# An Integral Result from Parseval’s Theorem

**Introduction:**

In this Insight article, Parseval’s theorem will be applied to a sinusoidal signal that lasts a finite period of time. It will be shown that it necessarily follows that ## (\frac{\sin(2 x_o)}{x_o})( \frac{\pi}{2})=\int\limits_{-\infty}^{+\infty} \frac{\sin(x-x_o) \sin(x+x_o)}{(x-x_o)(x+x_o)} \, dx ##.

Note: This integral result was computed by this author in 2009. I have not done a complete literature search, but I anticipate this result is probably already known.

**Parseval’s Theorem:**

Parseval’s theorem says ## \int\limits_{-\infty}^{+\infty} V^2(t) \, dt=\frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty} |\tilde{V}(\omega)|^2 \, d \omega ##.

**Calculations:**

Let ## V(t)=\sin(\omega_o t) ## for ## 0 \leq t \leq T ##, (where ## T ## is some completely arbitrary time interval), and ## V(t)=0 ## otherwise.

Let’s compute the Fourier transform ## \tilde{V}(\omega) ## :

## \tilde{V}(\omega)=\int\limits_{-\infty}^{+\infty} V(t)e^{-i \omega t} \, dt ##.

## \tilde{V}(\omega)=\int\limits_{0}^{T} \frac{1}{2i} (e^{i \omega_o t}-e^{-i \omega_o t})e^{-i \omega t} \, dt ##.

## \tilde{V}(\omega)=\frac{1}{2i} \int\limits_{0}^{T} [e^{-i (\omega-\omega_o)t}-e^{-i (\omega+\omega_o)t}] \, dt ##.

## \tilde{V} (\omega)=\frac{1}{2} (\frac{(e^{-i(\omega-\omega_o)T}-1)}{\omega-\omega_o}-\frac{(e^{-i(\omega+\omega_o)T}-1)}{\omega+\omega_o} ) ##.

The expression for ## \tilde{V}^*(\omega) \tilde{V} (\omega) ## is quite lengthy, and parts of the expansion separates into terms whose integrals are readily evaluated. The result is ## \int\limits_{-\infty}^{+\infty} |\tilde{V}(\omega)|^2 \, d \omega= \int\limits_{-\infty}^{+\infty} \tilde{V}^*(\omega)\tilde{V}(\omega) \, d \omega=\pi T-\cos(\omega_o T) \int\limits_{-\infty}^{+\infty} \frac{2 \sin[(\omega-\omega_o)T/2]\sin[(\omega+\omega_o)T/2]}{(\omega-\omega_o)(\omega+\omega_o)} \, d \omega ##, where the result for the integral on the right side still needs to be determined.

(Note: The numerator of the integral resulted from a term of the form ## \cos(\omega T)-\cos(\omega_o T) ##).

Meanwhile ## \int\limits_{-\infty}^{+\infty} V(t)^2 \, dt=\int\limits_{o}^{T} \sin^2(\omega_o t) \, dt=\frac{T}{2}-\frac{\sin(2 \omega_o T)}{4 \omega_o} ##.

Using Parseval’s theorem,

## \int\limits_{-\infty}^{+\infty} V^2(t) \, dt=\frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty} |V(\omega)|^2 \, d \omega ##, and using the identity ## \sin(2 \omega_o T)=2 \sin(\omega_o T) \cos(\omega_o T) ##, the result is that the following equality must hold:

## (\frac{\sin(\omega_o T)}{\omega_o})(\frac{\pi}{2})=\int\limits_{-\infty}^{+\infty} \frac{\sin[(\omega-\omega_o)T/2] \sin[(\omega+\omega_o)T/2]}{(\omega-\omega_o)(\omega+\omega_o)} \, d \omega ##.

**Parseval’s Theorem thereby gives this Integral Result:**

If we let ## T=2 ##, and change ## \omega ## to ## x ##, we get the integral result stated at the beginning of the article:

## (\frac{\sin(2 x_o)}{x_o})(\frac{\pi}{2})=\int\limits_{-\infty}^{+\infty} \frac{\sin(x-x_o) \sin(x+x_o)}{(x-x_o)(x+x_o)} \, dx ##.

**Comparing to another somewhat well-known integral:**

In the limit that ## x_o \rightarrow 0 ##, this gives the somewhat well-known result that

## \pi=\int\limits_{-\infty}^{+\infty} \frac{\sin^2(x)}{x^2} \, dx ##.

**Conclusion:**

It is not known whether this integral has any immediate applications, but it is hoped the reader finds the result of interest. Parseval’s theorem turned out to be quite useful for generating this result.

**Additional remarks:**

Perhaps there is a way to get this same result for this integral by an application of the residue theorem or some other similar technique. Any feedback is welcome, and if there is another way to get this same result, this author would welcome seeing the computation.

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.

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. :) :)

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 ##. A Taylor expansion shows the only pole is at ## z=-x_o ## (from the ## (z+x_o) ## term in the denominator). The contour will go along the x-axis with an infinitesimal hemisphere loop over the pole and will be closed in the upper half complex plane. (I needed to consult my complex variables book for this next part:). The infinitesimal hemisphere loop in the clockwise direction results in ## -B_o pi i ## where ## B_o ## is the residue of the function at ## z=-x_o ##. This results in ## -pi frac{sin(2x_o)}{2x_o} ## for the portion containing the infinitesimal hemisphere 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.