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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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Just try to use definition of step function H and insert it into integral dividing interval from [0,y] into appropriate subintervals. –  kakaz Mar 10 '11 at 12:31
    
@kakaz But what are the appropriate subintervals when there are products of step functions? –  Jane Mar 10 '11 at 13:07
    
@Jane: please take a look in answer below. –  kakaz Mar 10 '11 at 13:13
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@Jane: 1) this is not physics, 2) I don't see why you have any problem with this. $H(x)$ is a perfectly normal function and so is a product of any number of them. What you obtain is a piece-wise smooth function (if the $\alpha$, $\beta$, $\mu$, $\nu$ coefficients are smooth enough) so you can certainly integrate it piece by piece... –  Marek Mar 10 '11 at 14:20
    
@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). –  Jane Mar 10 '11 at 14:29
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2 Answers

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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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@Roy Simpson @kakaz So generally (since you wouldn't always know whether $\tau_1 > \tau_3$ or not), would you get $\int_0^{\tau_3} d\tau_2 bc +\int_{\tau_3}^{\tau_1} d\tau_2 ac + \int_{\tau_1}^\beta d\tau_2 ad +\int_0^{\tau_1} d\tau_2 bc +\int_{\tau_1}^{\tau_3} d\tau_2 bd + \int_{\tau_3}^\beta d\tau_2 ad$ –  Jane Mar 10 '11 at 14:02
    
@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. –  Roy Simpson Mar 10 '11 at 14:32
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It is usually done by transforming first to "Fourier Space". Very common in Physics.

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