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I'm reading a monograph where the following improper integral has infinite limits of integration (see pg. 125 of Seismic Inverse Q Filtering for $h(t)$ as Equation 7.19, and pg. 123 as the window function $w(t)$ as Equation 7.12):

$h(t) = {\left[ {\int\limits_{ - \infty }^\infty w (t - \tau )d\tau } \right]^{ - 1}}$

In the expression above, $t$ is the time (in seconds), and $\tau$ (in seconds) is the center of a window function that is slid across a time series. This implies that when $w(t - \tau)$ is numerically evaluated, for a given timestep $\Delta t$ (in seconds), and a given number $N = 11$ of elements, $t - \tau$ will always be a sequence with 11 elements: $$\text{targ} = \quad -0.0210\quad -0.0168\quad -0.0126\quad -0.0084\quad -0.0042\quad 0 \quad 0.0042 \quad 0.0084\quad 0.0126\quad 0.0168\quad 0.021$$

Using Matlab, I've computed $w(t-\tau)$ as the following discrete sequence (with 11 elements): $$w_t = 0.9841 \quad 4.1538 \quad 12.7307 \quad 28.3327 \quad 45.7877\quad 53.7323\quad 45.7877 \quad 28.3327 \quad 12.7307\quad 4.1538 \quad 0.9841$$

What I don't understand here is what is meant by the infinite limits of integration in the context of numerical implementation. Does $(-\infty, \infty)$ imply that the integration is over the 11 elements in the discrete sequence, or is the integration over the entire time series (with a number of elements much greater than $N=11$)?

The answer to this question can be better understood by looking at the window function:

$w(t) = \left\{ {\begin{array}{*{20}{c}} {\frac{2}{{T\sqrt \pi }}\exp \left[ { - {{\left( {\frac{{2t}}{T}} \right)}^2}} \right],{\rm{ for }} - T \le t \le T}\\ {0,{\rm{ otherwise}}} \end{array}} \right.$

In the above expression, the area under the curve of $w(t)$ will be zero outside of the half-width interval $[-T,T]$. This area will contribute nothing to the indefinite integration. I wrote the following Matlab code to numerically demonstrate this concept. I use trapezoidal integration to perform the numerical integration.

deltat = 0.1;
t = deltat * 5;
N = 50;
tau = linspace(-10,10, N);
arg = t - tau;

w_t = zeros(N, 1);
T = 5;

term1 = 2 / (T * sqrt(pi));
for i = 1:N

    if(arg(i) >= -T && arg(i) <= T)
        term2 = -((2 * arg(i)) / T)^2;
        term3 = exp(term2);
        term4 = term1 * term3;
        w_t(i) = term4;
    else
        w_t(i) = 0;
    end

end

plot(arg, w_t);
h_t = trapz(tau, w_t); 
h_tinv = 1 / h_t;

In this case, I changed the limits of integration to be sufficiently "big":

$h(t) = {\left[ {\int\limits_{ -10 }^{10} w (t - \tau )d\tau } \right]^{ - 1}}$

To me, this makes more sense than the infinite limits of integration as written in the original monograph.

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1 Answer 1

up vote 1 down vote accepted

Well, in your case the infinite limits make small sense: $$ \int\limits_{-\infty}^\infty w(t-\tau)\mathrm d\tau = \left/\begin{split}s &= t-\tau, \\\mathrm ds& = -\mathrm d\tau\end{split}\right/ = -\int\limits_\infty^{-\infty}w(s)\mathrm ds = \int\limits_{-\infty}^\infty w(s)\mathrm ds $$ which does not depend on the center of the window $t$. So, you should just implement numerical integration once, cutting limits and $\pm E$ for $E$ sufficiently big - maybe if you have some additional information, you can even upper-bound the computational error.

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Thank you; this is an interesting and good observation. But what I still don't understand is the intent of the monograph author. Perhaps this can be inferred from the original text? (I've provided a link in my question above.) Does the author want integration over the 11 elements, or over the entire time series? How do I make $E$ sufficiently "big"? –  Nicholas Kinar Jun 26 '12 at 16:39
    
I've now updated my question to demonstrate making the limits sufficiently "big" –  Nicholas Kinar Jun 27 '12 at 21:53

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