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I see many results concerning the asymptotics of Fourier transforms. These link in particular the regularity/continuation properties of the function to the polynomial/exponential decay of its Fourier transform. However, these results often hold only in the real variable. I am interested in the Fourier transform "along the imaginary axis" instead.

Let us be more precise. I am interested by the digamma function $\psi = \frac{\Gamma'}{\Gamma}$, and in the function $$h(\nu) = \exp\left(-\alpha \psi \left( \frac14 \pm \frac{i\nu}{2} \right)\right),$$

where $\alpha$ is a fixed parameter, say $\alpha > 1$. I am interested in the asymptotic behavior of the Fourier transform of $h$ at $+\infty$. More precisely, $$\widehat{h}(x) = \int_{\mathbb{R}} h(\nu) e^{ix\nu} d\nu.$$

How to get asymptoptics when $x \to +\infty$ in this situation? I have no feeling about what determines it: size? variations? only asymptotics of $h$?

I had many trials, not convincing. Typically, just changing variables, I can get an expression of the shape $$e^{-\frac{x}{2}} \int_{i\mathbb{R}} e^{-\alpha \psi(u)} e^{2xu} du$$

which looks more like a Laplace (?) transform than a Fourier transform. I was motivated by the fact that I am expecting for other reasons an exponential decay as above, so that I am hoping for a polynomial behavior in $x$ for the remaining integral. However, is the growth/decay estimate of this last integral easier to understand than the original one?

So my question could be synthzised into

Do we have $\int_{i\mathbb{R}} e^{-\alpha \psi(u)} e^{xu} du \ll x^A$ for a certain $A$?

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  • $\begingroup$ @TheSimpliFire Yes, thanks. Post edited $\endgroup$ Oct 22 '20 at 14:18
  • $\begingroup$ Your transformed integral does converge at $u=+\infty$ if $x>0$. Shouldn't you integrate along the imaginary axis? $\endgroup$
    – Gary
    Oct 22 '20 at 15:00
  • $\begingroup$ @Gary You are right, thanks, as a consequence of the change of variables. So this is not really anymore a genuine Laplace transform. This change of variables is just moving the original integration line from $(1/4)$ to $(0)$ $\endgroup$ Oct 23 '20 at 0:15
  • $\begingroup$ I think this change of variable is not simplifying the problem. Without doing the change of variable, the decay of the Fourier transform is linked to the regularity of your function, which can be proved by integrating by parts. It seems to me your function $h$ is analytic (the gamma function is nice on this line right?), so it's Fourier transform should decay faster than any polynomial. $\endgroup$
    – LL 3.14
    Oct 24 '20 at 19:37
  • $\begingroup$ @LL3.14 Yes I expect the exponential decay I indicated, but I would like to prove it and possibly to have more precisions (typically, and equivalence with $e^{-x/2}x^A$ for some explicit $A$ $\endgroup$ Oct 25 '20 at 2:24
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Happy new year! Here's a late answer.

It will be shown that, when $h(v)=\exp\left(-\alpha\psi\left(\frac14+\frac{iv}{2}\right)\right)$,

$$\widehat{h}(x)=(2\alpha)^{1/4}\sqrt{\pi}e^{\alpha\gamma}\cdot\frac{e^{-x/2+2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$ where $\gamma$ is the Euler-Mascheroni constant.

On the other hand, it is trivial that when $h(v)=\exp\left(-\alpha\psi\left(\frac14-\frac{iv}{2}\right)\right)$, $\widehat{h}(x)=0$ for $x>0$.


