Learn 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.