How does $ \frac{m\sinh\phi}{\sinh\tau}\,\Re[e^{ims\cosh(\phi-\tau)}]$ become $\frac{m\sinh(\phi+\tau)}{\sinh\tau}\,\Re[e^{ims\cosh(\phi)}]$?

Question

I'm stuck on the following steps:

$$ \frac{m\sinh\phi}{\sinh\tau}\,\Re[e^{ims\cosh(\phi-\tau)}] = \frac{m\sinh(\phi+\tau)}{\sinh\tau}\,\Re[e^{ims\cosh(\phi)}] $$ I'm not quite sure how we can get the right-hand side from the left. I know $\cosh(a-b) = \cosh a\cosh b-\sinh a\sinh b$, but how can we take out the '$-\tau$' from the real part of this exponent?

Answer 1

Since “We are actually integrating with respect to $\phi$”, let’s use $\Phi=\phi-\tau\implies \phi=\Phi+\tau,d\Phi=d\phi$:

$$ \int_a^b\frac{m\sinh\phi}{\sinh\tau}\Re[e^{ims\cosh(\phi-\tau)}] d\phi=\int_{a-\tau}^{b-\tau} \frac{m\sinh(\Phi+\tau)}{\sinh\tau}\Re[e^{ims\cosh(\Phi)}] d\Phi$$

since it is a definite integral, we just transformed the original one into an equivalent integral meaning $\Phi$ is a dummy variable, so:

$$ \int_a^b\frac{m\sinh\phi}{\sinh\tau}\Re[e^{ims\cosh(\phi-\tau)}] d\phi=\int_{a-\tau}^{b-\tau} \frac{m\sinh(\phi+\tau)}{\sinh\tau}\Re[e^{ims\cosh(\phi)}] d\phi$$

As for integrating the rest, $\Re(e^{iy})=\cos(y)$, so $\Re(e^{i ms\cosh(\phi)})=\cos(ms\cosh(\phi));m,s,\phi\in\Bbb R$