Re: Evaluating a convolution integral in Mathematica

• To: mathgroup at smc.vnet.net
• Subject: [mg80269] Re: Evaluating a convolution integral in Mathematica
• From: Randall Beer <rdbeer at indiana.edu>
• Date: Thu, 16 Aug 2007 04:46:06 -0400 (EDT)
• References: <f9ud4q\$a5q\$1@smc.vnet.net> <46C32C19.3090203@gmail.com>

```Thanks for your response.  I defined B as 1/2 (-Sign[-max + x] + Sign
[-min + x]). Yes, I did provide Mathematica with the assumptions I
stated in my message.  However, the result Mathematica gives seems to
produce incorrect answers when specific values of n, m, and the mins
and maxs are substituted in.

I will take a look at the MathSource notebook that you suggested.

Thanks,
Randy Beer

On Aug 15, 2007, at 12:38 PM, Jean-Marc Gulliet wrote:

> rdbeer at indiana.edu wrote:
>> I am having trouble evaluating a convolution integral in Mathematica.
>> Define B[x,{min,max}] to be a function that takes on the value 1 when
>> x is between min and max and 0 otherwise.  I need to find the
>> convolution of x^n B[x, {min1, max1}] with x^m B[x, {min2, max2}],
>> where x is Real, n and m are nonnegative integers, and the mins and
>> maxs are Real with the constraints that min1<=max1 and min2<=max2.
>> The resulting convolution integeral is Integrate[t^n B[t, {min1,
>> max1}] (x-t)^m B[x-t, {min2, max2}], {t, -Infinity, Infinity}].
>> Mathematica 6.0.1 has no problem evaluating this integral when
>> constant values are substituted for n, m, and the mins and maxs.
>> However, I need the general value of this integral.  Mathematica also
>> claims to be able to evaluate this general integral, returning a
>> complicated peicewise expression involving gamma functions and
>> hypergeometric functions.  However, when specific values for n, m and
>> the mins and maxs are then substituted into this general expression,
>> it always returns either 0 or Indeterminate.
>> Any help with evaluating this integral in general would be greatly
>> appreciated.
>
> It is not clear, at least for me, how you coded your function B?
> Did you use Piecewise, If, Which, Boole, ... ? Here is a simple
> example with Boole,
>
> B[x_, {min_, max_}] := Boole[min <= x <= max]
>
> Also, it is not clear whether you have tried to set up your
> assumptions in Integrate itself. For instance,
>
> Integrate[
>  t^n B[t, {min1, max1}] (x - t)^m B[
>    x - t, {min2, max2}], {t, -Infinity, Infinity},
>  Assumptions -> {{x, min1, mmax1, min2, max2} \[Element]
>     Reals, {n, m} \[Element] Integers, n >= 0, m >= 0, min1 <= max1,
>    min2 <= max2}]
>
> returns a long piecewise function with some hypergeometric
> functions (I  did not try to simplify nor to check the validity of
> the expression since I did not know what the function B really was).
>
> Also, it may be worthwhile to try Maxim Rytin's "Integration of
> Piecewise Functions with Applications" package available at
>
> http://library.wolfram.com/infocenter/MathSource/5117/
>
> Especially, you should check the function *PiecewiseIntegrate*.
> From the notebook piecewise.nb, we can read,
>
> "PiecewiseIntegrate[f,{x,xmin,xmax},{y,ymin,ymax},\[Ellipsis]]
> gives the definite integral of function f. It is intended for
> integrating piecewise continuous functions, and also generalized
> functions. It handles integrands and integration bounds involving
> the following expressions:
>
> . UnitStep, Sign, Abs, Min, Max
> . Floor, Ceiling, Round, IntegerPart, FractionalPart, Quotient, Mod
> . DiracDelta and its derivatives, DiscreteDelta, KroneckerDelta
> . If, Which, Element, NotElement
> . Piecewise, Boole, Clip"
>
> For instance,
>
> PiecewiseIntegrate[
>  t^n B[t, {min1, max1}] (x - t)^m B[
>    x - t, {min2, max2}], {t, -Infinity, Infinity},
>  Assumptions -> {{x, min1, mmax1, min2, max2} \[Element]
>     Reals, {n, m} \[Element] Integers, n >= 0, m >= 0, min1 <= max1,
>    min2 <= max2}]
>
> returns a long answer made of nested If constructs that return some
> simple expressions such as
>
> If[m == 0 && min1 < max1 && min2 == -max1 + max2 + min1 &&
>   x == max1 + min2,
>  If[min1 > 0, (-min1^(1 + n) + max1^n (-min2 + x))/(1 + n), [...]
>
> (Again, I have not try to simplify nor to check the validity of the
> expression.)
>
> HTH
> --
> Jean-Marc
>

```

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