Fourier transform of incomplete gamma function

Question

Ultimately I am interested in the Fourier transform of $$ e^{-i\zeta}(-i\zeta)^{-2\epsilon}\Gamma(2\epsilon,-i\zeta) $$ in a series expansion around $\epsilon=0$, so to first order in $$ \lim_{\epsilon\to 0}\int_{-\infty}^\infty \frac{\mathrm{d}\zeta}{2\pi}\, e^{i\zeta(x-1)} (-i\zeta)^{-2\epsilon}\Gamma(2\epsilon,-i\zeta). $$

Assuming that limit and integral can be exchanged, the first order term would be $$ \int_{-\infty}^\infty \frac{\mathrm{d}\zeta}{2\pi}\, e^{i\zeta(x-1)} \Gamma(0,-i\zeta)\,. $$ This can be computed with Mathematica either using Integrate or via FourierTransform and the result is $0$.

On the other hand, getting the antiderivative $F(\zeta)$ and computing $$ \lim_{L\to\infty}(F(L)-F(-L)) $$ seems to suggest a function $\mathrm{sign}(-x)/(1-x)$ (by using the Limit function and testing some $x$ values). Here clearly $x=1$ is a singular point, where a distributional result might become necessary, but let's ignore that for now.

A further possibility is to use the integral representation of the incomplete gamma function $$ \Gamma(n,-i\zeta) = \int_0^\infty\mathrm{d}\tau\, e^{-(-i\zeta+\tau)} (-i\zeta+\tau)^{n-1} $$ and then exchange the Fourier integration over $\zeta$ with the $\tau$ integration. The result of this procedure is $\theta(-x)(1-x)$.

For some context, this is a problem that appears in nuclear/particle physics and $\epsilon$ is the shift from four dimensions $D=4-2\epsilon$.

From a physics point of view there is some motivation for the result to be $\theta(-x)(1-x)$, but maybe there are problems in this being well-defined. There are singular points, for example $\epsilon$ regulates the gamma function for $\zeta\to0$. So the exchange of limit and integration in the first step might already be one of the problems.

I would hope for a discussion of the problems here, or at least pointers in the right direction. In particular: why does Mathematica integrate this to zero; in what theory of the integral would this be the case?

In[1] := Integrate[ Exp[I*(x - 1)*zeta]*Gamma[0, -I*zeta], {zeta, -Infinity, Infinity}]
Out[1] = ConditionalExpression[0, Element[x, Reals]]

In fact, even when regulating the Gamma function via Gamma[2\epsilon, -I*zeta], Mathematica finds $0$ assuming that $-1/2 < \epsilon < 1/2$.

Answer 1

I believe your integral can be evaluated as

$$F_T(x)=\mathcal{F}_{\zeta}[\Gamma(0,-i \zeta )](x-1)=\underset{T\to\infty}{\text{lim}} \left(\int\limits_{-T}^T \Gamma(0,-i \zeta)\, e^{i (x-1) \zeta} \, d\zeta\right)\\=\underset{T\to\infty}{\text{lim}}\left(\frac{-2\, \text{Ci}(T) \sin(T (x-1))+2\, \text{Si}(T x)+(\pi-2\, \text{Si}(T)) \cos (T (x-1))-\pi }{x-1}\right)\tag{1}$$

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Figure (1) below illustrates formula (1) above for $F_T(x)$ evaluated at $T=1,000,000$. I believe $F_T(x)=0$ for $x>0$, $F_T(0)=\pi$, and $\underset{\epsilon \to 0^+}{\text{lim}}F_T(-\epsilon )=2 \pi$.

Figure (1): Illustration of formula (1) for $F_T(x)$

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I believe

$$F(x)=\lim\limits_{T\to\infty} F_T(x)=\left\{\begin{array}{cc} \frac{\pi}{1-x} (\text{sgn}(-x)+1) & x\le 0 \\ 0 & x>0 \\ \end{array}\right.=\left\{\begin{array}{cc} \frac{2 \pi}{1-x} & x<0 \\ \pi & x=0 \\ 0 & x>0 \\ \end{array}\right.\tag{2}$$

since Mathematica indicates

$$\mathcal{F}_x^{-1}[F(x+1)](\zeta )=\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(x+1)\, e^{-i \zeta x} \, dx=\Gamma (0,-i \zeta)\,,\quad\zeta\in\mathbb{R}\tag{3}.$$