Fourier components of $\cos(2\pi f_1t)$ I have the signal $s(t) = \cos(2\pi f_1t)$ and I am looking for its components vs the Fourier basis, over the interval $[0, T]$. The formula for computing the coefficients is
$$
s_n = \int_{t_0}^{t_1} s(t) \frac{e^{-j\frac{2\pi}{T} nt}}{\sqrt{T}} dt
$$
and in this case $t_0 = 0$ and $t_1 = T$. The solution seems to be
$$
\frac{\sqrt{T}}{2}(-1)^n\bigg[\mathrm{sinc}\big(f_1T+n\big)e^{-j\pi f_1T} + \mathrm{sinc}\big(f_1T-n\big)e^{j\pi f_1T}\bigg]
$$
but i can get no further than this
$$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)$$
HINT:

Going from your last equation, 
$$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)$$
This can be simplified further down by considering the following:

\begin{align}
e^{j2\pi (f_1T-n)} &amp;= e^{j\pi (f_1T-n)}\cdot e^{j\pi (f_1T-n)}\\
1 &amp;=e^{j\pi (f_1T-n)}\cdot e^{-j\pi (f_1T-n)}\\
e^{-j\pi n} &amp;= (-1)^n\\
\end{align}
And that
\begin{align}
\mathrm{sinc}(x) &amp;= \frac{\sin(x)}{x}\\
\sin(x) &amp;= \frac{e^{jx} - e^{-jx}}{2j}
\end{align}