Author 
Comment/Response 
Nicola

01/15/13 09:24am
I have to solve a double integral with parametrized functions, that is (written as a latex equation)
$$
\int_{\infty}^\infty \int_0^\infty u(t)u(t\Delta)\frac{1+e^{2\int_{t\tau\Delta}^{t\tau} y(s)^2 ds }}{2} d\Delta \dt
$$
that depends upon:
 the function u(t), that in my case is a squared pulse between 0 and 1, defined in Mathematica as
u[t_] = Piecewise[{{1, 0 <= t <= 1}}, 0];
 the function y(s), that in my case is a piecewise constant function between 0 and 1 with a discontinuity in t1, defined in Mathematica as
y[t_, y1_, y2_, t1_] =
Piecewise[{{y1, 0 <= t <= t1}, {y2, t1 < t <= 1}}, 0];
 the parameter \tau.
Consider the assumptions
$Assumptions =
1 > \[Tau] > 0 && y1 \[Element] Reals && y2 \[Element] Reals && 1 > t1 > 0
and also t1 + \[Tau] < 1. I'd like to solve the integral to have a function of \tau, y1, y2, t1.
Now, if I just substitute the definition of u(t) and y(s) and use Mathematica, I get I can't get the integral solved, as you can check with
Assuming[t1 + \[Tau] < 1,
Integrate[
Integrate[
u[t] u[t  \[CapitalDelta]] (1 +
Exp[2 Integrate[
y[s, y1, y2, t1]^2, {s, t  \[CapitalDelta]  \[Tau],
t  \[Tau]}]])/2, {\[CapitalDelta],
0, +Infinity}], {t, Infinity, +Infinity}]] // Simplify.
So, I have done "by hand" part of the integral, in particular the integral at the exponential argument, defining the funtion
IntY2[a_, b_, y1_, y2_, t1_] = Piecewise[{
{y1^2 b, a < 0 && 0 <= b < t1}, {y1^2 t1 + y2^2 (b  t1),
a < 0 && t1 <= b <= 1}, {y1^2 t1 + y2^2 (1  t1), a < 0 && 1 < b},
{y1^2 (b  a),
0 <= a < t1 && 0 <= b < t1}, {y1^2 (t1  a) + y2^2 (b  t1),
0 <= a < t1 && t1 <= b <= 1}, {y1^2 (t1  a) + y2^2 (1  t1),
0 <= a < t1 && 1 < b},
{y2^2 (b  a), t1 <= a <= 1 && t1 <= b <= 1}, {y2^2 (1  a),
t1 <= a <= 1 && 1 < b}}, 0]
and calculating
Assuming[t1 + \[Tau] < 1,
Integrate[
Integrate[
u[t] u[t  \[CapitalDelta]] (1 +
Exp[2 IntY2[t  \[CapitalDelta]  \[Tau], t  \[Tau], y1, y2,
t1]])/2, {\[CapitalDelta],
0, +Infinity}], {t, Infinity, +Infinity}]] // Simplify
This elaboration works, and I get a solution. But doing by hand the substitution of u(t) and y(t) with some algebra I get
Integrate[(1 + Exp[2 y1^2 \[CapitalDelta]])/2, {\[CapitalDelta], 0,
t  \[Tau]}, {t, \[Tau], t1 + \[Tau]}] +
Integrate[(1 + Exp[2 y1^2 (t  \[Tau])])/2, {\[CapitalDelta],
t  \[Tau], t}, {t, \[Tau], t1 + \[Tau]}] +
Integrate[(1 + Exp[2 y2^2 (t  \[Tau]  t1)  2 y1^2 t1])/
2, {\[CapitalDelta], t  \[Tau], t}, {t, t1 + \[Tau], 1}] +
Integrate[(1 +
Exp[2 y2^2 (t  \[Tau]  t1) 
2 y1^2 (t1  (t  \[Tau]  \[CapitalDelta]))])/
2, {\[CapitalDelta], t  t1  \[Tau], t  \[Tau]}, {t,
t1 + \[Tau], 1}] +
Integrate[(1 + Exp[2 y2^2 \[CapitalDelta]])/2, {\[CapitalDelta], 0,
t  t1  \[Tau]}, {t, t1 + \[Tau], 1}] // Simplify
that is different from the one calculated by Mathematica. Of course, probably I made a mistake somewhere that I cannot find. Anyway, I tried to substitute some parts of the original integral to get my formula, for example sustituting the constraint given by u(t), then u(t\Delta), ecc checking each step of Mathematica. For example, with
Assuming[t1 + \[Tau] < 1,
Integrate[
Integrate[u[t]
u[t  \[CapitalDelta]] (1 +
Exp[2 IntY2[t  \[CapitalDelta]  \[Tau], t  \[Tau], y1, y2,
t1]])/2, {\[CapitalDelta], 0, +Infinity}], {t, 0,
1}]] // Simplify
I get the same (Mathematica's, not mine) solution.
When I get to
Assuming[t1 + \[Tau] < 1,
Integrate[
Integrate[
u[t  \[CapitalDelta]] (1 +
Exp[2 IntY2[t  \[CapitalDelta]  \[Tau], t  \[Tau], y1, y2,
t1]])/2, {\[CapitalDelta], 0, +Infinity}], {t, 0,
1}]] // Simplify
and I tried to make that solved, Mathematica cannot solve it (or at least it takes a lot of time, the previous integral it takes 20 s but in this one after 5 minutes it's not solved). Note that I just delete u[t] in the integral argument, but from the mathematical point of view it should be indifferent since t is integrated between 0 and 1.
I'd like to understand why Mathematica cannot solve this last formulation of the integral (and how can I fix it), because I want to check my solutions and also I have some more complex integrals that Mathematica cannot solve and I think the problem is the same.
Sorry for the long post,
thanks,
Nicola
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