I am working through the exercises in Gallagher, "Principles of Digital Communication", specifically exercise 4.1. This is not homework for a class, but purely self-study.
The text defines the Fourier series for a time-limited function that is non-zero for $[-\frac{T}{2}, \frac{T}{2}]$ as: $$u(t) = \begin{cases} \sum_{k=-\infty}^{\infty} \hat{u}_k \exp{\frac{j2\pi k t}{T}} & \text{for} -\frac{T}{2} \leq t \leq \frac{T}{2} \\ 0 & \text{otherwise.} \end{cases} $$
Each coefficient $\hat{u}_k$ is given by: $$\hat{u}_k = \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} u(t) \exp{\frac{-j2\pi k t}{T}} dt$$
From this definition, I understand the interval for the Fourier series to be $[-\frac{T}{2}, \frac{T}{2}]$, so I am assuming that if I were to compute the coefficients over some interval $T=L$, I would substitute $L$ in the expression above.
Now, problem 4.1 in the text asks us to consider $u(t) = \text{rect}(2t)$, and determine a general expression for the Fourier series coefficients for the Fourier series over $[-\frac{1}{2}, \frac{1}{2}]$. From this, I am determining that $T=1$, and the function $u(t) = \text{rect}(2t)$ is: $$u(t) = \text{rect} (2t) = \begin{cases} 1 & \text{for} -\frac{1}{2} \leq 2t \leq \frac{1}{2} \\ 0 & \text{otherwise.} \end{cases} \Rightarrow \begin{cases} 1 & \text{for} -\frac{1}{4} \leq t \leq \frac{1}{4} \\ 0 & \text{otherwise.} \end{cases} $$
Effectively, I am constraining the limits of integration; however, I do not change the value of $T$ itself. So, for the coefficients, I would be evaluating:
$$ \begin{aligned} \hat{u}_k &= \frac{1}{\mathbf 1} \int_{-\frac{1}{2}}^{\frac{1}{2}} \text{rect} (2t) \exp{\frac{-j2\pi k t}{\mathbf 1}} dt \\ &= \int_{-\frac{1}{4}}^{\frac{1}{4}} \exp({-j2\pi k t}) \, dt \\ &= \frac{1}{\pi k} \sin { \frac {\pi k} {2}} \end{aligned} $$
The question I have is - am I going about this the right way? The next step is to show that this series converges to $\frac{1}{2}$ at the endpoints $\pm\frac{1}{4}$, which I am having a little trouble with, so I want to make sure I am not missing something important.
Thanks!
~~~
Edit: continuing the question...
Assuming that the method in the previous part is correct, I am having trouble showing convergence of the Fourier series at the endpoints $t= -\frac{1}{4}$ and $t=+\frac{1}{4}$. My approach is to first substitute the coefficients $\hat{u}_k$ into the expression for $u(t)$ given previously: $$ u(t)= \begin{cases} \displaystyle \sum_{k=-\infty}^\infty \hat{u}_k \exp {\frac{j2\pi k t} {T}}, & \text{for} -\frac{T}{2} \le t \le \frac{T}{2} \\ \displaystyle 0 & \text{otherwise} \end{cases} $$ Since the Fourier coefficients were defined over $[-\frac{1}{2} , \frac{1}{2}]$, I take $T=1$. This results in: $$ \begin{aligned} u(t) &= \begin{cases} \displaystyle \sum_{k=-\infty}^\infty \hat{u}_k \exp {j2\pi k t}, & \text{for} -\frac{1}{2} \le t \le \frac{1}{2} \\ \displaystyle 0 & \text{otherwise} \end{cases} \\ &= \begin{cases} \displaystyle \sum_{k=-\infty}^\infty \frac{1}{\pi k} \sin {\frac{\pi k} {2} } \exp {j2\pi k t}, & \text{for} -\frac{1}{2} \le t \le \frac{1}{2} \\ \displaystyle 0 & \text{otherwise} \end{cases} \end{aligned} $$
I am to evaluate these at the endpoints $t= -\frac{1}{4}$ and $t=+\frac{1}{4}$. So, I substitute each endpoint into the expression above. For example, $t = \frac{1}{4}$ yields $$ \begin{aligned} u\left( \frac{1}{4} \right) &= \sum_{k=-\infty}^\infty \frac{1}{\pi k} \sin {\frac{\pi k} {2} } \exp {j2\pi k \frac{1}{4} } \\ &= \sum_{k=-\infty}^\infty \frac{1}{\pi k} \sin {\frac{\pi k} {2} } \exp {\frac{j\pi k}{2} } \end{aligned} $$
In this expression, for $k$ even, the terms in the summand are zero, since the sine becomes zero. For $k$ odd, when I work through it:
$$ \begin{aligned} k = +1: &\frac{1}{\pi \cdot(+1)} \sin {\frac{\pi \cdot(+1)} {2} } \exp \left( {j \frac{\pi}{2} \cdot(+1) } \right) = +\frac{1}{\pi} \sin {\frac{+\pi} {2} } \exp {j \frac{+\pi}{2} } = +\frac{1}{\pi} \cdot (+1) \cdot(+j) = +\frac{j}{\pi} \\ k = -1: &\frac{1}{\pi \cdot(-1)} \sin {\frac{\pi \cdot(-1)} {2} } \exp \left( {j \frac{\pi}{2} \cdot(-1) } \right) = -\frac{1}{\pi} \sin {\frac{-\pi} {2} } \exp {j \frac{-\pi}{2} } = -\frac{1}{\pi} \cdot (-1) \cdot(-j) = -\frac{j}{\pi} \\ \\ k = +3: &\frac{1}{\pi \cdot(+3)} \sin {\frac{\pi \cdot(+3)} {2} } \exp \left( {j \frac{\pi}{2} \cdot(+3) } \right) = +\frac{1}{3\pi} \sin {\frac{+3\pi} {2} } \exp {j \frac{+3\pi}{2} } = +\frac{1}{3\pi} \cdot (-1) \cdot(-j) = +\frac{j}{3\pi} \\ k = -3: &\frac{1}{\pi \cdot(-3)} \sin {\frac{\pi \cdot(-3)} {2} } \exp \left( {j \frac{\pi}{2} \cdot(-3) } \right) = -\frac{1}{3\pi} \sin {\frac{-3\pi} {2} } \exp {j \frac{-3\pi}{2} } = -\frac{1}{3\pi} \cdot (-1) \cdot(-j) = -\frac{j}{3\pi} \\ \\ \end{aligned} $$
And so on. I see cancellations on either side, resulting in zero.
What am I missing? According to the text, it is supposed to converge to $\frac{1}{2}$.