# Simpler code evaluating Dickman's function?

I've read here that Mathematica 10 can Obtain symbolic solutions to delay differential equations.

Would that help in numericaly evaluating Dickman's function $\rho(u)$ ?

It is a delay differential equation defined for $u\in\mathbb R^+$ by $$\begin{cases} \rho(u)=1 & \text{if }0\le u\le1\\ u\cdot\rho'(u)+\rho(u-1)=0& \text{if }u>1 \end{cases}$$

Otherwise said, can Mathematica 10 (or something) help improve/simplify this (working) code?

Dickm$pr=40 (*desired ABSOLUTE (rather than relative) accuracy*); Dickm$ex=Ceiling[Dickm$pr*Log[3,10]]+1 (*order of expansions; empirical*); Dickm$ma=Ceiling[Dickm$ex/Log[Dickm$ex]] (*domain limit*);

(*Dickm$dv[n] approximates DickmanRho[t+n-1/2] for -1/2\[LessEqual]t\[LessEqual]1/2*) Dickm$dv[1]=N[1,Dickm$pr+1]; Dickm$dv[n_]:=Dickm$dv[n]=Module[{u=Dickm$dv[n-1],v},v=Integrate[(u+O[\.b2]^Ceiling[Dickm$ex-n])/(\.b2+n-1/2),\.b2];(Normal[u]/.\.b2->1/2)+(Normal[v]/.\.b2->-1/2)-v]; DickmanRho[y_?NumberQ]:=Which[y<=0,1,y>Dickm$ma,N[1,y Log[10.,y]]-1,True,Normal[Dickm$dv[Ceiling[y]]]/.\.b2->y-Ceiling[y]+1/2]; (* Example use *) DickmanRho[15]  • Ack. I was going to suggest this: "Might want to check this old MathGroup post." Then I realized that was your work. Sep 11, 2014 at 20:31 ## 1 Answer Yes! fun = NDSolve[{u ρ'[u] + ρ[u - 1] == 0, ρ[u /; u < 1] == 1}, ρ, {u, 0, 20}, WorkingPrecision -> 40, PrecisionGoal -> ∞, AccuracyGoal -> 40, Method -> "StiffnessSwitching"][[1, 1, 2]]; fun[15] (* 7.5899080042980595046528227779709126741*10^-20 *) DickmanRho[15] (* 7.5899080042980595047*10^-20 *)  It works not only in V10 (in V8 at least). • @fgrieu, I tune the options for NDSolve and now it works much better. However, it takes several seconds which is ~100 times slower than your clever method! Sep 12, 2014 at 7:41 • Looks like Method made a difference.$\;\$ I wonder if we could make use the new DSolve capabilities. Sep 12, 2014 at 8:06
• @fgrieu I tried DSolve, but it produces very huge formulas so we have to manually simplify and rounding them. But it is what you have already done... Sep 12, 2014 at 8:14