Solving a tough integral I am studying telecommunications theory and I was doing an exercise where it's required to find the (infinite) taps of a zero forcing equalizer. Here's the point where I am stuck at:
$$
p_\ell=T\int_{-\frac{1}{2T}}^{\frac{1}{2T}}\frac{e^{j2\pi f\ell T}}{1+\alpha e^{-j2\pi fT}}df
$$
Where:  


*

*$\ell\in \mathbb{Z}$  

*$0<\alpha<\frac{1}{2}$  

*$T>0$  

*$T,\alpha\in\mathbb{R}$


That comes out because the channel time domain response is:
$$
g(t)=\delta(t)+\alpha\delta(t-T)
$$
And its fourier transform of course is:
$$
G(f)=1+\alpha e^{-j2\pi fT}
$$
In a ZF equalizer it is required that the total f-response of the channel and equalizer is unity, i.e. $ P(f)\cdot G(f)=1 $, so to find the $p_\ell$ sequence one has to anti-transform $\frac{1}{G(f)}$.
It doesn't look to me I've done any errors before the integral but I don't have a clue on how to solve it, if possible. Some help/hints would be very appreciated.
Thanks to PhoemueX answer:
$$
\frac{1}{1 + \alpha e^{-2\pi i f T}} = \frac{1}{1 - (- \alpha e^{-2\pi i f T})} = \sum_{n=0}^{\infty} (-\alpha \cdot e^{-2\pi i f T})^n,
$$
So let's start rocking:
$$
p_\ell=T\int_{-\frac{1}{2T}}^{\frac{1}{2T}}e^{j2\pi f\ell T}\sum_{n=0}^{\infty} (-\alpha \cdot e^{-2\pi i f T})^ndf=\\
=T\sum_{n=0}^{\infty}\int_{-\frac{1}{2T}}^{\frac{1}{2T}}(-\alpha)^ne^{j2\pi f T(\ell-n)}df=\\
=\frac{T}{j2\pi T}\sum_{n=0}^{\infty}\frac{(-\alpha)^n}{\ell-n}
\left(e^{j\pi(\ell-n)}-e^{-j\pi(\ell-n)}\right)=\\
=\frac{2j}{2j\pi}\sum_{n=0}^{\infty}(-\alpha)^n\frac{\sin[\pi(\ell-n)]}{\ell-n}=\\
=\sum_{n=0}^{\infty}(-\alpha)^n\text{sinc}(\ell-n)
$$
That last line equals zero whenever $\ell\neq n$, while when $\ell=n$ the sinc is not defined. We can not compute the limit because that is nonsense in $\mathbb{Z}$ but looking at the second equation we can see that when $\ell=n$ the integral becomes trivial and that sum equals $(-\alpha)^\ell$
To sum up:
$$
p_\ell=(-\alpha)^\ell
$$
Math is awesome.
 A: Expand the denominator into a geometric series like this:
$$
\frac{1}{1 + \alpha e^{-2\pi i f T}} = \frac{1}{1 - (- \alpha e^{-2\pi i f T})} = \sum_{n=0}^{\infty} (-\alpha \cdot e^{-2\pi i f T})^n,
$$
where the series converges uniformly as long as $|\alpha|<1$ (this is the case in your question).
Hence, we can interchange summation and integration.
Why does that help you?
A: You can also solve this problem with the standard tools of complex analysis. If you make a substitution $z=\exp(2\pi i fT)$, then the integral becomes
$$p_\ell=\frac{1}{2\pi i}\int dz\ \frac{z^\ell}{z+\alpha}$$
where the contour of integration is the unit circle oriented counter-clockwise.
Because $\alpha\in(0,\tfrac{1}{2})$, there is a pole at $z=-\alpha$ contained in the contour. Note that, when $\ell<0$, there is a second pole at $z=0$. The integral therefore evaluates to
$$p_\ell={\rm Res}\left(\frac{z^\ell}{z+\alpha};z=-\alpha\right)+{\rm Res}\left(\frac{z^\ell}{z+\alpha};z=0\right).$$
These residues can be evaluated easily:
$${\rm Res}\left(\frac{z^\ell}{z+\alpha};z=-\alpha\right)=(-\alpha)^\ell,$$
$${\rm Res}\left(\frac{z^\ell}{z+\alpha};z=0\right)=\begin{cases}0,&\ell\ge 0\\-(-\alpha)^\ell,&\ell<0\end{cases}.$$
The result is therefore slightly different from what you found. Namely, the integral vanishes for $\ell<0$:
$$p_l=\begin{cases}0,&\ell<0\\(-\alpha)^\ell,&\ell\ge 0\end{cases}.$$
