computing an integral with binomial coeficients Let $f$ be an integrable function such that $\int_\mathbb{R} {f(u)du = 1} $ 
We define 
$h(u) = \sum\limits_{k = 1}^l \binom{l}{k}{{{( - 1)}^{k + 1}}\frac{1}{k}f\left( {\frac{u}{k}} \right)}$ , $l$ integer
Show that $\int_\mathbb{R} {h(u){u^i}du = {\delta _{0,i}}}$ for $i=0,1,...,l-1$
This is what I did:
$\int_\mathbb{R} {h(u){u^i}du = } \sum\limits_{k = 1}^l \binom{l}{k}{{{( - 1)}^{k + 1}}} \int_\mathbb{R} {\frac{{{u^i}}}{k}} f\left( {\frac{u}{k}} \right)du$
Now, let $y=\frac{u}{k}$, then $u=ky$  and $du=kdy$
so,
$\int_\mathbb{R} {h(u){u^i}du = } \sum\limits_{k = 1}^l \binom{l}{k}{{{( - 1)}^{k + 1}}} \int_\mathbb{R} {{k^i}} {y^i}f(y)dy$
$\int_\mathbb{R} {h(u){u^i}du = } \sum\limits_{k = 1}^l \binom{l}{k}{{{( - 1)}^{k + 1}}} {k^i}\int_\mathbb{R} {{y^k}f(y)} dy$
I am stuck here
Some help would be appreciated
 A: First, let me disambiguate slightly by using j as the index of summation:
$$
\int_R h(u) u^k du = \sum_{j=1}^l \binom{l}{j} (-1)^{j+1} j^k \int_R y^k f(y) dy
= \left( \int_R y^k f(y) dy \right)\left( \sum_{j=1}^l \binom{l}{j} (-1)^{j+1} j^k \right)
$$
The latter sum can be evaluated explicitly.  For k=0, it is:
$$
\sum_{j=1}^l \binom{l}{j} (-1)^{j+1} = 1-\left[\sum_{j=0}^l \binom{l}{j} (-1)^{j} x^j\right]_{x=1} = 1-\left[(1-x)^k\right]_{x=1} = 1
$$
In this case, the integral $\int_R y^k f(y) dy = \int_R f(y) dy = 1$, which shows that $\int_R h(u) u^k du = 1 = \delta_{0,k}$, as wanted. 
For $k>0$, the sum is:
$$
\sum_{j=1}^l \binom{l}{j} (-1)^{j+1} j^k = - \left[\left(x \frac{d}{dx}\right)^k \sum_{j=0}^l \binom{l}{j} (-1)^{j} x^j \right]_{x=1} = -\left[\left(x\frac{d}{dx}\right)^k (1-x)^l\right]_{x=1}
$$
The important point here is that $\left[\left(x\frac{d}{dx}\right)^k (1-x)^l\right]$ has a factor of (1-x) for all $k<l$ (you can prove this by induction: $x\frac{d}{dx}(1-x)^l = $ (some polynomial)$\times (1-x)^{l-1}$, and $x\frac{d}{dx}$(some polynomial)$\times (1-x)^{l-m} = $(some other polynomial)$\times (1-x)^{l-m-1}$.  As long as you apply $x\frac{d}{dx}$ less than $l$ times, you still have a factor of $(1-x)$ left.)
Hence the sum for $0<k<l$ evaluates to zero, meaning the integral $\int_R h(u) u^k du=0 = \delta_{0,k}$, as wanted. 
