Show that $$f(x) = \sum_{k=1}^\infty \frac{\sin kx}{k^3} \tag1$$ converges, pointwise and uniformly on $\mathbb R$, to a differentiable function that satisfies $$\int_0^{\pi/2} f(x)\,dx = \sum_{k=1}^\infty \left(\frac{(-1)^{k-1}}{(2k)^4}+\frac{1}{k^4}\right)\tag{a}$$ and $$ |f'(x)|\le \pi/8 \tag{b} $$
I would be able to do it if it wasn't for the part "$f$ that satisfies" (a) and (b). If someone could explain it to me, that would be helpful.