Re: graphing traces of complicated evaluations (improved)
- To: mathgroup at smc.vnet.net
- Subject: [mg79251] Re: graphing traces of complicated evaluations (improved)
- From: dimitris <dimmechan at yahoo.com>
- Date: Sun, 22 Jul 2007 04:20:21 -0400 (EDT)
- References: <f7si68$28u$1@smc.vnet.net>
On 21 , 12:04, chuck009 <dmili... at comcast.com> wrote:
> Suppose I'm tenacious. Really I'm just not satisfied with Integrate returning Pi^2/6 for this integral. I know it's a bug but aren't they just curious why it's giving this answer?
>
> In[15]:=
> Integrate[Log[1 + Exp[-z]/Sqrt[z]], {z, 0, Infinity}]
>
> Out[15]=
> Pi^2/6
>
> I tried using your code to trace it but to no avail. Any suggestions?
>
>
>
> > Previously, I released code that could help with the
> > tracing of
> > evaluations,- -
>
> - -
Since
In[28]:=
Integrate[Log[1 + Exp[-z]/Sqrt[z]], z]
Out[28]=
Integrate[Log[1 + 1/(E^z*Sqrt[z])], z]
the integral is evaluated with the convolution method.
(BTW, the integral is not a trivial one for the Risch algorithm.
Professor R. Fateman some time ago wrote to me...
"While it may not seem hard to you, it turns out that there are three
components to the Risch integration algorithm: exponential, log and
algebraic.
this integrand, if it is done using the Risch algorithm, requires that
all of them work, together.
Since there is not, so far as I know, a complete implementation of the
algebraic case in Mathematica, there may be problems with
the hacks taken to bypass this lack of implementation.
So I don't find the issue to be surprising."
)
Somewhere during the process something went wrong.
I don't see how you could detect where it happens this failure
(bear in mind that it might be one place that the code fails; or
even more than one place!)
Only someone with a good knowledge of symbolic algebra could
do such a task...
Dimitris