Re: Fourier Transform
- To: mathgroup at smc.vnet.net
- Subject: [mg93584] Re: Fourier Transform
- From: "Nasser Abbasi" <nma at 12000.org>
- Date: Sun, 16 Nov 2008 07:03:41 -0500 (EST)
- References: <gf8rgp$poq$1@smc.vnet.net> <gfbur0$43e$1@smc.vnet.net> <gfgqhc$dkv$1@smc.vnet.net> <gfh3cl$g1l$1@smc.vnet.net> <gfimic$g0f$1@smc.vnet.net>
- Reply-to: "Nasser Abbasi" <nma at 12000.org>
"Nikolaus Rath" <Nikolaus at rath.org> wrote in message news:gfimic$g0f$1 at smc.vnet.net... > > > But what is the point of that? There is nothing special about the > integral in a Fourier transform, so any functionality that helps to > evaluate integrals is best added to Integrate[]. > > -Nikolaus > my 2 cents on this good question is the following: Using FourierTransform and using the equivlant integeration command is not always the same. ------------------- In[6]:= func = Cos[w0 t]; Integrate[func* Exp[(-I)*w*t], {t, -Infinity, Infinity}] Integrate::idiv: Integral of E^(-I t w) Cos[t w0] does not converge on {-\[Infinity],\[Infinity]}. FourierTransform[func, t, w] Sqrt[Pi/2] DiracDelta[w-w0]+Sqrt[Pi/2] DiracDelta[w+w0] ------------------------ Cos[] is not square integrable (its average power is not zero, the average power of cos[] is 1/2) So, Cos[] does not have a FourierTransform (one of the conditions to have FourierTransform is for the function to be square integrable). So, for non-square integrable functions, FourierTransform does not exist. Yet, you see that FourierTransform does generate a FourierTransform for the Cos[]. Since there are many useful functions which have infinite energy, but we want to find its fourier transform, a more generalized definition is used for the integral which uses a dirac delta function to allow one to integrate such functions, and that is why sometimes FourierTransform gives different answer than direct use of the Intgeration function. reference Delta Functions: An Introduction to Generalised Functions by R. F. Hoskins Nasser