Evaluation of a sinc integral

[chroot]

Can anyone show me how to evaluate an integral like this by hand? I believe such integrals have an analytic solution, but I can't figure out how to find them. Mathematica seems unable to help (the Integrate command runs forever) but I believe this can be done by hand. It's a sort of integral commonly found in communications theory. I actually don't think it's supposed to be very hard...


\int_{ - \infty }^\infty {\left[ {\frac{1}
{{\sqrt T }}\operatorname{sinc} \left( {\frac{t}
{T}} \right) - \frac{1}
{{2\sqrt T }}\operatorname{sinc} \left( {\frac{{t - T}}
{T}} \right)} \right]^2 dt}


where


\operatorname{sinc} \left( t \right) \triangleq \frac{{\sin \left( {\pi t} \right)}}
{{\pi t}}


Obviously I can expand the binomial out, but I'm left with products of sinc's with different arguments, and I don't know how to continue.

- Warren

[chroot]

Actually, the identity


\operatorname{sinc} \left( {\frac{{t + mT}}
{T}} \right) \cdot \operatorname{sinc} \left( {\frac{{t + nT}}
{T}} \right) = T\operatorname{sinc} \left( {\frac{{t + (m + n)T}}
{T}} \right)


makes life much sweeter. I got 1/4. Anyone else care to check my work?

- Warren

[Tom Mattson]

Using the identities you posted I got \pi T / 4. I squared out the integrand and split up the integral into 3 integrals. The first two exactly canceled each other out, leaving only the third one:

\frac{1}{4}\int_{ - \infty }^\infty \operatorname{sinc}\left(\frac{t-2T}{T}\right)dt

U-substitution yields:

\frac{T}{4}\int_{ - \infty }^\infty \operatorname{sinc}(u)du

The integral (sans outside coefficient) evaluates to \pi.

Regarding Mathematica, MathWorld has this to say in its article on the sinc function:


This function will be implemented in a future version of Mathematica as Sinc[x].


So I guess you'll have to wait a bit longer to get the computer to do it for you.

[chroot]

You can define your own piecewise Sinc function in Mathematica quite easily.

As it turns out, though, Tom, we're both wrong, but I can't exactly say why. I followed your method similarly, finding two terms which cancelled, leaving the third.

However, there's a much simpler way, subject to fewer mistakes. Note that all terms of the form


\frac{a}
{{\sqrt T }}\operatorname{sinc} \left( {\frac{{t - nT}}
{T}} \right)


are normalized, and integrate to a.

You can literally read the value of this integral directly off the coefficients a, if you consider the integrand to be a vector multiplication like this:


\frac{{a_1 }}
{{\sqrt T }}\operatorname{sinc} \left( {\frac{t}
{T}} \right) + \frac{{a_2 }}
{{\sqrt T }}\operatorname{sinc} \left( {\frac{{t - T}}
{T}} \right) \Leftrightarrow \left[ {\begin{array}{*{20}c}
{a_1 } & {a_2 } \\

\end{array} } \right] \cdot \left[ {\begin{array}{*{20}c}
1 \\
1 \\

\end{array} } \right]


The value of the integral, then, is the squared norm of this vector. In this case, the result is 5/4. I'm still not exactly sure why integrating this long-hand produces a different value, but I'm pretty sure I just flubbed some algebra.

I've learned something in this communications class: if anything looks hard, you're doing it wrong. Thanks for the help though!

- Warren

[Tom Mattson]

As it turns out, though, Tom, we're both wrong, but I can't exactly say why. I followed your method similarly, finding two terms which cancelled, leaving the third.


I am certain that I integrated correctly. That being the case I decided to take a closer look at the identity you posted, since I started from that.


\operatorname{sinc} \left( {\frac{{t + mT}}
{T}} \right) \cdot \operatorname{sinc} \left( {\frac{{t + nT}}
{T}} \right) = T\operatorname{sinc} \left( {\frac{{t + (m + n)T}}
{T}} \right)


Let m=0 and n=-1. Then you get:

sinc(t/T)sinc((t-T)/T)=Tsinc((t-T)/T)

Now let t=0:

sinc(0)sinc(-1)=Tsinc(-1)

Since T is arbitrary and sinc(-1) does not equal zero, the equation above is not an identity.

[chroot]

Whoops! I posted my identity incorrectly.

It should be


\operatorname{sinc} \left( {\frac{{t + mT}}
{T}} \right) \,\star\, \operatorname{sinc} \left( {\frac{{t + nT}}
{T}} \right) = T\operatorname{sinc} \left( {\frac{{t + (m + n)T}}
{T}} \right)


Convolution, not multiplication. :blushing:

- Warren