# integration of a function

I found this explanation in a journal paper but I could not understand it. Can someone give me an explanation or possibly a proof that:

If $$\frac{\mathrm{d}V(t)}{\mathrm{d}t}=\sqrt{2}\sum_{h=1}^{H}h\omega V_{h}\cos\left(h\omega t+\frac{\pi }{2}\right),$$ then why integration over whole period is: $$\frac{1}{T}\int_{0}^{T} \left( \frac{\mathrm{d} V(t)}{\mathrm{d} t} \right)^{2}dt=\omega \sum_{h=1}^{H}h^{2}V_{h}^{2}.$$

I have problem with the power of $\omega$; my solution returns $\omega^2$, while the power of $\omega$ in answer is one. Here is my solution: $$\frac{1}{T}\int_{0}^{T}\ \left( \frac{dV}{dt} \right)^{2}dt=\frac{2\omega ^{2}}{T}\int_{0}^{T}\sum_{h=1}^{H}h^{2}V_{h}^{2}\sin^{2}(h\omega t)dt$$ and over whole period: $$\frac{1}{T}\int_{0}^{T}\sin^{2}(h\omega t)dt=\frac{1}{2}$$ then we will have $$\omega ^{2}\sum h^{2}V_{h}^{2}$$ not
$$\omega \sum h^{2}V_{h}^{2}$$

Why?

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+1 for thinking about what you read and showing your work –  Ross Millikan Feb 10 '11 at 14:12
Is the fact that $\frac{1}{T}\int_0^T\sin^2(h\omega t)\,dt = \frac{1}{2}$ a given, or something you computed? If the latter, how? –  Arturo Magidin Feb 10 '11 at 14:28
@Arturo Magidin, e.g. in electrical engineering $\omega=\frac{2\pi}{T}$, where $T$ is the period and $\omega$ the angular frequency (in radians/s) of a sinus wave. –  Américo Tavares Feb 10 '11 at 16:39
@Americo: Again, thank you. Context is everything; my first thought when I see $\omega$ is "the first infinite ordinal", my second is "it's just a variable, then". Don't usually think of angular frequency. (-: –  Arturo Magidin Feb 10 '11 at 16:51
@user6856, Your formula is correct because if $m\neq n$, then $\int_{0}^{2\pi }\sin nx\sin mxdx=0$ –  Américo Tavares Feb 10 '11 at 18:38

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

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

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Thank you for confirmation, but I don't think that it is a typo, because he substituted this result into another equation which is correct. is it possible that I send the paper for you? –  user6856 Feb 11 '11 at 5:50
@ user6856: Yes. Please see the email-address shown on my profile page. –  Américo Tavares Feb 11 '11 at 9:56
@ user6856: Is $h$ dimensionless? If it is, the formula makes sense only in case it has $\omega ^2$ on the RHS (assuming $\omega$ is an angular frequency, $V$ a voltage and $t$ the time). –  Américo Tavares Feb 11 '11 at 11:04
@Américo: Dear Américo, only moderators can see the email address entered in your profile. –  Akhil Mathew Feb 11 '11 at 14:33
@Akhil: Dear Akhil, thanks for the information. –  Américo Tavares Feb 11 '11 at 18:44

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: $\sum (h\omega V_h)^2$.

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Somewhat related stuff: math.stackexchange.com/q/7391/1242 and math.stackexchange.com/q/19876/1242. –  Hans Lundmark Feb 10 '11 at 20:07