By the substitution $u=\frac14+\frac{iv}{2}$,

$$\begin{align} \widehat{h}(x) &:=\int_{\mathbb R}\exp\left(-\alpha\psi\left(\frac14+\frac{iv}{2}\right)\right)e^{ixv}dv \\ &=-2ie^{-x/2}\int_{\frac14+i\mathbb R}\underbrace{e^{-\alpha\psi(u)}e^{2xu}}_{:=f(u)}du \\ H(x)&:=\frac i2e^{x/2}\cdot\widehat{h}(x)=\int_{\frac14+i\mathbb R}e^{-\alpha\psi(u)}e^{2xu}du \\ \end{align} $$

By residue theorem and considering the exponential decay of $f(z)$, it can be shown that $$H(x)-\int_{-\frac12+i\mathbb R}f(u)du=2\pi i\operatorname*{Res}_{z=0}f(z)$$ $$H(x)=2\pi i\operatorname*{Res}_{z=0}f(z)+\underbrace{\int_{-\frac12+i\mathbb R}f(u)du}_{:=J_0}$$


Lemma 1: $J_0=O(e^{-x})$.

Proof:

$$\begin{align} \left|\int_{-\frac12+i\mathbb R}f(u)du\right| &=\left|\int_{\mathbb R}e^{-\alpha\psi(-1/2+iu)}e^{-x+2xiu}du\right| \\ &\le\int_{\mathbb R}\left|e^{-\alpha\psi(-1/2+iu)}e^{-x+2xiu}\right|du \\ &=e^{-x}\int_{\mathbb R}\left|e^{-\alpha\psi(-1/2+iu)}\right|du \\ &=e^{-x}\int_{\mathbb R}\left|e^{-\alpha\psi(3/2-iu)-\alpha\pi i\tanh(\pi u)}\right|du \quad (1)\\ &=e^{-x}\int_{\mathbb R}\left|e^{-\alpha\psi(3/2-iu)}\right|du \\ &=Ce^{-x} \qquad (2) \end{align} $$

$(1)$: By the reflection formula $\psi(1-x)-\psi(x)=\pi\cot(\pi x)$.

$(2)$: The last integral can be considered as a constant $C$ because it converges (as $e^{-\alpha\psi(3/2-iu)}\approx u^{-\alpha}$ for large $|u|$ and $\alpha>1$) and is independent of $x$.


Now, let's focus on the residue at $0$. Trivially, $$2\pi i\operatorname*{Res}_{z=0}f(z)=\oint_{|z|=R}f(z)dz \qquad R<1$$

The trick here is to take $R=\sqrt{\frac{\alpha}{2x}}$ (I will explain how to come up with this choice of contour on request).

Define $\phi(z)=\psi(z)+\frac1z+\gamma$. We have $\phi(z)=O(|z|)$ as $z\to 0$.

$$\begin{align} 2\pi i\operatorname*{Res}_{z=0}f(z) &=\oint_{|z|=R}f(z)dz \\ &=\oint_{|z|=R}\exp\left(-\alpha\left(-\frac1z-\gamma+\phi(z)\right)+2xz\right)dz \\ &=e^{\alpha\gamma}\oint_{|z|=R}\exp\left(\frac{\alpha}{z}+2xz\right)\left(e^{-\alpha\phi(z)}-1+1\right)dz \\ &=e^{\alpha\gamma}\underbrace{\oint_{|z|=R}\exp\left(\frac{\alpha}{z}+2xz\right)dz}_{:=J_1} \\ &+e^{\alpha\gamma}\underbrace{\oint_{|z|=R}\exp\left(\frac{\alpha}{z}+2xz\right)\left(e^{-\alpha\phi(z)}-1\right)dz}_{:=J_2} \\ \end{align} $$

