# Twisted convolution - simplifying multiple integral via substitution

I need help with understanding a proof from a paper. Following is the description:

Let

$$r_{1}(t) = \int \int h_{1}\left(\tau, \nu\right) e^{j 2\pi \nu (t - \tau)} s\left(t - \tau\right) d\tau d\nu$$

$$r(t) = \int \int h_{2}\left(\tau, \nu\right) e^{j 2\pi \nu (t - \tau)} r_{1}\left(t - \tau\right) d\tau d\nu$$

Substituting $$r_{1}(t)$$ in $$r_{2}(t)$$ we obtain after some algebraic manipulations,

$$r(t) = \int \int h\left(\tau, \nu\right) e^{j 2\pi \nu (t - \tau)} s\left(t - \tau\right) d\tau d\nu$$

where, $$f(\tau, \nu)$$ is given by

$$h(\tau, \nu) = \int \int h_{2}\left(\tau', \nu' \right) h_{1}\left(\tau - \tau', \nu-\nu' \right) e^{j 2\pi \nu' \left(\tau - \tau' \right)} d\tau' d\nu'$$

I am unable to use appropriate algebraic manipulations to arrive at the final expression for $$r(t)$$. Can anyone help me? The first step is arrived at by substituting $$r_{1}(t-\tau)$$ in $$r_{2}(t)$$. But I am unable to proceed further. (Incorporated corrections from Tanya Choudhary's answer)

$$r(t) = \int \int h_2(\tau,\nu)e^{j 2\pi\nu(t-\tau)} \int \int h_1(\tau_1,\nu_1)e^{j 2\pi\nu(t-\tau-\tau_1)}s(t-\tau-\tau_1)d\tau_1 d\nu_1 d\tau d\nu$$

Substituting $$\tau = \tau', \tau_1 = \tau - \tau', \nu = \nu', \nu_1 = \nu - \nu'$$,

$$r(t) = \int \int h_2(\tau',\nu')e^{j 2\pi\nu'(t-\tau')} \int \int h_1(\tau - \tau',\nu - \nu')e^{j 2\pi\nu(t-\tau)}s(t-\tau)d\tau' d\nu' d\tau d\nu$$

Since these are two different integral equations, during substitution you cannot use the same variables for both the integrals. On substituting $$r_1(t-\tau)$$ to $$r(t)$$ we get, $$r(t) = \int \int h_2(\tau,\nu)e^{j 2\pi\nu(t-\tau)} \int \int h_1(\tau_1,\nu_1)e^{j 2\pi\nu(t-\tau-\tau_1)}s(t-\tau-\tau_1)d\tau_1 d\nu_1 d\tau d\nu$$ Now substitute $$\tau = \tau', \tau_1 = \tau - \tau', \nu = \nu', \nu_1 = \nu - \nu'$$
• Thank you for pointing the flaw in the first step. I'm unable to figure out how to take $h_{2}(\tau, \nu)$ inside the inner integral. Can you please help me with that? (I've edited my answer). – MaxFrost Aug 2 at 6:43