Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 14A (Further Complex Methods)

Differentiate under the contour integral, with one consistent branch and convergent endpoint behavior. ThenThe stated zero boundary term therefore makes the integral a solution. On a closed contour this means returning to the same branch; on an open contour it requires the endpoint values to cancel or vanish.

For $\mathrm{Re}k>0$, choose the straight path from $-i$ to $i$ and put $t=is$, $-1\le s\le 1$. Choose the power of $1-s^2$ continuously, positive for real positive $k$. The boundary factor vanishes at both endpoints. At zero,using the symmetric beta function integral. This choice also gives a nonzero solution.

For positive integers $n$, the finite open segment from part (i) works because $k=n/2>0$. For $n\le 0$, take a finite closed contour enclosing both branch points $i,-i$, with the integrand analytic in the exterior of the joining cut. Near infinity choose $(t^2+1{)}^{n/2-1}=t^{n-2}(1+t^{-2}{)}^{n/2-1}$. Since $n$ is integral, a circuit around both branch points returns to the initial branch, and the endpoint boundary term cancels. Thus this is an admissible finite contour, including $n=0$.

For $n\le -1$, at $z=0$ that exterior branch is $O(t^{n-2})=O(t^{-3})$. Deforming the contour to a large circle makes its integral tend to zero; equivalently its Laurent series has no $t^{-1}$ coefficient. Hence $\boxed{w(0)=0(n\le -1)}$. An individual loop around one branch point would not generally return to the same branch when $n$ is odd, which is why both are enclosed.