Integrating binomial coefficients

Problem: Let \(C(z)\) = coefficient of \(x^{2010}\) in the expansion of \((1+x)^z\) about \(x=0\). What then is the value of

\(\int_{0}^{1}dy\) \(C(-y-1)\) \(\sum_{m=1}^{2010}\frac{1}{y+m}\) ?

Solution:

Let \(k=2010\).  The value of the binomial coefficient desired is

\(C(z)=\binom{z}{k}=\frac{z(z-1)(z-2) \ldots (z-k+1)}{k!}\).

It then follows that

\(C(-y-1)=\frac{(-1)^k}{k!}(y+1)(y+2) \ldots (y+k)\).

If we let \(\frac{(-1)^k}{k!}p(y)=C(-y-1)\), then it turns out that

\(\sum_{m=1}^{k}\frac{1}{y+m}=\frac{p'(y)}{p(y)}\).

Therefore

\(I(k)=\int_{0}^{1}dy \: C(-y-1)\sum_{m=1}^{k}\frac{1}{y+m}=\int_{0}^{1}dy \: p'(y)\). \(I(k)=\frac{(-1)^k}{k!}(p(1)-p(0))=\frac{(-1)^k}{k!}(2)(3) \ldots (k+1)\) \(I(k)=\frac{(-1)^k}{k!}(k+1)!=(-1)^k (k+1)\)

The answer is therefore 2011.