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Walter Roberson
Walter Roberson 2021-9-21
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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)
phi(cpd) = 
f = int(exp(-p)*phi(c*p+d), p, u, inf)
f = 
simplify(f)
ans = 

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Murali Krishna AG
Murali Krishna AG 2021-9-21
Thanks Mr. walter for your answer. If I take f= int(exp(-p)*phi(c*sqrt(p)+d), p, u, inf), what will the solution? I have tried as above but couldn't get the solution.
Walter Roberson
Walter Roberson 2021-9-23
I am not finding any closed-form integral for that.
Walter Roberson
Walter Roberson 2021-9-23
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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))')
ans = 
Murali Krishna AG
Murali Krishna AG 2021-9-23
Thats great. I tried using Wolfram alpha but couldn't get the result. Then I tried manually to get the solution. Thanks lot. It is perfect

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