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Integrate[Abs[Cos[u]], u] for u real
*To*: mathgroup at smc.vnet.net
*Subject*: [mg73753] Integrate[Abs[Cos[u]], u] for u real
*From*: "David W. Cantrell" <DWCantrell at sigmaxi.net>
*Date*: Tue, 27 Feb 2007 05:52:05 -0500 (EST)
*References*: <errl49$80t$1@smc.vnet.net> <erugbv$8o$1@smc.vnet.net>
In a thread with the informative title "Fw: Re: Fw: 2",
"dimitris" <dimmechan at yahoo.com> wrote:
> Hi Jerry.
>
> Someone could say this conversation is a little pointless since I
> don't have version 4.2.
>
> As I said I could find many occasions that 5.2 shows superior
> performance than 4.2. But as I understood is a matter of what
> applications someone is interested in.
>
> Anyway...
> I want you just to execute the following in your version 4.2:
>
> In[183]:=
> Integrate[Abs[Cos[u]], {u, 0, x*Pi}]
>
> Out[183]=
> Integrate[Abs[Cos[u]], {u, 0, Pi*x}]
>
> (*in version 4.0 you get the incorrect result Sqrt[Cos[Pi x]^2] Tan[Pi x];
> the correct symbolic result
> is Sqrt[Cos[Pi x]^2] Tan[Pi x] + 2 Floor[x + 1/2] see here
[snip]
I presume you meant for us to look specifically at Paul Cally's
<http://groups.google.com/group/comp.soft-sys.math.mathematica/msg/b1daeacee8a40791>.
Besides the fact that x was implicitly assumed to be real, it should be
noted that Cally's "correct" result is not fully correct, being
Indeterminate when x = (1/2 + an integer).
A result which is fully correct follows from the fact, not obtainable by
Mathematica AFAIK, that
for real u, Abs[Cos[u]] has a continuous antiderivative
Sin[u] (-1)^Floor[u/Pi + 1/2] + 2 Floor[u/Pi + 1/2].
Hence, the definite integral considered by Cally is correctly (and also
more simply)
Sin[Pi x] (-1)^Floor[x + 1/2] + 2 Floor[x + 1/2].
Perhaps a result equivalent to the above already appears at the Wolfram
Functions site, but I was unable to find it.
It's also interesting to see what the current version does if we try to
restrict x so that things should be easy:
In[5]:=
Integrate[Abs[Cos[u]], {u, 0, Pi x}, Assumptions-> -1/2 <= x <= 1/2]
Out[5]=
Piecewise[{{1, x == 1/2}, {Sin[Pi*x], -1/2 <= x < 0 || 0 < x < 1/2}}]
Strangely, Mathematica's answer is mute for x == 0 and it considers
x == 1/2 separately. The simple complete answer is merely Sin[Pi x].
David W. Cantrell
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