Lemma 2: $$J_1=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$

$$\begin{align} J_1 &=\int^\pi_{-\pi}\exp\left(\frac{\alpha}{R}e^{-i\theta}+2xRe^{i\theta}\right)iRe^{i\theta}d\theta \\ &=i\sqrt{\frac{\alpha}{2x}}\int^\pi_{-\pi}\exp\left(\alpha\sqrt{\frac{2x}{\alpha}}e^{-i\theta}+2x\sqrt{\frac{\alpha}{2x}}e^{i\theta}\right)e^{i\theta}d\theta \\ &=i\sqrt{\frac{\alpha}{2x}}\int^\pi_{-\pi}\exp\left(2\sqrt{2\alpha x}\cos\theta\right)e^{i\theta}d\theta \\ &=2i\sqrt{\frac{\alpha}{2x}}\int^\pi_{0}\cos\theta \, e^{2\sqrt{2\alpha x}\cos\theta} d\theta \qquad (1)\\ &=2i\sqrt{\frac{\alpha}{2x}}\cdot\pi I_1\left(2\sqrt{2\alpha x}\right) \qquad (2)\\ &=2\pi i\sqrt{\frac{\alpha}{2x}}\cdot \frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{\sqrt{2\pi}\sqrt{2\sqrt{2\alpha x}}}\left(1+O\left(\frac1{\sqrt x}\right)\right) \qquad (3) \\ &=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right) \\ \end{align} $$

$(1)$: The imaginary part cancels out due to oddness, and the extra factor of $2$ is due to the evenness of the real part.

$(2)$: $I_1$ is the first order modified Bessel function of the first kind.

$(3)$: Due to the well-known asymptotic expansion $I_1(z)=\frac{e^z}{\sqrt{2\pi z}}\left(1+O\left(\frac1z\right)\right)$ for $z\to\infty$.

Lemma 3: $$J_2=O\left(\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{5/4}}\right)$$

Proof:

$$\begin{align} |J_2| &=\left|\oint_{|z|=R}\exp\left(\frac{\alpha}{z}+2xz\right)\left(e^{-\alpha\phi(z)}-1\right)dz\right| \\ &=\left|\int^\pi_{-\pi}e^{2\sqrt{2\alpha}\sqrt{x}\cos\theta}\left(\exp\left(-\alpha\phi(Re^{i\theta})\right)-1\right)iRe^{i\theta}d\theta\right| \\ &\le R\int^\pi_{-\pi}e^{2\sqrt{2\alpha}\sqrt{x}\cos\theta}\left|\exp\left(-\alpha\phi(Re^{i\theta})\right)-1\right|d\theta \\ &\le R\int^\pi_{-\pi}e^{2\sqrt{2\alpha}\sqrt{x}\cos\theta}C|Re^{i\theta}|d\theta \qquad (1)\\ &=CR^2\int^\pi_{-\pi}e^{2\sqrt{2\alpha}\sqrt{x}\cos\theta}d\theta \\ &=CR^2\cdot 2\pi I_0(2\sqrt{2\alpha}\sqrt{x}) \\ &=C\cdot\frac{\alpha}{2x}\cdot 2\pi \cdot \frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{\sqrt{2\pi}\sqrt{2\sqrt{2\alpha x}}}\left(1+O\left(\frac1{\sqrt x}\right)\right) \qquad (2)\\ &=O\left(\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{5/4}}\right) \end{align} $$

$(1)$: As $\phi(z)=O(|z|)$, $\exp\left(-\alpha\phi(z)\right)-1=\exp(O(|z|))-1=1+O(|z|)-1=O(|z|)$.

$(2)$: Due to the well-known asymptotic expansion $I_0(z)=\frac{e^z}{\sqrt{2\pi z}}\left(1+O\left(\frac1z\right)\right)$ for $z\to\infty$.

Therefore, $$2\pi i\operatorname*{Res}_{z=0}f(z)=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}e^{\alpha\gamma}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)+O\left(\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{5/4}}\right)$$ $$\implies 2\pi i\operatorname*{Res}_{z=0}f(z)=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}e^{\alpha\gamma}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$


In conclusion, $$H(x)=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}e^{\alpha\gamma}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)+O(e^{-x})$$

$$\implies H(x)=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}e^{\alpha\gamma}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$

As we defined $H(x)=\frac i2e^{x/2}\cdot\widehat{h}(x)$, it can be concluded, eventually, $$\widehat{h}(x)=(2\alpha)^{1/4}\sqrt{\pi}e^{\alpha\gamma}\cdot\frac{e^{-x/2+2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$

