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Integral evaluation fails

Welling Oei, UMC Utrecht
Posted 10 years ago
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POSTED BY: Welling Oei
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Cp is indeed the complete (not all relevant) conditions for various settings we consider in the calculation. The full formulas for Cp was derived by Mathematica when I specified:

I'm still not confident with the model, for example it could be written equivalently as

In[19]:= Clear[infectable, preinfect, Di, Dv, D0, Inf, Ip]
{$Assumptions = Di < Dv && Di > 0 && D0 > 0};
infectable[t_] := UnitStep[t]*UnitStep[D0 - t]
preinfect[t_] := infectable[-t]
(* NRp=Inf/D0*FullSimplify[Integrate[infectable[ti],{ti,te,te+Dv}]]+\
Ip/D0*FullSimplify[Integrate[preinfect[ti],{ti,te,te+Dv}]] *)
NRp = Integrate[
   Inf infectable[ti] + Ip preinfect[ti], {ti, te, te + Dv}, 
   Assumptions -> te \[Element] Reals]/D0

Out[23]= (1/D0)(Inf UnitStep[
    Dv + te] (UnitStep[
       D0 - te] (D0 - 
        te + (-D0 + Dv + te) UnitStep[D0 - Dv - te]) UnitStep[
       te] - (-1 + 
        UnitStep[
         te]) ((Dv + te) UnitStep[-Dv - te] UnitStep[
          D0 - Dv - 
           te] + (D0 + (-D0 + Dv + te) UnitStep[
             D0 - Dv - te]) UnitStep[Dv + te])) + 
  Ip UnitStep[
    D0 + Dv + 
     te] ((-te + (Dv + te) UnitStep[-Dv - te]) UnitStep[-te] UnitStep[
       D0 + te] - (-1 + 
        UnitStep[
         D0 + te]) ((D0 + Dv + te) UnitStep[-Dv - te] UnitStep[-D0 - 
           Dv - te] + (D0 + (Dv + te) UnitStep[-Dv - te]) UnitStep[
          D0 + Dv + te])))

if this NRp is inverted, one gets a clear statement where the NRp is zero (there 1/NRp equals to ComplexInfinity) which was absent in other formulations ...

1/NRp reformulated
POSTED BY: Udo Krause

You're welcome. The specific setting where Do>Dv>Di and te<0

In[7]:= (* First setting: Do>Dv>Di and te<0 *)
Clear[Cp, infectable]
Cp[te_,Di_,Dv_,D0_]:=Which[D0>=Dv+te&&te<0&&D0+te>=0&&Di<=Dv+te,(Di Inf)/(Dv Inf+Inf te-Ip te),D0<Dv+te&&D0+Di>=Dv+te&&te<0&&D0+te>=0&&Di<=Dv+te,(Inf (D0+Di-Dv-te))/(D0 Inf-Ip te),D0>=Dv+te&&D0+te>=0&&Di>Dv+te&&Dv+te>0,(Inf (Dv+te))/(Dv Inf+(Inf-Ip) te),True,0]
infectable[t_]:=UnitStep[t]*UnitStep[D0-t]
In[6]-= {$Assumptions=Di<Dv&&Dv<D0&&Di>0&&Di\[Element]Reals&&D0\[Element]Reals&&Inf>0&&Inf\[Element]Reals&&Ip>0&&Ip\[Element]Reals};
In[10]:= A=FullSimplify[(Inf/N/D0)*phi*tau*Integrate[Integrate[Integrate[infectable[ti] Cp[te,Di,Dv,D0],{tx,te+Dv,ti+Di}],{ti,te+Dv-Di,te+Dv}],{te,-Dv,D0-Dv+Di}]]

has a result (shown as picture because of the formatting )

first setting

because of the factor in the denominator (Inf - Ip)^4 it's clear that a condition appears. That's solved by Mathematica. If one uses one Integrate only, one integral remains.
POSTED BY: Udo Krause

In the second case for the setting where Do > Di > Dv and te < 0 the conditional expression under the remaining integral works into your hands stating

-Inf (Dv + te)^2 (-2 Di + Dv + te)/(2 (Dv Inf + (Inf-Ip) te)) for D0>= Dv + te && Di > Dv + te && Dv + te > 0
0 elsewhere

because you say te < 0 the upper integration limitD0 +Di -Dv for te must be smaller than 0

second setting

there is a contradiction: If Di > Dv and D0 > Dv then D0 + Di - Dv > 0 so te > 0 (not te < 0) on the upper limit .... okay, let's send this back to you and continue with the integration limits:

The lower limit is fullfilled: D0 > 0 && Di > 0 && 0 >= 0 because D0> Dv in the assumptions and if te < 0 at least on the lower limit one has Dv > 0.

