integration of erf function
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Murali Krishna AG
am 21 Sep. 2021
Kommentiert: Murali Krishna AG
am 23 Sep. 2021
f=
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/745289/image.png)
where
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/745284/image.png)
How to find the f?
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Walter Roberson
am 21 Sep. 2021
syms c p d cpd u x real
Pi = sym(pi);
phi(cpd) = 1/sqrt(2*Pi) * int(exp(-x^2/2), x, -inf, cpd)
f = int(exp(-p)*phi(c*p+d), p, u, inf)
simplify(f)
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Walter Roberson
am 23 Sep. 2021
When I use the Maple programming package, and tell it to expand the integral (which splits the erf), the Maple is able to integrate the split in terms of a limit as p approaches infinity. If you then ask to simplify under the assumption that all of the variables involved are real-valued, then MATLAB produces a closed-form output,
str2sym('(exp(-u)*sqrt(c^2 + 2)*(erf(((sqrt(u)*c + d)*sqrt(2))/2) + 1) - exp(-d^2/(c^2 + 2))*c*(erf(sqrt(2)*(sqrt(u)*c^2 + d*c + 2*sqrt(u))/(2*sqrt(c^2 + 2))) - 1))/(2*sqrt(c^2 + 2))')
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