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For example, two Green functions:

\begin{equation} G_1(\tau_1 - \tau_2, x) = \alpha_1(\tau_1 - \tau_2)\beta_1(x) H(\tau_1 - \tau_2) + \mu_1(\tau_1 - \tau_2) \nu_1 (x) H(\tau_2 - \tau_1) \end{equation}

\begin{equation} G_2(\tau_2 - \tau_3, x') = \alpha_2(\tau_2 - \tau_3)\beta_2(x') H(\tau_2 - \tau_3) + \mu_2(\tau_2 - \tau_3) \nu_2 (x') H(\tau_3 - \tau_2) \end{equation}

where the H is a step function, $H(\tau_a - \tau_b)=1$ when $\tau_a > \tau_b$ and zero otherwise. Can a product of these two Green functions be integrated? -

\begin{equation} \int_0^y d\tau_2 G_1(\tau_1 - \tau_2, x) G_2(\tau_2 - \tau_3, x') \end{equation}

If they can be, is there a general method to do so?

Edit: For an easier comment below, I'll use

\begin{equation} G_1(\tau_1 - \tau_2, x) = c H(\tau_1 - \tau_2) + d H(\tau_2 - \tau_1) \end{equation}

\begin{equation} G_2(\tau_2 - \tau_3, x') = a H(\tau_2 - \tau_3) + b H(\tau_3 - \tau_2) \end{equation}

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    @Marek I classified it as physics because Green functions are used in so many areas of physics. And it's not clear to me which integrals are appropriate when the relationship between $\tau_1$ and $\tau_3$ is unknown (please see my comment on Roy's answer).2011-03-10

2 Answers 2

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Yes I think so. Expand the integrand into the 4 components like:

\begin{equation} [\alpha_1(\tau_1 - \tau_2)\beta_1(x) H(\tau_1 - \tau_2)][ \alpha_2(\tau_2 - \tau_3) \beta_2 (x) H(\tau_2 - \tau_3)] + .... \end{equation}

giving

\begin{equation} [\alpha_1(\tau_1 - \tau_2)\beta_1(x) \alpha_2(\tau_2 - \tau_3) \beta_2 (x) H(\tau_1 - \tau_2) H(\tau_2 - \tau_3)] + .... \end{equation}

to proceed we need to determine the relation between $\tau_1$ and $\tau_3$, say $\tau_1> \tau_3$. Now the conditions from the 4 double H terms will expand onto conditions between the three $\tau$s. The fourth case is $\tau$ inconsistent so we get three such cases. Then proceed with each $\alpha$ $\tau_2$ integral. You may need to determine the relation between the $\tau$s and 0 and y as well.

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    @Jane: Some comments. Since we dont know we need the first three integrals times H(t1,t3) plus the second three times H(t3,t1) since both conditions are never true together. Also that beta was a y in the original example. Recheck the algebra with an intermediate H step, but I think the integrals themselves are correct.2011-03-10
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It is usually done by transforming first to "Fourier Space". Very common in Physics.