Let \(C\) be a positively oriented simple closed contour enclosing the points \(0,1,\dots,k\). Evaluate, for \(k=0,1,2,\dots\), \[ I_k=\int_C\frac{dz}{z(z-1)\cdots(z-k)} \] and \[ J_k=\int_C\frac{(z-1)\cdots(z-k)}{z}\,dz. \]
Solution
1. For \(k=0\),\[ \boxed{I_0=2\pi i}. \]
When \(k=0\), \[ I_0=\int_C\frac{dz}{z}. \] The contour contains the simple pole at \(0\) with residue \(1\), so the residue theorem gives \(I_0=2\pi i\).
2. For \(k\geq1\) and \(0\leq j\leq k\), the residue of the integrand of\(I_k\) at \(z=j\) is\[ \operatorname{Res}_{z=j} \frac{1}{z(z-1)\cdots(z-k)} = \frac{(-1)^{k-j}}{j!(k-j)!}. \]
The pole at \(j\) is simple, so its residue is \[ \frac{1}{ \displaystyle\prod_{\substack{0\leq m\leq k\\m\neq j}}(j-m) }. \] The factors with \(m<j\) multiply to \[ j!, \] while the factors with \(m>j\) multiply to \[ (-1)^{k-j}(k-j)!. \] Taking the reciprocal gives the claimed residue.
3. For every \(k\geq1\),\[ \boxed{I_k=0}. \]
By the residue theorem and step 2, \[ \begin{aligned} I_k &= 2\pi i \sum_{j=0}^k \frac{(-1)^{k-j}}{j!(k-j)!}\\ &= \frac{2\pi i}{k!} \sum_{j=0}^k \binom{k}{j}(-1)^{k-j}\\ &= \frac{2\pi i}{k!}(1-1)^k\\ &=0. \end{aligned} \]
4. For every \(k\geq0\), the only pole of the integrand of \(J_k\) is\(z=0\), and its residue is\[ (-1)^k k!. \]
The numerator \[ (z-1)\cdots(z-k) \] is a polynomial, so the factor \(1/z\) supplies the only possible pole. At \(z=0\) the residue is the numerator evaluated at \(0\): \[ (-1)(-2)\cdots(-k) = (-1)^k k!. \] For \(k=0\), the product is empty and equals \(1=0!\), so the same formula holds.
5. Therefore, for every \(k\geq0\),\[ \boxed{J_k=2\pi i\,(-1)^k k!}. \]
The contour encloses \(0\) and the integrand of \(J_k\) has no other poles. Apply the residue theorem to the residue computed in step 4.
6. Thus the two requested families are\[ \boxed{ I_k= \begin{cases} 2\pi i,&k=0,\\ 0,&k\geq1, \end{cases} \qquad J_k=2\pi i\,(-1)^k k! }. \]
Combine steps 1, 3, and 5.
7. Q.E.D.
Step 6 gives both evaluations.