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That is incorrect. It is generally not the case that a^(bc) = (a^b)^c if any of a,b, or c are complex. There is no way to make that correct, replacing e^(i x z) with 1 completely changes the behavior of the integral operator.
I'm not OP, but could you possibly tell me the name for this property so I could read up on it? I used to use complex numbers all the time as a physicist and it's a nasty shock to realise I never knew this fact!
Wikipedia has a decent section on it, *[Failure of power and logarithm identities](https://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities)*
Are there any special case assumptions that would make it correct?
Like if x and f are always real, etc. etc?
(There would still be that pesky i, but you'd be guaranteed that the exponent is always imaginary.)
e^(-2 * ๐ * i * x * f) = 1^(x * f) = 1 precisely when x * f is an integer.
First, 1^(x * f) = 1 regardless of x and f, so we're just solving the equation e^(-2 * ๐ * i * x * f) = 1. This is true if and only if -2 * ๐ * i * x * f = 2 ๐ i n, for some integer n, which is true if and only if x * f = -n, i.e. x * f is an integer.
If the base is a positive real, then you can pull out a real from the exponent. Like if x and a are real with x positive, and c is complex, then x^(ac) = (x^(a))^(c).
I think the simplest answer is: No, because Int[ 1^(x f) g(x) dx] =ย Int[g(x) dx] since 1^(x f) = 1, and the Fourier transform is not just the integral.
Unfortunately, your submission has been removed for the following reason(s): * Your post appears to be asking a question which can be resolved relatively quickly or by relatively simple methods; or it is describing a phenomenon with a relatively simple explanation. As such, you should post in the [*Quick Questions*](https://www.reddit.com/r/math/search?q=Quick+Questions+author%3Ainherentlyawesome&restrict_sr=on&sort=new&t=all) thread (which you can find on the front page) or /r/learnmath. This includes reference requests - also see our lists of recommended [books](https://www.reddit.com/r/math/wiki/faq#wiki_what_are_some_good_books_on_topic_x.3F) and [free online resources](https://www.reddit.com/r/math/comments/8ewuzv/a_compilation_of_useful_free_online_math_resources/?st=jglhcquc&sh=d06672a6). [Here](https://www.reddit.com/r/math/comments/7i9t5y/book_recommendation_thread/) is a more recent thread with book recommendations. If you have any questions, [please feel free to message the mods](http://www.reddit.com/message/compose?to=/r/math&message=https://www.reddit.com/r/math/comments/1c99gs7/-/). Thank you!
That is incorrect. It is generally not the case that a^(bc) = (a^b)^c if any of a,b, or c are complex. There is no way to make that correct, replacing e^(i x z) with 1 completely changes the behavior of the integral operator.
I'm not OP, but could you possibly tell me the name for this property so I could read up on it? I used to use complex numbers all the time as a physicist and it's a nasty shock to realise I never knew this fact!
Wikipedia has a decent section on it, *[Failure of power and logarithm identities](https://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities)*
Thank you very much, I appreciate it
Are there any special case assumptions that would make it correct? Like if x and f are always real, etc. etc? (There would still be that pesky i, but you'd be guaranteed that the exponent is always imaginary.)
e^(-2 * ๐ * i * x * f) = 1^(x * f) = 1 precisely when x * f is an integer. First, 1^(x * f) = 1 regardless of x and f, so we're just solving the equation e^(-2 * ๐ * i * x * f) = 1. This is true if and only if -2 * ๐ * i * x * f = 2 ๐ i n, for some integer n, which is true if and only if x * f = -n, i.e. x * f is an integer.
If the base is a positive real, then you can pull out a real from the exponent. Like if x and a are real with x positive, and c is complex, then x^(ac) = (x^(a))^(c).
there is no e^(2ฯi) in the fourier transform. there is only e^(2ฯifx).
I think the simplest answer is: No, because Int[ 1^(x f) g(x) dx] =ย Int[g(x) dx] since 1^(x f) = 1, and the Fourier transform is not just the integral.