4 Removing obsolete sentence edited Jun 6 '18 at 19:04 Sven B 3,41044 silver badges2424 bronze badges Expanding the sum will result in \left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\left[2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})\right] \end{align} If $$\n_1 = n_2 = n\$$, then we simply get $$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$ For the mixed terms where $$\n_1 \neq n_2\$$, we can calculate the integral using the following equality: $$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$ \begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align}\begin{align} & \int_0^T \sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_{n_1} - \gamma_{n_2}) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_{n_1} + \gamma_{n_2}) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_{n_1} - \gamma_{n_2})) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_{n_1} + \gamma_{n_2})) dt \end{align} Both of these integrals are of the form: $$\int_0^T cos(m\omega t + \phi)dt$$ with $$\m\$$ a non-zero integer (else $$\n_1 = n_2\$$), meaning that $$\T\$$ is a multiple of the period of this cosine. The latter means that the integral will result to 0. So your formula can be simplified to $$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$ The last integral should yield 1/2, leading you to the final result. Expanding the sum will result in \left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\left[2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})\right] \end{align} If $$\n_1 = n_2 = n\$$, then we simply get $$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$ For the mixed terms where $$\n_1 \neq n_2\$$, we can calculate the integral using the following equality: $$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$ \begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} Both of these integrals are of the form: $$\int_0^T cos(m\omega t + \phi)dt$$ with $$\m\$$ a non-zero integer (else $$\n_1 = n_2\$$), meaning that $$\T\$$ is a multiple of the period of this cosine. The latter means that the integral will result to 0. So your formula can be simplified to $$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$ The last integral should yield 1/2, leading you to the final result. Expanding the sum will result in \left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\left[2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})\right] \end{align} If $$\n_1 = n_2 = n\$$, then we simply get $$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$ For the mixed terms where $$\n_1 \neq n_2\$$, we can calculate the integral using the following equality: $$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$ \begin{align} & \int_0^T \sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_{n_1} - \gamma_{n_2}) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_{n_1} + \gamma_{n_2}) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_{n_1} - \gamma_{n_2})) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_{n_1} + \gamma_{n_2})) dt \end{align} Both of these integrals are of the form: $$\int_0^T cos(m\omega t + \phi)dt$$ with $$\m\$$ a non-zero integer (else $$\n_1 = n_2\$$), meaning that $$\T\$$ is a multiple of the period of this cosine. The latter means that the integral will result to 0. So your formula can be simplified to $$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$ 3 Added extra brackets [] possiblty redundant in this case but adds clarity. edited Jun 6 '18 at 18:57 Warren Hill 3,8651212 silver badges2727 bronze badges Expanding the sum will result in \left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2}) \end{align}\left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\left[2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})\right] \end{align} If $$\n_1 = n_2 = n\$$, then we simply get $$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$ For the mixed terms where $$\n_1 \neq n_2\$$, we can calculate the integral using the following equality: $$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$ \begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} Both of these integrals are of the form: $$\int_0^T cos(m\omega t + \phi)dt$$ with $$\m\$$ a non-zero integer (else $$\n_1 = n_2\$$), meaning that $$\T\$$ is a multiple of the period of this cosine. The latter means that the integral will result to 0. So your formula can be simplified to $$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$ The last integral should yield 1/2, leading you to the final result. Expanding the sum will result in \left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2}) \end{align} If $$\n_1 = n_2 = n\$$, then we simply get $$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$ For the mixed terms where $$\n_1 \neq n_2\$$, we can calculate the integral using the following equality: $$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$ \begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} Both of these integrals are of the form: $$\int_0^T cos(m\omega t + \phi)dt$$ with $$\m\$$ a non-zero integer (else $$\n_1 = n_2\$$), meaning that $$\T\$$ is a multiple of the period of this cosine. The latter means that the integral will result to 0. So your formula can be simplified to $$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$ The last integral should yield 1/2, leading you to the final result. Expanding the sum will result in \left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\left[2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})\right] \end{align} If $$\n_1 = n_2 = n\$$, then we simply get $$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$ For the mixed terms where $$\n_1 \neq n_2\$$, we can calculate the integral using the following equality: $$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$ \begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} Both of these integrals are of the form: $$\int_0^T cos(m\omega t + \phi)dt$$ with $$\m\$$ a non-zero integer (else $$\n_1 = n_2\$$), meaning that $$\T\$$ is a multiple of the period of this cosine. The latter means that the integral will result to 0. So your formula can be simplified to $$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$ The last integral should yield 1/2, leading you to the final result. 2 Clarification for the different cases edited Jun 6 '18 at 14:28 Sven B 3,41044 silver badges2424 bronze badges Any mixed product between two different sines will have an integral that results to 0. You can useExpanding the following equality for that:sum will result in $$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$\left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2}) \end{align} Like so:If $$\n_1 = n_2 = n\$$, then we simply get Assume$$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$ For the mixed terms where $$\n_1 \neq n_2\$$, thenwe can calculate the integral using the following equality: \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ \begin{align} &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align}$$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$ \begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} Both of these integrals are of the form: $$\int_0^T cos(m\omega t + \phi)dt$$ with $$\m\$$ a non-zero integer (else $$\n_1 = n_2\$$), meaning that $$\T\$$ is a multiple of the period of this cosine. The latter means that the integral will result to 0. So your formula can be simplified to $$I_0^2 + \sum_{n=1}^{\infty} 2I_n^2 \int_0^T \sin^2(n\omega t + \gamma_n)dt$$$$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$ The last integral should yield 1/2, leading you to the final result. Any mixed product between two different sines will have an integral that results to 0. You can use the following equality for that: $$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$ Like so: Assume $$\n_1 \neq n_2\$$, then \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ \begin{align} &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} Both of these integrals are of the form: $$\int_0^T cos(m\omega t + \phi)dt$$ with $$\m\$$ a non-zero integer (else $$\n_1 = n_2\$$), meaning that $$\T\$$ is a multiple of the period of this cosine. The latter means that the integral will result to 0. So your formula can be simplified to $$I_0^2 + \sum_{n=1}^{\infty} 2I_n^2 \int_0^T \sin^2(n\omega t + \gamma_n)dt$$ The last integral should yield 1/2, leading you to the final result. Expanding the sum will result in \left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2}) \end{align} If $$\n_1 = n_2 = n\$$, then we simply get $$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$ For the mixed terms where $$\n_1 \neq n_2\$$, we can calculate the integral using the following equality: $$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$ \begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} Both of these integrals are of the form: $$\int_0^T cos(m\omega t + \phi)dt$$ with $$\m\$$ a non-zero integer (else $$\n_1 = n_2\$$), meaning that $$\T\$$ is a multiple of the period of this cosine. The latter means that the integral will result to 0. So your formula can be simplified to $$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$ The last integral should yield 1/2, leading you to the final result. 1 answered Jun 6 '18 at 13:42 Sven B 3,41044 silver badges2424 bronze badges