## 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

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