Integration time period

By the problem statement, this integral must have the same integral for all $n$ . If we substitute $n=1$ , the integrand becomes 0 so the answer is 0.

Let $f(x) = \cos(\frac{2\pi}{T}x), F(x) = \frac{T}{2\pi} \sin(\frac{2\pi}{T}x)$ . The above definite integral evaluates to:

$\Large \int_{0}^{T} F(nx) - F(x) - f(x) \cdot \frac{(n-1)T}{2} dx = \int_{0}^{T} \frac{T}{2\pi} \sin(\frac{2\pi}{T}nx) - \frac{T}{2\pi}\sin(\frac{2\pi}{T}x) - \cos(\frac{2\pi}{T}x )\cdot \frac{(n-1)T}{2} dx$ ;

or $\Large -\frac{T}{2\pi n}\cos(\frac{2\pi}{T} nx) - \frac{T}{2\pi}\cos(\frac{2\pi}{T}x)- \frac{T}{2\pi} \sin(\frac{2\pi}{T}x) \cdot \frac{(n-1)T}{2}|_{0}^{T};$

or $\Large -\frac{T}{2\pi n}[\cos(2n\pi) - \cos(0)] - \frac{T}{2\pi}[\cos(2\pi) - \cos(0)] - \frac{(n-1)T^{2}}{4\pi}[\sin(2\pi) - \sin(0)]$ ;

or $\Large \boxed{0}.$