So first I am showing that the series below converges uniformly: $$\sum_{n=1}^\infty \frac{\sin(n^2x)}{n^4}$$
I can do this by using Weierstrass' M-test i.e. I let $\sum_{n=1}^\infty \ \frac{\sin(n^2x)}{n^4} = f_n$ where $f_n : A \to R$ for $n \ \epsilon \ N$ and we suppose that $M_n$ exists such that $$ |f_n(x)| \leq M_n $$ for all $ x \ \epsilon \ A$
Using this result and that $|\sin(n^2x)| \leq 1 $, I have that: $$ \biggl| \frac{\sin(n^2x)}{n^4}\biggr|= \frac{|\sin(n^2x)|}{n^4} \leq \frac{1}{n^4}$$
(I am unsure if the above calculation that I have done is correct, can anyone tell me if I have made any errors in this?)
Continuing on from here I can say that $\frac{1}{n^4} = M_n$ and we know that $\sum^\infty_{n=1} \frac{1}{n^4}$ converges which means that $\sum_{n=1}^\infty \ \frac{\sin(n^2x)}{n^4}$ does in fact converge uniformly on $R$
If we call $\sum_{n=1}^\infty \ \frac{\sin(n^2x)}{n^4} = f_n(x)$ I now want to define $f(x)$ as the sum of that series. After that I plan on determining whether $f(x)$ is integrable or not, this is not a problem however I am unsure of how to work out $f(x)$ (i.e. find the sum of the series). Can anyone help with this?
First of all, $f_n(x)$ is the term of the series: $f_n(x)=\frac{\sin(n^2x)}{n^4}$. And $f(x)$ is the sum: $f(x)=\sum_{n=1}^{\infty}f_n(x)$. You've correctly showed that the series is uniformly convergent on $(-\infty,\infty)$ using the Weierstrass M-test.
As for integrability, you don't need the explicit form of $f(x)$ (in fact, there may be no nice closed-form). In any interval $[a,b]$, $f_n(x)$ is continuous for all $n=1,2,3,\dots$. This implies $f(x)$ is continuous on $[a,b]$ and therefore integrable. Furthermore, $$\int_a^bf(x)\,\mathrm{d}x=\sum_{n=1}^{\infty}\int_a^bf_n(x)\,\mathrm{d}x $$ So you can integrate termwise and then sum.