Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 14C (Further Complex Methods)

Parametrize the continued branch by $t=r{e}^{i\theta }$, $0\le \theta \le 2\pi$. Then $t^{z-1}dt=i{r}^z{e}^{iz\theta }d\theta$, givingThe apparent singularity at zero is removable, so $f$ is an entire function. The factor $r^z=e^{z\log r}$ shows its dependence on the contour radius; for instance $f(1/2)=-4\sqrt{r}$. The integrand is continued along the contour and generally returns on a different branch, so this is not an ordinary integral of a single-valued holomorphic function around a closed loop.

There is a branch-phase discrepancy in the printed identity. Set $K(z)={\int }_{-1}^1{e}^t(1-t^2{)}^{z}dt$ using the positive real base. Shrink the anticlockwise loop at $1$. The outward interval carries phase $e^{-i\pi z}$ and its return phase $e^{i\pi z}$; their difference gives $-2i\sin (\pi z){\int }_0^1{e}^t(1-t^2{)}^{z}dt$. The clockwise loop at $-1$ starts with phase $e^{i\pi z}$ and returns with $e^{-i\pi z}$, giving the same coefficient on $(-1,0)$. Small endpoint circles vanish for $\mathrm{Re}z>-1$. Thus the branch phase in a figure-eight analytic continuation integral isIf the printed interval integral uses the prescribed argument $-\pi$, then $I(z)=e^{-i\pi z}K(z)$, and the corrected formula is insteadFor example at $z=1/2$, $K>0$, $I=-iK$ and $J=-2iK$; the printed right side without the phase would be the real number $-2K$. Interpreting its interval integrand as $(1-t^2{)}^z$ restores the intended identity.

On the fixed contour away from its branch points, the continued logarithm is bounded and $e^t\exp [z\log (t^2-1)]$ is entire in $z$, uniformly on compact sets. Hence $J$ is an entire function. Division by the sine gives a meromorphic continuation of $K$, and multiplication by $e^{-i\pi z}$ gives the continuation of the specified $I$. The only candidate singularities are integers, at most simple. All $0,1,2,\dots$ are removable because the interval integral is already holomorphic on $\mathrm{Re}z>-1$.

The negative integers $z=-n$, $n\ge 1$, are genuine simple poles. At such a point the contour integrand is single-valued andThe first residue is $e$ times a rational number. The second is $e^{-1}$ times a nonzero rational number: expand $e^s(-2+s{)}^{-n}$ at $s=0$, whose degree-$n-1$ coefficient has sign $(-1{)}^n$ and nonzero magnitude. They cannot be equal because $e^2$ is irrational. Therefore $J(-n)\ne 0$, and the simple sine zero creates a simple pole. In particular ${\mathrm{Res}}_{z=-n}K=-J(-n)/[2\pi i(-1{)}^n]$.

A circle enclosing both branch points has total logarithmic change $4\pi i$, rather than zero as for the figure eight. For a circle based at $t=2$ with initial real logarithm, shrinking it leaves a connector from $1$ to $2$ as well as the two banks of the interval. Explicitly its integral isThus it is not just a known sine factor times $K$: an additional endpoint-singular integral remains. The figure-eight contour eliminates precisely that connector term. This explains why replacing it by the large circle does not give the same direct continuation method.