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No result for integral?

Posted 3 years ago

I am trying to solve the integral resulting from a convolution between a stretched single exponential decay and a gaussian function.

In the simpler case using just a single exponential decay function:

F[t] = exp^(-t/τ)

and a Gaussian function:

R[t] = 1/(σ*Sqrt(2*π)*exp^(-t^2/(2*σ^2))

I obtain the convolution integral:

enter image description here

with Mathematica the solution of this integral is:

enter image description here

which I know is the correct result.

Now, if I try to change a bit the exp function importing the beta parameter and therefore make it a streched exponential,

F[t] = exp^[(-t/τ)^b] , 0<=b<=1

the convolution integral is:

enter image description here

for this integral I do not get a solution

enter image description here

but, the same expression. Is it a matter of defining the parameters? e.g. 0<=b<=1 I am new in Wolfram, thus I couldn't get over this issue and any help will be appreciated.

POSTED BY: E. Nico
3 Replies

Yes, you should always define the range of your parameters. First you should use the normal expression of Integrate, check the documentation Integrate

Then you add in it the option Assumptions where you will be able to define the range of b and sigma.

POSTED BY: Ahmed Elbanna
Posted 3 years ago

Thank you for your answer.

I defined the the range of the stochastic parameters with Assumptions:

0 < σ  && 0 <= b <= 1 && 0 < τ

and I use the expression Integrate to solve the integral for: {x, 0, +[Infinity]}

Integrate[
1/(\[Sigma] Sqrt[2 \[Pi]]) E^(-(x^b)/\[Tau]^b) E^(-(y - x)^2/(
2 \[Sigma]^2)), {x, 0, +\[Infinity]}, 
Assumptions -> 0 < \[Sigma] && 0 <= b <= 1 && 0 < \[Tau]]

but, again it doesn't give an output solution, but the initial expression. If I replace b = 1, then it give me the correct solution for the simpler case with the one function to be just an exponential decay. enter image description here

POSTED BY: E. Nico

When Mathematica returns the input as the output, it means that the calculation returned unevaluated. This often means that the function does not have the methods available to solve the problem symbolically, or it is mathematically impossible to obtain a symbolic solution (not all sums, integrals, or differential equations have symbolic solutions after all).

POSTED BY: Mariusz Iwaniuk
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