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MathGroup Archive 2007

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Re: bug in Integrate (5.2) (Fullani's integral)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74967] Re: bug in Integrate (5.2) (Fullani's integral)
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Fri, 13 Apr 2007 02:00:16 -0400 (EDT)
  • References: <evhtn4$2sr$1@smc.vnet.net><evksfg$pu1$1@smc.vnet.net>

Ooops!

I mean Frullani integral. Not Fullani.
In the title is the misprint; inside the message I typed correctly the
name.

Dimitris

=CF/=C7 dimitris =DD=E3=F1=E1=F8=E5:
> > By hand the integral can be obtained easily by integration by parts.
>
> I think I was fooling myself!
>
> Anyway...
>
> The integral in the current thread is an example of a Frullani
> integral.
>
> If f'[x] is continuous and the integral converges, we have
>
> Integrate[(f[a*x] - f[b*x])/x, {x, 0, Infinity}] = (f[0] -
> f[Infinity])*Log[b/a]
>
> In our case
>
> In[54]:=
> (Simplify[#1, a > 0 && b > 0] & )[Log[b/a]*(Limit[ArcTan[x/(a)] -
> ArcTan[x/b], x -> 0] -
>     Limit[ArcTan[x/(-a)] - ArcTan[x/b], x -> Infinity])]
>
> Out[54]=
> Pi*Log[b/a]
>
> See also here
>
> http://www.artofproblemsolving.com/Forum/viewtopic.php?start&t=103776
>
>
> Dimitris
>
> =CF/=C7 dimitris =DD=E3=F1=E1=F8=E5:
> > Consider the integral of the following function in [0,Infinity).
> >
> > In[710]:=
> > f[x_, a_, b_] := (ArcTan[x/a] - ArcTan[x/b])/x
> >
> > The integrand can be proved that converges for a, b>0.
> >
> > In[723]:=
> > f[x, a, b] + O[x]^4
> > (Simplify[#1, a > 0 && b > 0] & )[f[x, a, b] + O[x, Infinity]^4]
> >
> > Out[723]=
> > SeriesData[x, 0, {a^(-1) - b^(-1), 0, -1/(3*a^3) + 1/(3*b^3)}, 0, 4,
> > 1]
> > Out[724]=
> > SeriesData[x, Infinity, {-a + b}, 2, 4, 1]
> >
> > However Mathematica returns
> >
> > In[725]:=
> > Block[{Message}, Integrate[f[x, a, b], {x, 0, Infinity}]]
> >
> > Out[725]=
> > Infinity
> >
> > result which came from a very bad performance of Mathematica's
> > Integrate algorithm.
> >
> > (The ommited message is Integrate::idiv, generated when Mathematica
> > considers an integral
> > divergent.)
> >
> > However it must be pointed out that giving specific values for a and b
> > Mathematica returns correct results.
> >
> > E.g.
> >
> > In[735]:=
> > lst = Table[{Random[Integer, {1, 10}], Random[Integer, {1, 10}]},
> > {10}]
> > (Integrate[f[x, #1[[1]], #1[[2]]], {x, 0, Infinity}] & ) /@ lst
> > N[%]
> > (NIntegrate[f[x, #1[[1]], #1[[2]]], {x, 0, Infinity}] & ) /@ lst
> >
> > Out[735]=
> > {{6, 10}, {9, 10}, {7, 3}, {10, 8}, {3, 1}, {4, 2}, {7, 3}, {6, 5},
> > {6, 2}, {4, 1}}
> >
> > Out[736]=
> > {(1/2)*Pi*Log[5/3], (1/2)*Pi*Log[10/9], (-(1/2))*Pi*Log[7/3], (-
> > (1/2))*Pi*Log[5/4], (-(1/2))*Pi*Log[3], (-(1/2))*Pi*Log[2], (-
> > (1/2))*Pi*Log[7/3], (-(1/2))*Pi*Log[6/5], (-(1/2))*Pi*Log[3], (-
> > Pi)*Log[2]}
> >
> > Out[737]=
> > {0.8024030134443301, 0.16549991098452982, -1.3309323667973947,
> > -0.35051307075232924, -1.7256961476116015, -1.088793045151801,
> > -1.3309323667973947, -0.286390031707471, -1.7256961476116015,
> > -2.177586090303602}
> >
> > Out[738]=
> > {0.802403013444152, 0.1654999109844223, -1.3309323667973953,
> > -0.35051307075216187, -1.7256961476116057, -1.088793045151836,
> > -1.3309323667973953, -0.28639003170546784, -1.72569614760899,
> > -2.1775860903036426}
> >
> >
> > Also, (and more importantly) specifying explicitly assumptions for a,b
> > give correct symbolic results.
> >
> > E.g.
> >
> > In[764]:=
> > Integrate[f[x, a, b], {x, 0, Infinity}, Assumptions -> a > 0 && b > 0]
> > Out[764]=
> > (1/2)*Pi*Log[b/a]
> >
> > In[766]:=
> > Block[{Message}, Integrate[f[x, a, b], {x, 0, Infinity}, Assumptions -
> > > a < 0 && b > 0]]
> > Out[766]=
> > Infinity
> >
> > Note that the option GenerateConditions does not modify above
> > mentioned buggy performance
> > of Integrate
> >
> > In[775]:=
> > Block[{Message}, (Integrate[f[x, a, b], {x, 0, Infinity},
> > GenerateConditions -> #1] & ) /@ {Automatic(*default*), True(*does
> > more extensive searching*), False}]
> > Out[775]=
> > {Infinity, Infinity, Integrate[(ArcTan[x/a] - ArcTan[x/b])/x, {x, 0,
> > Infinity}, GenerateConditions -> False]}
> >
> >
> > If Mathematica uses (in the curent integral) the Newton Leibniz
> > formula for obtaining the definite integral from the indefinite one,
> > below may be is one explanation why it fails if we don't specify
> > ranges for the parameters a,b.
> >
> > In[795]:=
> > Integrate[f[x, a, b], x]
> > Limit[%, x -> Infinity] - Limit[%, x -> 0](*no specification*)
> > Limit[%%, x -> Infinity, Assumptions -> a > 0 && b > 0] - Limit[%%, x -
> > > 0, Assumptions -> a > 0 && b > 0](*specification*)
> >
> > Out[795]=
> > (1/2)*I*(PolyLog[2, -((I*x)/a)] - PolyLog[2, (I*x)/a]) -
> > (1/2)*I*(PolyLog[2, -((I*x)/b)] - PolyLog[2, (I*x)/b])
> > Out[796]=
> > DirectedInfinity[I*Sign[Log[-(I/a)] - Log[I/a] - Log[-(I/b)] + Log[I/
> > b]]]
> > Out[797]=
> > (1/2)*Pi*Log[b/a]
> >
> >
> > Another way to get the definite integral is
> >
> > In[804]:=
> > Integrate[f[x, Catalan, EulerGamma], {x, 0, Infinity}] /. {Catalan ->
> > a, EulerGamma -> b}
> > Out[804]=
> > (1/2)*Pi*Log[b/a]
> >
> >
> > By hand the integral can be obtained easily by integration by parts.
> >
> >
> > Dimitris



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