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- Jan 31, 2012

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I want to evaluate $ f(z) = z^{a-1} (z + z^{-1})^{b} \ \ (a > b >-1) $ around the right half of the circle $|z|=1$ (call it $C$).

I closed the contour with the vertical segment from $z=i$ to $z=-i$.

The integrand has branch points at $z=i, z=-i$, and $z=0$. So I made a cut along the positive imaginary axis from $i$ to $i \infty$, a cut along the negative imaginary axis from $-i$ to $-i \infty$, and a cut along the negative real axis (including the origin).

Under the restrictions placed on the parameters, all of the branch points are weak.

And it should be that $ \int_{C}z^{a-1}\left( z+z^{-1}\right)^{b}\ dz = 2i \sin\left(\frac{\pi(a-b)}{2}\right)\int_{0}^{1}t^{a-b-1}\left(1-t^{2}\right)^{b}\ dt$.

What I get is $\int_{C}z^{a-1}\left( z+z^{-1}\right)^{b}\ dz + \int_{1}^{0}(it)^{a-1}\left( it+(it)^{-1}\right)^{b}i \ dt - \int_{0}^{1}(-it)^{a-1}\left( -it+(-it)^{-1}\right)^{b}i \ dt = 0 $

or $\int_{C}z^{a-1}\left( z+z^{-1}\right)^{b}\ dz = i \ i^{a-b}\int_{0}^{1}t^{a-b-1}\left(1-t^{2}\right)^{b}\ dt + (-1)^{a+b-1} i \ i^{a-b}\int_{0}^{1}t^{a-b-1}\left(1-t^{2}\right)^{b}\ dt$

$ = i e^{i \pi \frac{(a-b)}{2}} \Big( \int_{0}^{1}t^{a-b-1}\left(1-t^{2}\right)^{b}\ dt + (-1)^{a+b-1} \int_{0}^{1}t^{a-b-1}\left(1-t^{2}\right)^{b}\ dt \Big)$

Now comes the part that has me completely confused. On that part of the contour does $-1= e^{i \pi}$ or does $-1 = e^{- i \pi}$? It can't be both due to the cut along the negative real axis. The latter leads to the correct answer. But I don't know why it's necessarily the latter.

EDIT: I think this strange issue is the result of not being careful with the parametrization of the contour.

I closed the contour with the vertical segment from $z=i$ to $z=-i$.

The integrand has branch points at $z=i, z=-i$, and $z=0$. So I made a cut along the positive imaginary axis from $i$ to $i \infty$, a cut along the negative imaginary axis from $-i$ to $-i \infty$, and a cut along the negative real axis (including the origin).

Under the restrictions placed on the parameters, all of the branch points are weak.

And it should be that $ \int_{C}z^{a-1}\left( z+z^{-1}\right)^{b}\ dz = 2i \sin\left(\frac{\pi(a-b)}{2}\right)\int_{0}^{1}t^{a-b-1}\left(1-t^{2}\right)^{b}\ dt$.

What I get is $\int_{C}z^{a-1}\left( z+z^{-1}\right)^{b}\ dz + \int_{1}^{0}(it)^{a-1}\left( it+(it)^{-1}\right)^{b}i \ dt - \int_{0}^{1}(-it)^{a-1}\left( -it+(-it)^{-1}\right)^{b}i \ dt = 0 $

or $\int_{C}z^{a-1}\left( z+z^{-1}\right)^{b}\ dz = i \ i^{a-b}\int_{0}^{1}t^{a-b-1}\left(1-t^{2}\right)^{b}\ dt + (-1)^{a+b-1} i \ i^{a-b}\int_{0}^{1}t^{a-b-1}\left(1-t^{2}\right)^{b}\ dt$

$ = i e^{i \pi \frac{(a-b)}{2}} \Big( \int_{0}^{1}t^{a-b-1}\left(1-t^{2}\right)^{b}\ dt + (-1)^{a+b-1} \int_{0}^{1}t^{a-b-1}\left(1-t^{2}\right)^{b}\ dt \Big)$

Now comes the part that has me completely confused. On that part of the contour does $-1= e^{i \pi}$ or does $-1 = e^{- i \pi}$? It can't be both due to the cut along the negative real axis. The latter leads to the correct answer. But I don't know why it's necessarily the latter.

EDIT: I think this strange issue is the result of not being careful with the parametrization of the contour.

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