# erf function and solving for two unknowns

My question is motivated by this psychology terminology heavy question about signal detection theory. I believe the question can be boiled down to: given the system of equations

$$a = 1+\int_{-\inf}^{x}e^{-u^2}du+\int_{-\inf}^{y}e^{-v^2}dv$$

and

$$b = x+y$$

and a known $$a$$ and $$b$$, is it possible to analyticalally solve for unknown $$x$$ and $$y$$?

My thinking is that it is not possible since while it is two equations and two unknowns, the equations are not linear so a solution is not guaranteed. Further, the integral of $$e^{-u^2}$$ is nasty and does not have an analytical solution.

• What do you mean by "analytically"? The error function itself is a non-elementary function. – callculus Nov 7 '18 at 17:02
• @callculus I am not sure. Maybe closed form is a better term? I guess I want $x=f(a,b)$ and $y=g(a,b)$ where $f$ and $g$ are things that I can "calculate" with something like matlab/python. – StrongBad Nov 7 '18 at 17:13
• Its trivial to write it as $const=erf(x)+erf(b-x)$ which any maths package would easily solve numerically. See Wolfram alpha's effort wolframalpha.com/input/?i=solve+1.5%3Derf(x)%2Berf(2-x) – user121049 Nov 7 '18 at 17:20
• @StrongBad Closed form has more or less the same meaning. What I meant is that you cannot calculate analytically $g(x)$ and $h(y)$ and consequently you cannot calculate the inverse functions. But if I understand you right now, you don´t reject using the erf-function (or the inverse)? – callculus Nov 7 '18 at 17:21
• @callculus yes the answer can have erf and erf-inverse. – StrongBad Nov 7 '18 at 18:08