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I am reading a monograph on seismic filtering techniques (link). The author rewrites Equation (7.3) in this monograph as Equation (7.4). By adding integrals to the equation as in Equation (7.4), what is the conceptual difference between the equations, and what is being implied by the use of the integrals? Is this associated in some way with the $\Delta \tau$ term?

Equation (7.3) is:

$U(\tau + \Delta \tau ,\omega ) = U(\tau ,\omega )\exp \left[ {\frac{\omega \Delta \tau }{{2Q(\tau)}}} \right]\exp \left[ {i{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{ - \gamma (\tau)}}\omega \Delta \tau } \right]$

Equation (7.4) is (note the similarity):

$U(\tau ,\omega ) = U(0,\omega )\exp \left[ {\int\limits_0^\tau {\frac{\omega }{{2Q(\tau ')}}d\tau '} } \right]\exp \left[ {i\int\limits_0^\tau {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{ - \gamma (\tau ')}}} \omega d\tau '} \right]$

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Denote $$\lambda(\tau)=\dfrac{\omega}{2Q(\tau)}+i\left(\dfrac{\omega}{\omega _h} \right)^{ - \gamma (\tau)}\omega{\,} \tag{*}$$ and rewrite Equation (7.3) changing $\tau$ by $\tau'$: $$ U(\tau' + \Delta \tau' ,\omega ) = U(\tau' ,\omega )\exp \left[\lambda(\tau') \Delta \tau' \right].$$ Taking logarithm, we have \begin{gather} \ln{U(\tau' + \Delta \tau' ,\omega )}=\ln{U(\tau' ,\omega )}+\lambda(\tau') \Delta \tau'; \\ \ln{U(\tau' + \Delta \tau' ,\omega )}- \ln{U(\tau' ,\omega )}=\lambda(\tau') \Delta \tau';\\ \frac{\ln{U(\tau' + \Delta \tau' ,\omega )}- \ln{U(\tau' ,\omega )}}{\Delta \tau'}=\lambda(\tau') \end{gather} By taking limit for $\Delta \tau' \rightarrow{0}:$ \begin{gather} \frac{d{\ \ln{U(\tau' ,\omega )}}}{d\tau'}=\lambda(\tau') \end{gather} or \begin{gather} {d{\ \ln{U(\tau' ,\omega )}}}=\lambda(\tau'){d\tau'}, \end{gather} thus, integrating by $\tau'$ from $0$ to $\tau$ we have \begin{gather} \ln{U(\tau' ,\omega )}\vert_{0}^{\tau}=\int\limits_{0}^{\tau}\lambda(\tau'){d\tau'}; \\ \ln{U(\tau ,\omega )}-\ln{U(0 ,\omega )}=\int\limits_{0}^{\tau}\lambda(\tau'){d\tau'}; \\ U(\tau ,\omega )=U(0 ,\omega )\exp\left[\int\limits_{0}^{\tau}\lambda(\tau'){d\tau'} \right]. \end{gather} Finally, substitution $(*)$ gives the desired result.

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Thank you; what a wonderful explanation! – Nicholas Kinar Oct 28 '12 at 22:10

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