Re: Re: Symbolic integration

*To*: mathgroup at smc.vnet.net*Subject*: [mg102696] Re: [mg102685] Re: Symbolic integration*From*: Leonid Shifrin <lshifr at gmail.com>*Date*: Wed, 19 Aug 2009 07:02:44 -0400 (EDT)*References*: <200908161039.GAA01165@smc.vnet.net> <h6bh5c$6pi$1@smc.vnet.net>

Because, as you could have seen from the rest of my post, in that case Mathematica (with our help) is able to compute the integral - this integral *can* be computed. If you just put the absolute value, this is correct, but useless - in that form Mathematica can not do the (indefinite) integral. If you rewrite it with UnitStep, you are basically using an identity: f[Abs[x]]=UnitStep[x]*f[x]+UnitStep[-x]*f[-x], (assuming that the point x = 0 does not matter, which is true for the integral with non-singular f) but then (at least in your case), Mathematica was able to get the answer. I was answering your initial post, where I had an impression that you needed an anti-derivative (indefinite integral). If you only need an integral in the infinite limits, as you mentioned later, then this might be easier and might not require this trick, although of course this case also follows from the general one, result for which is in my post). That's why. Regards, Leonid On Tue, Aug 18, 2009 at 2:11 PM, tzygmund <tzygmund at googlemail.com> wrote: > Thanks to all of you who replied, with the level of sophistication I > would expect on the Mathematica board. > > Leonid, why would I not want to have an absolute value function inside > the exponent? > > Thanks > > On Aug 17, 1:07 pm, tzygmund mcfarlane <tzygm... at googlemail.com> > wrote: > > I should have known better than to write to the mailing list within my > > first hour of starting to use a new software. I went and RTM. Here is > > what I intended: > > > > ****************************** > > \[Lambda] = (2^(-2/\[Nu]) Gamma[1/\[Nu]]/Gamma[3/\[Nu]])^(-1/2); > > > > Integrate[(\[Nu] * > > Exp[-0.5 *Abs[z/ \[Lambda]]^\[Nu]]/(\[Lambda]* 2^(1 + 1/\[Nu])* > > Gamma[1/\[Nu]])), {z, -\[Infinity], +\[Infinity]}] > > ****************************** > > > > Thanks for your comment. > > > > > > > > On Mon, Aug 17, 2009 at 4:54 AM, DrMajorBob<btre... at austin.rr.com> > wrote: > > > The integral can't be computed: > > > > > Integrate[\[Nu] E[-0.5 Abs[ > > > z/ \[Lambda]]^\[Nu]]/(\[Lambda] 2^(1 + 1/\[Nu]) Gamma[ > > > 1/\[Nu]]), z] > > > > > (2^(-1 - 1/\[Nu]) \[Nu] \[Integral]E[-0.5 Abs[ > > > z/\[Lambda]]^\[Nu]] \[DifferentialD]z)/(\[Lambda] Gamma[1/\[= > Nu= > > ]] > > > ) > > > > > Maybe you meant the first line to be > > > > > \[Lambda] = (2^(-2/\[Nu]) Gamma[1/\[Nu]]/ Gamma[3/\[Nu]])^(-1/2); > > > > > (parentheses for power, not brackets) and then the integral is still > > > undefined: > > > > > Integrate[\[Nu] E[-0.5 Abs[ > > > z/ \[Lambda]]^\[Nu]]/(\[Lambda] 2^(1 + 1/\[Nu]) Gamma[ > > > 1/\[Nu]]), z] > > > > > (2^(-1 - 1/\[Nu]) \[Nu] Sqrt[(2^(-2/\[Nu]) Gamma[1/\[Nu]])/ > > > Gamma[3/\[Nu]]] \[Integral]E[-0.5 2^(-\[Nu] Re[1/\[Nu]]) > > > Abs[z]^\[Nu] Abs[Gamma[1/\[Nu]]/Gamma[3/\[Nu]]]^(\[Nu]/ > > > 2)] \[DifferentialD]z)/Gamma[1/\[Nu]] > > > > > But I doubt that's what you intended, either. > > > > > Bobby > > > > > On Sun, 16 Aug 2009 05:39:45 -0500, tzygmund <tzygm... at googlemail.com> > > > wrote: > > > > >> Hi, > > > > >> I have a fairly simple question which I cannot solve. I want to assign > > >> a symbolic expression to a greek letter and then use this in a > > >> subsequent integral. So, > > >> ******************************** > > >> \[Lambda] -> [ > > >> \!\(\*SuperscriptBox["2", > > >> RowBox[{"[", > > >> FractionBox[ > > >> RowBox[{"-", "2"}], "\[Nu]"], "]"}]]\) Gamma[1/\[Nu]]/ > > >> Gamma[3/\[Nu]]]^(-1/2) > > > > >> Integrate[\[Nu] E[-0.5 Abs[z/ \[Lambda]]^\[Nu]]/(\[Lambda] 2^(1 + > > >> 1/\[Nu]) Gamma[1/\[Nu]]), z] > > >> ********************************** > > > > >> How can I get this to work? > > > > >> Thanks > > > > > -- > > > DrMajor... at bigfoot.com > > >

**References**:**Symbolic integration***From:*tzygmund <tzygmund@googlemail.com>