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An Integral Result from Parseval's Theorem

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In summary, Charles Link has submitted a new PF Insights post discussing an integral result from Parseval's Theorem. He explains that the result is similar to what would be obtained from residue theory, but it can also be solved using the Parseval method. Charles also welcomes alternative solutions and clarifies the statement of Parseval's theorem for real and complex-valued functions.

- #1

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An Integral Result from Parseval's Theorem

Continue reading the Original PF Insights Post.

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Thanks Charles! Some day I will understand this :)

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Hi @Greg BernhardtGreg Bernhardt said:Thanks Charles! Some day I will understand this :)

I will be very pleased if someone comes up with an alternative method to evaluate this integral. I am pretty sure I computed it correctly, but I would really enjoy seeing an alternative solution. :) :)

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I believe I have now also succeeded at evaluating this integral by use of the residue theorem: ## \\ ## The integral can be rewritten as ## I=-\frac{1}{2} \int Re[ \frac{e^{i2z}[1-e^{-i2(z-x_o)}]}{(z-x_o)(z+x_o)}] \, dz ##. ## \\ ## (Using ## sin(x-x_o)sin(x+x_o)=-\frac{1}{2}[cos(2x)-cos(2x_o)] ## and then writing ## cos(2x) ## as the real part of ## e^{i2x} ##). ## \\ ## A Taylor expansion shows the only pole is at ## z=-x_o ## (from the ## (z+x_o) ## term in the denominator. ## \\ ## (Note that ## e^{-i2(z-x_o)}=1-2i(z-x_o)+higher \, order \, terms \, in \, (z-x_o) ##. Thereby, there is no pole at ## z=x_o ##). ## \\ ## The contour will go along the x-axis with an infinitesimal semi-circle loop over the pole at ## z=-x_o ##, and will be closed in the upper half complex plane. (I needed to consult my complex variables book for this next part:). The infinitesimal semi-circle loop over the pole in the clockwise direction results in ## -B_o \pi i ## where ## B_o ## is the residue of the function at ## z=-x_o ##. ## \\ ## Upon evaluating the residue ## B_o ##, this results in ## -\pi \frac{sin(2x_o)}{2x_o} ## for the portion containing the infinitesimal semi-circle loop over the pole (which is not part of the integration of the function along the x-axis, where the entire function is considered to be well-behaved.). Since the contour doesn't enclose any poles, the complete integral around it must be zero, so the functional part along the x-axis must be equal to ## \pi \frac{sin(2x_o)}{2x_o} ##, which is the result that we anticipated.

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Delta2

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If ##V(t)## is real-valued (and in this article it is )then it is correct as it is stated

if ##V(t)## is complex-valued then in the statement of theorem it has to be put inside norm, that is ##\int_{-\infty}^{\infty}|V(t)|^2dt=\int_{-\infty}^{\infty}|\tilde{V}(\omega)|^2d\omega##

Parseval's Theorem is a mathematical result in Fourier analysis that relates the energy of a signal in the time domain to its frequency domain representation. It states that the sum of the squared magnitudes of the signal in the time domain is equal to the integral of the squared magnitudes of its Fourier transform.

In this integral result, Parseval's Theorem is used to simplify the integration of a function over a finite interval by using its Fourier transform. This allows for the calculation of the integral in a more efficient and accurate manner.

The integral result from Parseval's Theorem has many applications in various fields of science and engineering. It allows for the efficient calculation of integrals, which is an essential tool in data analysis, signal processing, and other areas of research. It also provides a deeper understanding of the relationship between signals in the time and frequency domains.

Yes, this integral result can be extended to higher dimensions through the use of multidimensional Fourier transforms. The concept of Parseval's Theorem remains the same, but the integration and calculation methods become more complex with increasing dimensions.

While Parseval's Theorem and this integral result are powerful tools, they do have some limitations. They are only applicable to functions that are square integrable, and they assume that the Fourier transform exists for the given function. Additionally, the accuracy of the result may be affected by the choice of sampling rate and the presence of noise in the signal.

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