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# What is the Fourier transform of k^x?

I'm sorry. I meant f(x)=k^(-x), k is a nonnegative integer

I want to evaluate int(f(x)*exp(-2*i*Pi*s*x),x,-infinity,infinity).

This means you integrate [k^(-x)*exp(-2*i*Pi*s*x)] with respect to x from minus infinity to positive infinity. My eternal gratitude I offer to those who answered.

### 2 Answers

- HelmutLv 71 decade agoFavorite Answer
k^x = e^xln(k) and has no Fourier transform

k^-x = e^-xln(k) has the transform

(√(2/π))[(ln(k))/(((ln(k))^2 + ω^2)] , k > 1, x > 0

- Sean HLv 51 decade ago
I posted previously and then realized I was wrong. While it's not defined for any value of k, it will be defined if there is a condition on k. If |k| < 1, it should be defined, and if |k| = 1 it also should be defined. I thought about how to calculate it, but couldn't really figure it out. I think you could get a start by decomposing:

k^x = e^{a*x} + e^{ib*x}

where a + ib = log k. You can find the fourier transform of the second term easily, but the first one is tougher, and I'm not sure how to do it.

Addendum: Doesn't really matter if your function is k^x or k^{-x} since you're integrating over the whole real line (if they did exist, the fourier transform of one could be found from the fourier transform of the other by a simple change of variables).

This integral does not exist unless k is 0. Additionally, the Fourier transform of k^x does not exist when k is an integer larger than 1. When k = 1, f(x) = 1^x = 1, and the fourier transform is the delta function.