The upper limit D0 >= D0 + Di && Di > D0 + Di && D0 + Di > 0 is not fullfilled because Di > 0; one has to determine where the integrand is 0 and where not. To get out of this mess one could plot it

With[{Dv = 1, Di = 4.5, D0 = 8},
 Plot[If[D0 >= Dv + te && Di > Dv + te && Dv + te > 0, 
   1, -1], {te, -Dv, D0 + Di + Dv}]
 ]

fortunately there is only one switch from 1 to -1 at te = Di - Dv leaving us behind with the integral

In[30]:= Clear[Dv, Di, D0]
Inf phi tau Integrate[-((Inf (Dv + te)^2 (-2 Di + Dv + te))/(
    2 (Dv Inf + (Inf - Ip) te))), {te, -Dv, Di - Dv}, 
   Assumptions -> Dv > 0]/(D0 N)

Out[31]= ConditionalExpression[-((
  Inf^2 phi tau (-Di (Inf - Ip) (4 Di^2 (Inf - Ip)^2 - 6 Dv^2 Ip^2 + 
        9 Di Dv Ip (-Inf + Ip)) + 
     6 Dv^2 Ip^2 (2 Di (Inf - Ip) + Dv Ip) Log[(Dv Ip)/(
       Di Inf - Di Ip + Dv Ip)]))/(12 D0 (Inf - Ip)^4 N)), 
 Di Ip^2 < Di Inf Ip || Di (Inf - Ip) (Di (Inf - Ip) + Dv Ip) < 0]

the integrals are done. Please check for errors.
POSTED BY: Udo Krause
POSTED BY: Daniel Lichtblau
POSTED BY: Udo Krause

Dear Udo, Thank you for your kind assessment and suggestions.

Cp is indeed the complete (not all relevant) conditions for various settings we consider in the calculation. The full formulas for Cp was derived by Mathematica when I specified:

{$Assumptions = Di < Dv && Di > 0 && D0 > 0};
infectable[t_] := UnitStep[t]*UnitStep[D0 - t]
preinfect[t_] := UnitStep[D0 + t]*UnitStep[-t]
NRp = Inf/D0*
    FullSimplify[Integrate[infectable[ti], {ti, te, te + Dv}]] + 
   Ip/D0*FullSimplify[Integrate[preinfect[ti], {ti, te, te + Dv}]];
FullSimplify[PiecewiseExpand[TR/NRp], 
 Assumptions -> 
  Di < Dv && Di > 0 && D0 > 0 && Inf > 0 && Ip > 0  &&  D0 > 0  && 
   te \[Element] Reals]

Following your suggestion, I tried to reduce the conditions to a specific setting where Do>Dv>Di and te<0:

Cp[te_, Di_, Dv_, D0_] := 
 Which[D0 >= Dv + te && te < 0 && D0 + te >= 0 && 
   Di <= Dv + te, (Di Inf)/(Dv Inf + Inf te - Ip te), 
  D0 < Dv + te && D0 + Di >= Dv + te && te < 0 && D0 + te >= 0 && 
   Di <= Dv + te, (Inf (D0 + Di - Dv - te))/(D0 Inf - Ip te), 
  D0 >= Dv + te && D0 + te >= 0 && Di > Dv + te && Dv + te > 0, (
  Inf (Dv + te))/(Dv Inf + (Inf - Ip) te), True, 0]
infectable[t_] := UnitStep[t]*UnitStep[D0 - t]

{$Assumptions = 
   Di < Dv && Dv < D0 && Di > 0 && Di \[Element] Reals && 
    D0 \[Element] Reals && Inf > 0 && Inf \[Element] Reals && Ip > 0 &&
     Ip \[Element] Reals};

A = FullSimplify[(Inf/N/D0)*phi*tau*
   Integrate[
    Integrate[
     Integrate[
      infectable[ti] Cp[te, Di, Dv, D0], {tx, te + Dv, ti + Di}], {ti,
       te + Dv - Di, te + Dv}], {te, -Dv, D0 - Dv + Di}]]

Or for the setting where Do>Di>Dv and te<0:

Cp[te_, Di_, Dv_, D0_] := 
 Which[D0 >= Dv + te && te < 0 && D0 + te >= 0 && 
   Di <= Dv + te, (Di Inf)/(Dv Inf + Inf te - Ip te), 
  D0 < Dv + te && D0 + Di >= Dv + te && te < 0 && D0 + te >= 0 && 
   Di <= Dv + te, (Inf (D0 + Di - Dv - te))/(D0 Inf - Ip te), 
  D0 >= Dv + te && D0 + te >= 0 && Di > Dv + te && Dv + te > 0, (
  Inf (Dv + te))/(Dv Inf + (Inf - Ip) te), True, 0]
infectable[t_] := UnitStep[t]*UnitStep[D0 - t]

{$Assumptions = 
   Di > Dv && Dv < D0 && Di > 0 && Di \[Element] Reals && 
    D0 \[Element] Reals && Inf > 0 && Inf \[Element] Reals && Ip > 0 &&
     Ip \[Element] Reals};

A = FullSimplify[(Inf/N/D0)*phi*tau*
   Integrate[
    Integrate[
     Integrate[
      infectable[ti] Cp[te, Di, Dv, D0], {tx, te + Dv, ti + Di}], {ti,
       te + Dv - Di, te + Dv}], {te, -Dv, D0 - Dv + Di}]]

But I still got a conditional expression as the output.

Do you have some suggestions how to solve this?

Kind regards, Welling
POSTED BY: Welling Oei
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