互质正整数p,q下(x-c)^(-p/q)的傅里叶变换求解问询

Great question—this is a classic extension of single-branch contour integrals to general rational powers, and the core ideas build directly on the q=1 and q=2 cases you've already worked through. Let's break down the step-by-step strategy:

1. Set Up the Branch Cut and Function Branch #

First, since $\mathrm{Re}(c)=0$, let $c=ib$ where $b\in\mathbb{R}$. The function $(z-c)^{-p/q}$ has a branch point at $z=c$, so we need to define a branch cut that keeps our real-axis integration path valid. The most convenient choice is a cut along the imaginary axis extending from $c$ to infinity (either up to $i\infty$ or down to $-i\infty$—we'll pick upward for $a>0$ below).

Define the branch by setting $z-c = re^{i\phi}$ where $\phi\in(-\pi/2, 3\pi/2)$. This ensures the real axis ($z=x\in\mathbb{R}$) lies entirely within the domain of the single-valued function, with $\phi=\arg(x-ib)\in(-\pi/2, \pi/2)$ for real $x$.

2. Build the Contour (for $a>0$) #

We construct a closed contour that encloses the real axis and avoids the branch cut:

• $\Gamma_1$: The real-axis segment $[-R, R]$ (this is our target integral, with $R\to\infty$).
• $\Gamma_2$: A large semicircular arc in the upper half-plane, $z=Re^{i\theta}$ for $\theta\in[0,\pi]$. By the Jordan Lemma, this integral vanishes as $R\to\infty$—the $e^{iaz}$ term decays exponentially in the upper half-plane, and $(z-c)^{-p/q}$ behaves like $R^{-p/q}$, so the whole integral goes to zero.
• $\Gamma_3$: The upper side of the branch cut, running from the end of $\Gamma_2$ (at $iR$) down to a small neighborhood around $c$.
• $\Gamma_4$: A small circular arc of radius $\varepsilon$ around $c$, connecting the upper and lower sides of the branch cut. For $p<q$, this integral vanishes as $\varepsilon\to0$ (since $1-p/q>0$, the integrand is bounded by $\varepsilon^{1-p/q}$).
• $\Gamma_5$: The lower side of the branch cut, running back from the small arc to the end of $\Gamma_2$.

3. Apply Cauchy's Theorem #

Since our contour encloses no analytic singularities (only the branch point on the boundary), the total contour integral is zero:
$$\int_{\Gamma_1} + \int_{\Gamma_2} + \int_{\Gamma_3} + \int_{\Gamma_4} + \int_{\Gamma_5} = 0$$
Taking $R\to\infty$ and $\varepsilon\to0$, the integrals over $\Gamma_2$ and $\Gamma_4$ disappear, leaving:
$$I = -\left(\int_{\Gamma_3} + \int_{\Gamma_5}\right)$$
where $I$ is our target integral.

4. Evaluate the Branch Cut Integrals #

The key here is the phase shift of the multi-valued function across the cut. When moving from the upper to lower side of the cut, the argument of $z-c$ decreases by $2\pi$, so $(z-c)^{-p/q}$ is multiplied by $e^{-2\pi ip/q}$.

For the upper cut ($z=c+it$, $t>0$), $z-c=te^{i\pi/2}$, so $(z-c)^{-p/q}=t{-p/q}e^{-ip\pi/(2q)}$. The integral over $\Gamma_3$ becomes:
$$\int_{\Gamma_3} = -i e^{iac} e^{-ip\pi/(2q)} \int_0^\infty t^{-p/q}e{-at}dt$$
For the lower cut, $z-c=te^{-3\pi i/2}$, so $(z-c)^{-p/q}=t{-p/q}e^{3ip\pi/(2q)}$, giving:
$$\int_{\Gamma_5} = i e^{iac} e^{3ip\pi/(2q)} \int_0^\infty t^{-p/q}e{-at}dt$$

Adding these two integrals and using Euler's formula to simplify the exponential terms, we get:
$$\int_{\Gamma_3} + \int_{\Gamma_5} = -2 e^{iac} e^{ip\pi/(2q)} \sin\left(\frac{\pi p}{q}\right) a^{1-p/q} \Gamma\left(1-\frac{p}{q}\right)$$
Here, we've used the Gamma function identity $\int_0^\infty t^{s-1}e{-at}dt = a^{-s}\Gamma(s)$ with $s=1-p/q$.

5. Final Result #

Substituting back into our earlier equation, the target integral is:
$$I = 2 e^{iac} e^{ip\pi/(2q)} \sin\left(\frac{\pi p}{q}\right) a^{1-p/q} \Gamma\left(1-\frac{p}{q}\right)$$

6. Extensions #

• For $a<0$: Use a lower-half-plane contour instead (Jordan Lemma applies here since $e^{iaz}$ decays in the lower half-plane), leading to a similar result with adjusted phase terms.
• For $p>q$: Write $p=kq+r$ (where $0<r<q$) to split the integrand into $(x-c)^{-k}(x-c){-r/q}$. You can then use integration by parts or extend the contour method to handle the higher-power term, treating it as a repeated singularity alongside the branch point.

内容的提问来源于stack exchange，提问作者user3763801