Re: HoldAll for Integrate
- To: mathgroup at smc.vnet.net
- Subject: [mg97997] Re: HoldAll for Integrate
- From: David Bailey <dave at removedbailey.co.uk>
- Date: Fri, 27 Mar 2009 05:37:18 -0500 (EST)
- References: <gqfknb$kht$1@smc.vnet.net>
Tammo Jan Dijkema wrote:
> The following commands yield unexpected output:
>
> x=10;
> Integrate[x^2, {x,0,1}]
>
> That is because Integrate does not have attributes HoldAll, so that the
> second command will be interpreted as Integrate[100, {10,0,1}] which is
> not a command that Integrate can work with (so a warning message is
> returned).
>
> However, I can add the attribute HoldAll to Integrate myself:
>
> SetAttributes[Integrate, HoldAll];
>
> I'm not very sure what to expect when now trying the same experiment
> (the correct answer 1/3 would be nice), but the actual output surprised
> me:
>
> x=10;
> Integrate[x^2, {x,0,1}]
>
> Yields as output: 0 (without any warnings). Could anyone explain why I
> should have expected this result?
>
> On a side note, I found a similar in the Tech Support column of the
> Mathematica Journal Volume 6, Issue 2, by Carl Roy. He tried the
> integral without limits:
>
> x=10;
> SetAttributes[Integrate, HoldAll];
> Integrate[x^2, x]
>
> In the journal, the output 1000/3 is mentioned, whereas in Mathematica
> 7.0.1 this outputs 1000.
>
> Again, does anyone understand this?
>
> Regards,
>
> Tammo Jan Dijkema
>
>
Changing an attribute for a function, is like making a change to its
code - the effect is completely unpredictable!
The purpose of the HoldAll attribute is to let the writer of the
function get access to the original expression, but that expression may
also get evaluated internally within the function, as the following
example illustrates:
SetAttributes[debugPrint,HoldAll];
debugPrint[x_]:=Print[Unevaluated[x],"=",x];
(try it - it is quite useful)
The correct way to perform your calculation, is
x = 10;
Integrate[y^2, {y, 0, x}]
(where y does not have a value)
David Bailey
http://www.dbaileyconsultancy.co.uk