>>10940569Use 1.20 in picture:

(1.1) dF(t,t)/dt = dF(t',t'')/dt'+dF(t',t'')/dt''

The differential of dF(t,t) is likewise

dF(t,t) = dF(t',a)+dF(b,t'') or expressed in terms of (1.1)

dF(t,t) = f(t',t'')dt'+f(t',t'')dt''

Integrate both sides to obtain

F(t,t) = Int (f(t',t'')dt' + Int f(t',t'')dt'' where Int stands for Integration.

Make the integrals dependent on x rather than on t. We get Int(f(x,x)) = Int(f(t',t'')dt' + Int(f(t',t'')dt''

Consider case (1.21).

Since the result of Int(dte^-xt^2) only depends on x, we may equate differentiation by dx with dt. Now f(t',t') here is dt' and e^-xt''^2. We already know that Intf(t,t)dt = Int(t',a)dt'+Int(b,t'')dt''. a and b are assumed t according to (1.20). They are actually not.

We get d(Int(dt*e^-xt^2))/dx = d(Int(dt'))e^-xa^2/dt'+d(Int(e^-xt''^2)db). We get d(Int(dt*e^-xt^2)/dx = e^-x^3-Int(dt*t^2*e^-xt^2) as expected.