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Stephen Crowley edited this page Sep 25, 2023 · 1 revision

Certainly, you can attempt to approach the problem using term-by-term integration if the series is absolutely convergent. The hypergeometric series for the $_2F_1$ function is indeed absolutely convergent within the range $-1 < x < 1$.

The Gauss Hypergeometric Function, $_2F_1(a, b; c; z)$, has the series representation:

$$_2F_1(a, b; c; z) = \sum_{k=0}^{\infty} \frac{(a)_k (b)_k}{(c)_k} \frac{z^k}{k!}$$

where $(a)_k$ represents the Pochhammer symbol, defined as $(a)_k = a(a+1)(a+2)\ldots(a+k-1)$.

Derivation for Fourier Transform:

Given that the Fourier transform of the function $f(x)$ is defined as:

$$\hat{f}(k) = \int_{-\infty}^{\infty} e^{-2\pi i k x} f(x) \,dx$$

For the series representation of $_2F_1$, if you substitute this series into the Fourier Transform and swap the sum and the integral, you will get term by term integration. Here, ensure to be meticulous with the conditions of the

This would involve expressing the result of each term’s integration in terms of hypergeometric functions again. Likely the terms would be more complicated due to the integral, which might be where the $_3F_1$ representation comes in. For instance, the more involved integrals might lead to representations involving more parameters, hence transitioning from a $_2F_1$ to a $_3F_1$ representation.

Procedure:

  1. Substitute the series representation of $_2F_1$ into the Fourier Transform integral.
  2. Perform term-by-term integration.
  3. Express each term’s result in terms of hypergeometric functions or other special functions.
  4. Identify the resulting series as a known hypergeometric function, such as $_3F_1$.

Final Result:

The meticulous execution of each step should help in understanding how the $_3F_1$ function emerges as the Fourier transform of the given $_2F_1$ representation. It is essential to ensure absolute convergence at each step, particularly when interchanging the order of summation and integration.

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