In comment 1 you said you had to compute the range yourself. What do you 
mean and why did you have to do so?

Regarding a later comment about handing convolutions where different 
functions have different ranges, that *should* be handled with Piecewise. 
Perhaps you could give my `lts` branch in PR #12587 a try. A major overhaul 
of Piecewise is presented there.

On Wednesday, June 7, 2017 at 10:35:16 AM UTC-5, [email protected] wrote:

> I'm trying to write a generic function to convolve two functions (in one 
> dimension, for now) and came across a few issues I don't know how to solve. 
>  Rather than give up, maybe I can start a discussion.  Some Sympy follows:
>
> a , b = symbols('a b', positive=True)
> f = exp(-a*x)
> g = exp(-b*x)
> integrate(f.subs(x,tau)*g.subs(x,t-tau), (tau,0,t))
>
>
> 1. I had to compute the range of integration myself (tau, 0, t).  This 
> isn't a fatal problem, but piecewise doesn't have enough information to 
> compute it simply.  (My long term solution is to propose a replacement for 
> the piecewise function, but I'm not there yet.)
>
> The answer is below:
>
> Piecewise((t*exp(-b*t), Eq(a, b)), (1/(a*exp(b*t) - b*exp(b*t)) - 
> 1/(a*exp(a*t) - b*exp(a*t)), True))
>
>
> The answer is correct, but there are problems trying to incorporate it 
> into a larger function.
>
> 2. It is not obvious in advance the answer will depend on whether a = b. 
>  Is there some way to assume in advance that a=b or a!=b?
>
> 3. The second condition, True, isn't helpful.  To understand what True 
> means, one has to keep track of the previous conditions.  Is there some way 
> to replace the True with a != b.
>
> 4. This is a simpler question, but I can't figure it out.  How do I 
> manipulate the second answer to my preferred form?
>
> (exp(-b*t)-exp(-a*t))/(a-b)
>
> Is there a simple way to do it programmatically?
>
> Thanks
>

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