ($z=\infty$)?
Apparent paradox from deforming a Hankel contour on the Riemann sphere
An integral representation of the modified Bessel function is $$I_\nu(t) = \frac1{2\pi i} \oint_C \exp(t (z+1/z))\frac{dz}{z^{1+\nu}}$$ 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 becomes a "longitude", say the meridian, and the contour 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 ($z=\infty$)?