Re: bug in Integrate
- To: mathgroup at smc.vnet.net
- Subject: [mg78306] Re: bug in Integrate
- From: dimitris <dimmechan at yahoo.com>
- Date: Thu, 28 Jun 2007 04:19:31 -0400 (EDT)
- References: <f5ivfs$avv$1@smc.vnet.net>
Veit Elser :
> In version 6.0, the Integrate function applied to
>
> Integrate[x Exp[-x] Log[x z], {x, 0, Infinity}, Assumptions -> z > 0]
>
> evaluates to 0. Version 5.2 gives the correct result,
>
> 1 - EulerGamma + Log[z]. Interestingly, back in version 6.0
>
> Integrate[x Exp[-x] Log[x z], {x, 0, Infinity}, Assumptions -> z == 1]
>
> does evaluate correctly to
>
> 1 - EulerGamma
>
> Let's hope this is fixed before the end of Euler's tercentennial.
>
>
> Veit Elser
Hi.
$VersionNumber->5.2
(I don't have version 6)
Note that if you give specific values in parameter(s)
you should use With instead of Assumptions->z==1...
That is I consider more suitable the following structure
In[76]:=
With[{z = 1}, Integrate[x*Exp[-x]*Log[x*z], {x, 0, Infinity}]]
Out[76]=
1 - EulerGamma
As you noticed Mathematica 5.2 gets correctly the integral.
Note however that,
In[99]:=
Integrate[x*Exp[-x]*Log[x*z], {x, 0, Infinity}]
Out[99]=
1 - EulerGamma + Log[z]
That is no need for assumptions.
(*check*)
In[101]:=
N[(1 - EulerGamma + Log[z] /. z -> #1 & ) /@ {2, 3, -4, I + 6, -3*I}]
Out[101]=
{1.1159315156584124, 1.5213966237665768, 1.8090786962183576 +
3.141592653589793*I, 2.228243291420579 + 0.16514867741462683*I,
1.5213966237665768 - 1.5707963267948966*I}
In[102]:=
(NIntegrate[x*Exp[-x]*Log[x*#1], {x, 0, Infinity}] & ) /@ {2, 3, -4, I
+ 6, -3*I}
Out[102]=
{1.1159315184569099, 1.5213966033406756, 1.809078675792414 +
3.141592653589326*I, 2.2282432709945725 + 0.1651486774146023*I,
1.5213966033406756 - 1.570796326794663*I}
What does the version 6 returns if you don't specify a range for the
parameter?
Here are some workarounds based on my experience with earlier versions
of Mathematica that
could work in Mathematica 6:
In[106]:=
f = HoldForm[Integrate[x*Exp[-x]*Log[x*z], {x, 0, Infinity}]]
ReleaseHold[f /. z -> Catalan] /. Catalan -> z
Out[106]=
HoldForm[Integrate[(x*Log[x*z])/E^x, {x, 0, Infinity}]]
Out[107]=
1 - EulerGamma + Log[z]
2)
In[92]:=
ReleaseHold[f /. Integrate[g_, h_] :> Integrate[g, x]]
Limit[%, x -> Infinity] - Limit[%, x -> 0, Direction -> -1]
Out[92]=
-E^(-x) + ExpIntegralEi[-x] - ((1 + x)*Log[x*z])/E^x
Out[93]=
1 - EulerGamma + Log[z]
Dimitris