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 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 URL: ,