integration of erf function

f=

where

How to find the f?

Answer 1

[Walter Roberson]

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 = 

[Murali Krishna AG]

Thank u so much

[Murali Krishna AG]

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]

I am not finding any closed-form integral for that.

[Walter Roberson]

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]

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