I believe fourier_transform does use integration. But this is what the
integral returns
In [15]: fourier_transform(f(x), x, z).rewrite(Integral).subs(f(x), 1)
Out[15]:
∞
⌠
⎮ -2⋅ⅈ⋅π⋅x⋅z
⎮ ℯ dx
⌡
-∞
In [16]: fourier_transform(f(x), x, z).rewrite(Integral).subs(f(x), 1).doit()
Out[16]:
⎧ │ ⎛ -ⅈ⋅π ⎞│
│ ⎛ ⅈ⋅π ⎞│
⎪ │ ⎜ ───── ⎟│
│ ⎜ ─── ⎟│
⎪ │ ⎜ 2 ⎟│
π │ ⎜ 2 ⎟│ π
⎪ 0 for │periodic_argument⎝ℯ ⋅polar_lift(z), ∞⎠│ <
─ ∧ │periodic_argument⎝ℯ ⋅polar_lift(z), ∞⎠│ < ─
⎪
2 2
⎪
⎨∞
⎪⌠
⎪⎮ -2⋅ⅈ⋅π⋅x⋅z
⎪⎮ ℯ dx otherwise
⎪⌡
⎪-∞
⎩
So it can't get the result in terms of the Dirac delta function. In
fact, if you pass noconds=False to fourier_transform, you get those
same conditions. Actually, if I plug 0 into the conditions, I get nan
< pi/2, and if I plug any other real number, I get False. So I'm not
really sure about the result anyway. I was never clear what the
periodic_argument at oo was. I CCd Tom. Maybe he can answer.
Aaron Meurer
On Wed, Oct 30, 2013 at 3:32 PM, Pablo Puente <[email protected]> wrote:
> Yes, this is a defect.
>
> I created in fact a Issue this week for it:
> https://code.google.com/p/sympy/issues/detail?id=4079
>
>
>
> On Wednesday, October 30, 2013 12:03:32 PM UTC+1, Harsh Gupta wrote:
>>
>> >>> from sympy.integrals import transforms
>> >>> FT = fourier_transform
>> >>> from sympy.abc import x, k
>> >>> FT(1,x,k)
>> 0
>>
>> Sympy evaluates the fourier transform of 1 as 0 though it is
>> dirac_delta(k).
>> Similarly sympy evaluates the fourier transform of powers of x as 0 as
>> well.
>>
>> http://mathworld.wolfram.com/FourierTransform1.html
>>
>> This might be arising because fourier transform of a 1 is not evaluated by
>> direct integration
>> rather it is evaluated using a form of the generalized unit function.
>>
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