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.

Be Sociable, Share!

Published by

Ron Gordon

Math nerd in his early 40’s who seems to have an opinion about everything and an inability to keep it to himself.