Asymptotic behavior of a Fourier/Laplace transform

Question

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 $A$?

Answer 1

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 $x>0$.

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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 $$H(x)-\int_{-\frac12+i\mathbb R}f(u)du=2\pi i\operatorname*{Res}_{z=0}f(z)$$ $$H(x)=2\pi i\operatorname*{Res}_{z=0}f(z)+\underbrace{\int_{-\frac12+i\mathbb R}f(u)du}_{:=J_0}$$

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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 $x$.

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

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

Answer 2

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.)

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.