I have the following equation:

$\tilde U(\tau ,\omega ) = \frac{1}{{\Lambda (\tau ,\omega )}}\exp \left[ {i\int\limits_0^\tau {\left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{1}{{\pi Q(\tau ')}}}} - 1} \right)\omega } d\tau '} \right]$

In the above,

$\Lambda (\tau ,\omega ) = \frac{{\beta (\tau ,\omega ) + {\sigma ^2}}}{{{{\left( {\beta (\tau ,\omega )} \right)}^2} + {\sigma ^2}}}$

$\beta (\tau ,\omega ) = \exp \left[ { - \int\limits_0^\tau {\frac{\omega }{{2Q(\tau ')}}{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{{ - 1}}{{\pi Q(\tau ')}}}}} d\tau '} \right]$

$\tilde U(\tau ,\omega )$ is a complex number, $\{\tau, \omega\}$ are real numbers, $i = \sqrt{-1}$ as expected, and $\{\omega, \omega_h, Q(\tau), \sigma \}$ are all real numbers. Notice how $i$ is only found inside the exponential function. The exponential function used here is the complex exponential (Wikipedia).

I would like to separate $\tilde U(\tau, \omega)$ into a real and complex part. Here's what I think would be the way to do this. Taking the complex logarithm (Wikipedia) of $\tilde U(\tau ,\omega )$, and noting that we are taking the logarithm of a product:

$\log \left( {\tilde U(\tau ,\omega )} \right) = \log \left( {\frac{1}{{\Lambda (\tau ,\omega )}}} \right) + \log \left( {\exp \left[ {i\int\limits_0^\tau {\left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{1}{{\pi Q(\tau ')}}}} - 1} \right)\omega } d\tau '} \right]} \right)$


${\mathop{\rm Re}\nolimits} \left( {\log \left( {\tilde U(\tau ,\omega )} \right)} \right) = \log \left( {\frac{1}{{\Lambda (\tau ,\omega )}}} \right)$

${\mathop{\rm Im}\nolimits} \left( {\log \left( {\tilde U(\tau ,\omega )} \right)} \right) = \frac{1}{i}\log \left( {\exp \left[ {i\int\limits_0^\tau {\left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{1}{{\pi Q(\tau ')}}}} - 1} \right)\omega } d\tau '} \right]} \right)$

Is it right to assume that $\log(\exp(ix))$, where $x$ is a real number, will always be a complex number? For example, I evaluate the following in Matlab (using 1i as an imaginary number in the program code), and get:

log(exp(30*1i)) = 0 - 1.4159i

Now what I want to do next is remove the integral from the ${\mathop{\rm Im}\nolimits} \left( {\log \left( {\tilde U(\tau ,\omega )} \right)} \right)$. I suspect that this can be done by taking the partial derivative of both sides with respect to $\tau$:

$\frac{\partial }{{\partial \tau }}\left[ {{\mathop{\rm Im}\nolimits} \left( {\log \left( {\tilde U(\tau ,\omega )} \right)} \right)} \right] = \frac{\partial }{{\partial \tau }}\left[ {\frac{1}{i}\log \left( {\exp \left[ {i\int\limits_0^\tau {\left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{1}{{\pi Q(\tau ')}}}} - 1} \right)\omega } d\tau '} \right]} \right)} \right] = \left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{1}{{\pi Q(\tau )}}}} - 1} \right)\omega $

Notice how the variable of integration $\tau'$ has been replaced by $\tau$.

After a rather exhaustive search, I don't think it is very easy to remove the integral from the real part since due to the chain rule, there is always an integral in the resulting expression after taking the derivative.

Am I thinking clearly when splitting the equation into real and complex parts?

Notice how we are working with complex exponential and logarithm functions, so $\log(\exp(x)) \neq x$, where $x$ is a complex number.

Once again, here is an example using Matlab:

log(exp(3 + 4*1i)) = 3.0000 - 2.2832i

Notice how the real part is the same, whereas the complex part is not. So for numerical computations, is it true that:

$\frac{\partial }{{\partial \tau }}\left[ {{\mathop{\rm Im}\nolimits} \left( {\log \left( {\tilde U(\tau ,\omega )} \right)} \right)} \right] \stackrel{?}{=} \left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{1}{{\pi Q(\tau )}}}} - 1} \right)\omega $

If not, what would I have to change in the above equation to make both sides equal?


Checking both the mathematics by hand and by writing simple test computer programs, it is apparent that:

${\mathop{\rm Re}\nolimits} \left( {\log \left( {\tilde U(\tau ,\omega )} \right)} \right) = \log \left( {\frac{1}{{\Lambda (\tau ,\omega )}}} \right)$

${\mathop{\rm Im}\nolimits} \left( {\log \left( {\tilde U(\tau ,\omega )} \right)} \right) = \frac{1}{i}\log \left( {\exp \left[ {i\int\limits_0^\tau {\left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{1}{{\pi Q(\tau ')}}}} - 1} \right)\omega } d\tau '} \right]} \right)$

Moreover, as posted in the original question above:

$\frac{\partial }{{\partial \tau }}\left[ {{\mathop{\rm Im}\nolimits} \left( {\log \left( {\tilde U(\tau ,\omega )} \right)} \right)} \right] = \left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{1}{{\pi Q(\tau )}}}} - 1} \right)\omega$

This holds even for numerical computations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.