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.


  • 1
    $\begingroup$ In my opinion this is pure nonsense, and the authors should be excluded from the academia for publishing it. Is it possible to have a link to this mathematical calamity? $\endgroup$ – Alex M. Jul 30 '15 at 16:18
  • $\begingroup$ Indeed the source is needed to make sense of this, if this is possible. $\endgroup$ – Did Jul 30 '15 at 16:34

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

  • $\begingroup$ Not so sure: each term produces a $K^i$, why isn't it found in the result? The area of one such rectangle would be $T \cdot \Big( \frac {TK} {2 \pi} \Big) ^i f(nT)^i$ - where is $y[n]$ then? No, I don't see how your hints come even close to an answer. $\endgroup$ – Alex M. Jul 30 '15 at 16:17
  • $\begingroup$ @AlexM....I have added further explanation to my answer. Please let me know if there is something wrong. Thanks $\endgroup$ – David Quinn Jul 30 '15 at 17:52
  • $\begingroup$ Indeed, with these clarifications your answer makes perfect sense. I had understood its first version differently. $\endgroup$ – Alex M. Jul 30 '15 at 18:08
  • $\begingroup$ OK thanks, no problem $\endgroup$ – David Quinn Jul 30 '15 at 18:13

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: $$\int\limits_{(n-1)T}^{nT}\left[\frac{2\pi}{T}\displaystyle\sum_{i=2}^{\infty}\left(\frac{TK}{2\pi}f(x)\right)^i\right]dx = \int\limits_{(n-1)T}^{nT}\frac{2\pi}{T}\left[\displaystyle\sum_{i=0}^{\infty}\left(\frac{TK}{2\pi}f(x)\right)^i - \frac{TK}{2\pi}f(x) - 1\right]dx$$ $$= \int\limits_{(n-1)T}^{nT}\frac{2\pi}{T}\left[\displaystyle\frac{1}{1 - \left(\frac{TK}{2\pi}f(x)\right)} - \frac{TK}{2\pi}f(x) - 1\right]dx = \int\limits_{(n-1)T}^{nT}\frac{2\pi}{T}\left[\frac{\left(\frac{TK}{2\pi}f(x)\right)^2}{1 - \frac{TK}{2\pi}f(x)}\right]dx$$

  • $\begingroup$ How does this lead to the answer? $\endgroup$ – David Quinn Jul 30 '15 at 23:44

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