Define two functions:

${\beta _1}(\tau ,\omega ) = \exp \left[ {\int\limits_0^\tau {\frac{\omega }{{2Q(\tau ')}}d\tau '} } \right]$

${\beta _2}(\tau ,\omega ) = \exp \left[ {\frac{{\omega \tau }}{{2Q(\tau )}}} \right]$

All variables in the above are real numbers, and $Q(\tau)$ is a real-valued function.

For what conditions are the two functions equal, such that $\beta_1(\tau, \omega) = \beta_2(\tau, \omega)$?

What can be said about equality between the two functions over an interval $[\tau_1,\tau_2]$, when the function $Q(\tau)$ is approximately constant over this interval?


1 Answer 1


For the functions in your questions to be equal, we must have

$\int\limits_0^\tau {\frac{\omega }{{2Q(\tau ')}}d\tau '} = \frac{{\omega \tau }}{{2Q(\tau )}}$

Taking the derivative of both sides with respect to $\tau$ we get

$\frac{\omega }{{2Q(\tau)}} = \frac{{\omega Q(\tau ) - \omega \tau Q'(\tau )}}{{2(Q(\tau ))^2}}$

Which must hold for any $\omega,\tau$.

Hence, following should be true $\frac{{Q(\tau ) - \tau Q'(\tau )}}{{Q(\tau )}} = 1$. This is only possible when $\tau Q'(\tau ) = 0$. Since $Q'(\tau ) \neq 0 $, for $\forall \tau$ the only valid value of $\tau$ is 0.

  • $\begingroup$ Thanks, Anvar. So the equality will only hold when $\tau = 0$? If the function $Q(\tau)$ is approximately constant over an interval, could the equality be approximately equal, or does $\tau$ have to be zero? $\endgroup$ Nov 9, 2012 at 16:31
  • $\begingroup$ Yes, I believe so. For approximate equality, I would add $\epsilon \to 0 $ to say first equation and solve the integral equation again. Note that $\epsilon = e^{\ln(\epsilon)}$. But in this case you may need to take care of the constants ($f'(x) = \int F(t) dt + C$). $\endgroup$
    – Anvar
    Nov 9, 2012 at 16:41
  • $\begingroup$ Why is approximate equality possible over a function interval $Q(\tau_1)$ to $Q(\tau_2)$? $\endgroup$ Nov 9, 2012 at 16:45

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