An Integral Result from Parseval's Theorem - Comments

[Charles Link]

Charles Link submitted a new PF Insights post

An Integral Result from Parseval's Theorem

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[Charles Link]

Additional comments: The result for this integral is similar to the type of result that would be obtained from residue theory if the integrand gets evaluated at the "pole" at $x=x_o$. Alternatively, with the transformations $y=x-x_o$ and $y=\pi u$, in the limit $T \rightarrow +\infty$, $\frac{\sin(\pi u T)}{\pi u }=\delta(u)$, but this only gives a result for large $T$ for an evaluation of the integral. So far, the Parseval method is the only way I have of solving it.

 

[Greg Bernhardt]

Thanks Charles! Some day I will understand this :)

 

[Charles Link]

Thanks Charles! Some day I will understand this :)

Hi @Greg Bernhardt
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. :) :)

 

[Charles Link]

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.

 

[Delta2]

Just a small addition regarding the statement of Parseval's theorem:
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 $\int_{-\infty}^{\infty}|V(t)|^2dt=\int_{-\infty}^{\infty}|\tilde{V}(\omega)|^2d\omega$