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We have a function as

e[t_] :=(E^(-t^2)) Cos[0.1 t]

and we must evaluate below integration (However I used the variable x here, I just want to calculate the integration and x is a local variable)

a[t_] := Integrate[e[x], {x, 0, t}]

the final results is a[t]. That it will be of the form:

(0. + 2.91163*10^-40 I) (1. Erfi[9.68227 - (0. + 0.00294353 I) t] -   1. Erfi[9.68227 + (0. + 0.00294353 I) t])

As it shown, in this result, Error function will appear. But we have to use of a[t] in next our calculation. Actually, we have to evaluate the p[t_, \[Tau]_]:=Integrate[a[x],{x,t-\[Tau],t}] .

For this reason we should have a more clear form of a[t] In which there should be no error function. How can we receive the a[t] in the absence of Error function? Are there any way to get rid of it?

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    $\begingroup$ The error function is a standard higher transcendental function, so it should be considered a "clear form." Further it cannot be written in terms of elementary functions. Unless the error functions cancel each other out (there are identities after all), I do not see how you are going to get rid of it. $\endgroup$ – Michael E2 Dec 21 '14 at 14:39
  • $\begingroup$ You are right, It is completely rational. The problem was not aroused of Error Function, it was related to some negligible imaginary parts. $\endgroup$ – Unbelievable Dec 21 '14 at 15:18
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It is unclear why you believe that there is a problem with the presence of the error function in a[t].

e[t_] = E^(-t^2) Cos[t/10];

a[t_] = Integrate[e[x], {x, 0, t}];

a[t] is readily integrated

 p[t_, tau_] = Integrate[a[x], {x, t - tau, t}]

(1/(80*E^(1/400)))* (20*(E^(1/20 - It)^2 + E^(1/20 + It)^2 - E^(1/20 + I*(t - tau))^2 - E^(1/20 - It + Itau)^2) + Sqrt[Pi]*((-1 + 20*It) Erfi[1/20 - I*t] + (-1 - 20*I*t)Erfi[ 1/20 + It] + (1 + 20*I*t - 20*Itau) Erfi[1/20 + I*(t - tau)] + (1 - 20*I*t + 20*Itau) Erfi[1/20 - It + Itau]))

p[t, tau] can be evaluated. Chop can be used to remove any negligible imaginary component due to numerical noise.

p[2., 1.]

0.8408887255235893 + 0.*I

% // Chop

0.840889

Table[
  {t, tau, p[t, tau] // Chop},
  {t, -4., 4., 2.}, {tau, -4., 4., 2.}] //
 Flatten[#, 1] &

{{-4., -4., 3.03855}, {-4., -2., 1.76719}, {-4., 0., 0}, {-4., 2., -1.76803}, {-4., 4., -3.53606}, {-2., -4., 0}, {-2., -2., 1.27137}, {-2., 0., 0}, {-2., 2., -1.76719}, {-2., 4., -3.53521}, {0., -4., -3.03855}, {0., -2., -1.27137}, {0., 0., 0}, {0., 2., -1.27137}, {0., 4., -3.03855}, {2., -4., -3.53521}, {2., -2., -1.76719}, {2., 0., 0}, {2., 2., 1.27137}, {2., 4., 0}, {4., -4., -3.53606}, {4., -2., -1.76803}, {4., 0., 0}, {4., 2., 1.76719}, {4., 4., 3.03855}}

p[t, tau] can also be readily plotted.

Plot3D[p[t, tau], {t, -10, 10}, {tau, -10, 10},
 AxesLabel -> (Style[#, 14, Bold] & /@
    {"t", "tau", "p(t, tau)"})]

enter image description here

As expected from looking at the plot

p[t, tau] == p[-t, -tau] // Simplify

True

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  • $\begingroup$ +1. But I think the 0. I is not from numerical noise, but because the numerical type used in the computation is a machine/floating-point Complex, which is displayed with + 0. I when the imaginary part is zero. $\endgroup$ – Michael E2 Dec 21 '14 at 14:45
  • $\begingroup$ OK. Thanks a bunch. I think I had not recognized that the source of the problem. Maybe it arouses of imaginary part of the function. I think 'chop' will help. $\endgroup$ – Unbelievable Dec 21 '14 at 15:13
  • $\begingroup$ If there has been remained a little question but there would be not a suitable space in this comment place, How can be asked? $\endgroup$ – Unbelievable Dec 23 '14 at 5:37
  • $\begingroup$ Initiate a new question, just make sure that it differs from the one above. $\endgroup$ – Bob Hanlon Dec 23 '14 at 5:45
  • $\begingroup$ Approximately, it needs to the above information. As a matter of fact, I have to use 'p' in other calculations. It means 'p' must be as a parameter not directly as a numeric. the final step is just an integration of a function of 'p' and 'a' (Which is absent here). Every thing is ok with 'p' and 'a'. But that Integration (containing functions of 'p' and 'a') will be gotten stuck. I so much tried to find what is the problem but I could not find any answer or key! Moreover, I can't understand should I substitute CHOP in a very step to the final or should I do that just to the final step? $\endgroup$ – Unbelievable Dec 23 '14 at 6:01

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