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  • $\begingroup$ Thank you very much for this detailed answer! I suppose that the choice of $R$ is made so that the cosine appears (it is a balancing to make the expression turn to be $z+z^{-1}$)? However it seems less natural to make appear the function $\phi(z)$, is it a standard way to write some local properties of $\psi$? Otherwise, every step is totally clear, thanks again. $\endgroup$ Jan 3 '21 at 10:07
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I'll provide a sketch answer, to illustrate a general process. It's all about poles! Admittedly I acted as though the function was the transform of something un smooth, like a counting function. Perhaps

$$h(z) := \psi\left(\frac 14 + \frac i2 z\right).$$

Contour integral approach

First consider the poles of $h(z)$, which can be deduced from those for the original Digamma function (in turn corresponding just to poles of $\Gamma$). These are simple poles at $$z_n := (2n + 1/2)i,\qquad n \in \{0,1,2,3,\ldots\},$$

each with residue $2/i = -2i$ (since those for the digamma function each have residue $1$, which we have "scaled" by a factor of $i/2$).

Fix $S$ such that $ \mathrm{Im}(z_N) < S < \mathrm{Im}z_{N+1}$ for some $N$, and let $T>0$. Consider the following contour integral over the rectangle:

(My bad! The labels on the $x_n$ should start from $0$ in the picture.)

Contour integral

Fix $x>0$ for now.

For $f(z) := h(z) e^{ixz}$, Cauchy's Residue theorem gives that

$$\int_{A_T} f(z) = \int_{B_T} f(z) + \left(\int_{C_T} f(z) + \int_{D_T} f(z)\right) + 2\pi i\sum_{n=0}^N \mathrm{res}(f,z_n)$$

This simplifies in the limit with the following claim, which should follow from a naive uniform estimate on $|h|$ on $D_T$ and $C_T$. Since I haven't proven it, I'll label it as an assumption:

Assumption: As $T → ∞$,

$$ \int_{C_T} f(z)\ dz + \int_{D_T} f(z)\ dz \to 0.$$

Also noting that

$$\mathrm{res}(f,z_n) = -2i e^{ixz_n} = -2i e^{-\frac{4n+1}2x},$$

in the limit we would have an asymptotic expansion, with the resonances given precisely by the residues:

$$\hat h(x) = \lim_{T→ ∞}\int_{A_T} f(z)\ dz= 4\pi \sum_{n=0}^N e^{-x(4n+1)/2} + \int_{\mathbb R + iS} h(z) e^{ixz}\ dz.$$

To finish off, one would have to show that the last integral decays at a faster rate than the other terms (in terms of $x$), which I shall not do. Does Paley–Wiener still apply?

Full asymptotic expansion?

Perhaps simpler is to ignore the unrigourousness above and investigate the natural conjecture which results from taking $ N → ∞ $:

$$\hat h(x) = 4\pi \sum_{n=0}^∞ e^{-x(4n+1)/2} = 4\pi \frac{ e^{3 x/2}}{e^{2 x}-1}.$$

This conjecture would also come from some formal-series argument; it should be enough to show that the inverse transform of this gives the original $h$. I'm not convinced that it is true, but it's worth a pop.

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  • $\begingroup$ In the question, the function $h$ was basically the exponential of your $h$. You are looking at a different problem. $\endgroup$
    – Gary
    Oct 31 '20 at 10:11
  • $\begingroup$ Apologies! I got ahead of myself there. It seems for the real $h$ the singularities are in the same place (the $z_n$) but are now essential — so no residue theorem applies, and not even Wiener–Ikehara. (That does explain why nobody answered until now.) $\endgroup$
    – Good Boy
    Oct 31 '20 at 12:26
  • $\begingroup$ @GoodBoy Yes this is the main issue with this exponential of $h$, we cannot pick resisues. However, thanks for having tried and posted the details. $\endgroup$ Nov 1 '20 at 7:56

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