Help needed with the integral of an infinite series

Question

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:

1. $T$ and $K$ are constants
2. $ \int\limits_{(n-1)T}^{nT}Kf(x)dx = y[n] $
3. $ 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!

Answer 1

Not a full answer, but too large for a comment:
If we assume that $\frac{TK}{2\pi}f(x) < 1$, you have the following: $$\int\limits_{(n-1)T}^{nT}\left[\frac{2\pi}{T}\displaystyle\sum_{i=2}^{\infty}\left(\frac{TK}{2\pi}f(x)\right)^i\right]dx = \int\limits_{(n-1)T}^{nT}\frac{2\pi}{T}\left[\displaystyle\sum_{i=0}^{\infty}\left(\frac{TK}{2\pi}f(x)\right)^i - \frac{TK}{2\pi}f(x) - 1\right]dx$$ $$= \int\limits_{(n-1)T}^{nT}\frac{2\pi}{T}\left[\displaystyle\frac{1}{1 - \left(\frac{TK}{2\pi}f(x)\right)} - \frac{TK}{2\pi}f(x) - 1\right]dx = \int\limits_{(n-1)T}^{nT}\frac{2\pi}{T}\left[\frac{\left(\frac{TK}{2\pi}f(x)\right)^2}{1 - \frac{TK}{2\pi}f(x)}\right]dx$$

Answer 2

This isn't intended to be a rigorous explanation, but it appears that

A) The order of integration and summation have been exchanged

Then B) Each integral has been approximated as a rectangle of base width $T$ because of point 3 given.

Just to clarify, from point 2, they are making the approximation: $$Kf(x)T\simeq y(n)$$ So that $$(Kf(x))^i\simeq \frac{(y(n))^i}{T^i}$$ Then each integral is approximated as $$\frac{(y(n))^i}{T^i}\times T$$

So now if we substitute this into the sum, we get$$\frac{2\pi}{T}\sum_{i=2}^{\infty}(\frac{T}{2\pi})^i\frac{(y(n))^i}{T^i}\times T$$

which simplifies to the result you stated