This question deals with finding the Nyquist Frequency of a given signal.
Suppose you have the signal $x(t)=\sum_{n=1}^{3}\sin(2\pi \frac{n}{8}\frac{t}{T})$ in the time domain where $T>0$ is some positive number.
Our goal is to find the nyquist frequency, or in other words, find $u_0$ such that $\forall u>u_0: \hat x(u)=0$ where $\hat x(u)$ is the fourier transform of $x(t)$.
If i understand correctly, to do that we need to find the fourier transform, and I have a few issues:
$\hat x(u) = \int_{\mathbb R}x(t)e^{-2\pi iut}dt=\int_{\mathbb R}\sum_{n=1}^{3}\sin(2\pi \frac{n}{8}\frac{t}{T})e^{-2\pi iut}dt=\sum_{n=1}^{3}\int_{\mathbb R}\sin(2\pi \frac{n}{8}\frac{t}{T})e^{-2\pi iut}dt$
We can use euler's identity:$sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$ to get:
$$\sum_{n=1}^{3}\int_{\mathbb R}\sin(2\pi \frac{n}{8}\frac{t}{T})e^{-2\pi iut}dt=\sum_{n=1}^{3}\int_{\mathbb R}\frac{e^{2\pi i\frac{n}{8}\frac{t}{T}}-e^{-2\pi i\frac{n}{8}\frac{t}{T}}}{2i}e^{-2\pi iut}dt$$
after multiplying the exponents, we will see the above integral is equal to $$\sum_{n=1}^{3}\int_{\mathbb R}\frac{e^{2\pi i t(\frac{n}{8t}-u)}-e^{-2\pi i t(\frac{n}{8t}-u)}}{2i}dt=\sum_{n=1}^{3}\int_{\mathbb R}\sin[2\pi t(\frac{n}{8T}-u)]dt$$
The last equality is true again from Euler's identity.
But the problem is that this integral does not converge. It is an improper integral over $\sin$ function...How do we evaluate this in order to find the fourier transform and find the nyquist frequency?
Unless your signal goes on forever, there is no frequency beyond which the FT is zero. Usually, the Nyquist frequency is the frequency beyond which there is no significant energy, where the definition of significant is to be determined by the user/designer/requirements. In the example you gave, this would just be $3/8T$, the largest frequency in the sum of sinusoids.
However, if you give bounds to your signal in the time domain such that $x(t)=0$, $t \notin [-T,+T]$, you'll find the integral is solvable and each $\sin$ function will transform to two $\operatorname{sinc}$ functions centered on the positive and negative frequency of the sinusoid. The larger $T$ is, the narrower the $\operatorname{sinc}$ functions will be and as $T\rightarrow \infty$, the $\operatorname{sinc}$'s will approach impulse functions.