integration of a function

Solution 1:

Your solution is right. It should be a typo in the paper. Here is my evaluation confirming yours. Since

$$\begin{eqnarray*} \frac{dV(t)}{dt} &=&\sqrt{2}\sum_{h=1}^{H}h\omega V_{h}\cos \left( h\omega t+% \frac{\pi }{2}\right) \\ &=&-\sqrt{2}\sum_{h=1}^{H}h\omega V_{h}\sin h\omega t, \end{eqnarray*}$$

and assuming $\omega$ is the angular frequency given by

$$\omega =\frac{2\pi }{T},$$

we have

$$\left( \frac{dV(t)}{dt}\right) ^{2}=2\omega ^{2}\left( \sum_{h=1}^{H}hV_{h}\sin \left( h\omega t\right) \right) ^{2}$$

and

$$\begin{eqnarray*} \frac{1}{T}\int_{0}^{T}\left( \frac{dV(t)}{dt}\right) ^{2}dt &=&\frac{\omega }{2\pi }\int_{0}^{2\pi /\omega }\left( \frac{dV(t)}{dt}\right) ^{2}dt \\ &=&\frac{\omega }{2\pi }\int_{0}^{2\pi /\omega }2\omega ^{2}\left( \sum_{h=1}^{H}hV_{h}\sin \left( h\omega t\right) \right) ^{2}dt \\ &=&\frac{\omega ^{3}}{\pi }\int_{0}^{2\pi /\omega }\left( \sum_{h=1}^{H}hV_{h}\sin \left( h\omega t\right) \right) ^{2}dt. \end{eqnarray*}$$

The integrand $\left( \sum_{h=1}^{H}hV_{h}\sin \left( h\omega t\right) \right) ^{2}$ is a sum of terms of two different types:

i) $h^{2}V_{h}^{2}\sin ^{2}\left( h\omega t\right) $ and

ii) $k\left( pV_{p}\sin \left( p\omega t\right) \cdot qV_{q}\sin \left( q\omega t\right) \right) \,$, with $p\neq q$ and $p,q,k\in\mathbb{N}$.

The second type terms do not contribute to the last integral, because the $\sin nx$ ($n\in\mathbb{N}$) functions form an orthogonal system over $[0,2\pi ]$:

$$\int_{0}^{2\pi /\omega }k\left( pV_{p}\sin \left( p\omega t\right) \cdot qV_{q}\sin \left( q\omega t\right) \right) dt=0\quad p\neq q$$

The sum of the first type ones is $\sum_{h=1}^{H}h^{2}V_{h}^{2}\sin ^{2}\left( h\omega t\right) $. Thus

$$\begin{eqnarray*} \frac{1}{T}\int_{0}^{T}\left( \frac{dV(t)}{dt}\right) ^{2}dt &=&\frac{\omega ^{3}}{\pi }\int_{0}^{2\pi /\omega }\sum_{h=1}^{H}h^{2}V_{h}^{2}\sin ^{2}\left( h\omega t\right) dt \\ &=&\frac{\omega ^{3}}{\pi }\sum_{h=1}^{H}h^{2}V_{h}^{2}\int_{0}^{2\pi /\omega }\sin ^{2}\left( h\omega t\right) dt \\ &=&\frac{\omega ^{3}}{\pi }\sum_{h=1}^{H}h^{2}V_{h}^{2}\cdot \frac{\pi }{% \omega } \\ &=&\omega ^{2}\sum_{h=1}^{H}h^{2}V_{h}^{2}, \end{eqnarray*}$$

because

$$\begin{eqnarray*} \int \sin ^{2}\left( h\omega t\right) dt &=&\frac{1}{h\omega }\left( -\frac{1% }{2}\cos h\omega t\sin h\omega t+\frac{1}{2}h\omega t\right) \\ \int_{0}^{2\pi /\omega }\sin ^{2}\left( h\omega t\right) dt &=&\frac{\pi }{% \omega }. \end{eqnarray*}$$

Solution 2:

I agree, it should be $\omega^2$. The whole thing is in fact just the Pythagorean theorem: the functions $\sqrt{2} \cos(\dots)$ are orthonormal in the space $L^2([0,T])$, and the integral is the square of the $L^2$ norm of $dV/dt$, hence the sum of the squares of the coefficients: $\sum (h\omega V_h)^2$.