Q:

Help needed with the integral of an infinite series

Can you please help me with the integral of this series? I came across it in a signal processing paper and haven't been able to figure out the solution myself. 
$$
\int\limits_{(n-1)T}^{nT}\left[\frac{2\pi}{T}\displaystyle\sum_{i=2}^{\infty}\left(\frac{TK}{2\pi}f(x)\right)^i\right]dx
$$
given that:

$T$ and $K$ are constants
$ \int\limits_{(n-1)T}^{nT}Kf(x)dx = y[n] $
$ f(x) $ does not change significantly between $ (n-1)T $ and $T$

The answer I have is:
$$
\displaystyle\sum_{i=2}^{\infty}\alpha_i(y[n])^i
$$
where:
$$
\alpha_i \cong \left(\frac{1}{2\pi}\right)^{(i-1)}
$$
I will really appreciate some brief explanation of how this answer is derived.
Thanks!

A:

The key in this development is to exploit that $f(x)$ does not change significantly for $x\in [(n-1)T,nT]$.  If we do this, then we can approximate $f(x)$ as constant within the interval of integration.  
To that end, we write
$$\begin{align}
\int_{(n-1)T}^{nT}\left(f(x)\right)^idx &\approx. \left(f(nT)\right)^i\,T \tag 1\\\\
&\approx. \left(\frac1{TK}\int_{(n-1)T}^{nT} Kf(x)dx\right)^i\,T \tag 2\\\\
&=\left(\frac{y[n]}{KT}\right)^i\,T\tag 3
\end{align}$$
where in going from $(1)$ to $(2)$ we again exploited the fact that $f$ does not significantly change in the interval of integration and approximated $f(nT)$ by the integral $f(nT)\approx. \frac{1}{KT}\int_{(n-1)T}^{nT} Kf(x)dx$.
Therefore, using $(3)$ we have
$$\begin{align}
\int_{(n-1)T}^{nT}\left[\frac{2\pi}{T}\sum_{i=2}^{\infty}\left(\frac{TK}{2\pi}f(x)\right)^i\right]\,dx&=\frac{2\pi}{T}\sum_{i=2}^{\infty}\left(\frac{TK}{2\pi}\right)^i\int_{(n-1)T}^{nT}\left(f(x)\right)^i\,dx\\\\
&\approx. \frac{2\pi}{T}\sum_{i=2}^{\infty}\left(\frac{TK}{2\pi}\right)^i\left(\frac{y[n]}{KT}\right)^i\,T\\\\
&=\sum_{i=2}^{\infty}\alpha_i\left(y[n]\right)^i
\end{align}$$
where $\alpha_i=\left(\frac{1}{2\pi}\right)^{i-1}$$
as was to be shown!

