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Re: question about the inverse li function


invLogInteg[y_?NumericQ]:=
    x/.FindRoot[y==LogIntegral[x],
          {x,y,y+2}][[1]];

Plot[invLogInteg[y],{y,-1,6},
    PlotRange->{{-1.1,6.1},{-0.4,10.2}}];

For comparison,

Show[Plot[LogIntegral[x],{x,0,10},
        DisplayFunction->Identity]/.
      {x_,y_}->{y,x},
    PlotRange->{{-1.1,6.1},{-0.4,10.2}},
    DisplayFunction->$DisplayFunction];

Symbolically,

Off[InverseFunction::ifun];

Solve[y==LogIntegral[x],x]

{{x -> InverseFunction[LogIntegral, 1, 1][y]}}

InverseFunction[LogIntegral,1,1][LogIntegral[x]]

x

LogIntegral[InverseFunction[LogIntegral,1,1][x]]

x


Bob Hanlon

> 
> From: "Capet Arthur" <Arthur.Capet at student.ulg.ac.be>
To: mathgroup at smc.vnet.net
> Subject: [mg65374] [mg65348] question about the inverse li function
> 
> the question...
> 
> i've measured C(x), wich is equal to li(I(x)), where li denotes the
> logarithmic integral function. I would like to compute I(x)= li^(-1)
> (C(x))
> 
> how can i compute the inverse of the logarithmic integral function ?
> Is there a function Inverse[_function] ?
> 
> thanx a lot
> 
> 
> Arthur Capet, ULG, Belgium
> 
> 
> 
> 


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