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Assume $\alpha > -1$. Prove that $$ \frac{1}{(i\xi + 0)^{1+\alpha}} \overset{\mathcal{D}'}{=} \lim_{\varepsilon \to 0^+} (i\xi + \varepsilon)^{-1-\alpha} $$ defines a distribution on $\mathbb{R}$. Where $\alpha > -1$.
Use $\Gamma(z)$ (the Gamma function) and $\frac{1}{(i\xi + 0)^{1+\alpha}}$ to represent the Fourier transform of $u = x_+^\alpha$;

For the first part, I got $$\frac{1}{(i\xi+\varepsilon)^{1+\alpha}}=\left(\frac{\varepsilon-i\xi}{\xi^2+\varepsilon^2}\right)^{1+\alpha}\overset{\varepsilon\to0^+}{==}\left(\pi\delta_0-i\text{p.v.}\frac1\xi\right)^{1+\alpha}$$ I'm not sure that it's right. But I don't have any thought to solve the second part. How can I find the relationship between it with $\Gamma(z)$?

TaD
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  • Hint: Perhaps it helps to realize the Fourier transform is of the form $\frac{\Gamma(\alpha+1)}{\xi^{\alpha+1}} f(\xi)$. – Steven Clark Nov 24 '24 at 20:59
  • @StevenClark, I think it's Laplace transform instead of Fourier transform. – TaD Nov 25 '24 at 01:39
  • Your question is about the Fourier transform and so was my comment. I'm not sure if by $x_+^\alpha$ you mean $$\mathcal{F}x\leftx^{\alpha}, \theta(x)\right=\int\limits_0^\infty x^\alpha, e^{-2 \pi i \xi x} , dx$$ instead of $$\mathcal{F}_x\leftx^{\alpha}\right=\int\limits{-\infty}^\infty x^\alpha, e^{-2 \pi i \xi x} , dx$$ for example, but in either case my comment is correct. – Steven Clark Nov 25 '24 at 04:27
  • If its the one-sided Fourier transform then it also might help to realize this is equivalent to the Mellin transform $$\mathcal{M}_x\left[e^{-2 \pi i \xi x}\right](\alpha +1)=\int\limits_0^\infty e^{-2 \pi i \xi x}, x^\alpha , dx.$$ – Steven Clark Nov 25 '24 at 04:52
  • @StevenClark, you can realize $x_+^\alpha$ as $\mathbf{pf}x_+^\alpha$, which gives $\langle x_+^\alpha,\varphi\rangle=\int_0^\infty x^\alpha \varphi dx$. – TaD Nov 25 '24 at 05:32

2 Answers2

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For the study of $\lim_{\epsilon\to0}(i\xi+\epsilon)^{-1-\alpha}$ perhaps polar form might be useful? $$ (i\xi+\epsilon)^{-\beta} = \left( \sqrt{\xi^2+\epsilon^2}\exp(i\arctan\frac{\xi}{\epsilon}) \right)^{-\beta} \\ = (\xi^2+\epsilon^2)^{-\beta/2} \exp(i\beta\arctan\frac{\xi}{\epsilon}) \\ \to |\xi|^{-\beta} \exp(i\beta\frac{\pi}{2}\operatorname{sign}(\xi)). $$ Here I have only studied the convergence pointwise.

For the case $\alpha=0$ it's however easiest to consider $$ (i\xi+\epsilon)^{-1} = -i\frac{d}{d\xi}\log(i\xi+\epsilon) = -i\frac{d}{d\xi}\left( \ln\sqrt{\xi^2+\epsilon^2} + i\arctan\frac\xi\epsilon \right) \\ \to -i\frac{d}{d\xi}\left( \ln|\xi| + i\frac\pi2\operatorname{sign}(\xi) \right) = -i \left( \operatorname{pv}\frac1\xi + i\delta(\xi) \right) . $$

For the Fourier transform we would like to write $$ \mathcal{F}\{x_+^\alpha\} = \int_0^\infty x^\alpha\,e^{-ix\xi}\,dx $$ but for this isn't convergent. Therefore we add a damping factor: $$ \mathcal{F}\{x_+^\alpha\} = \lim_{\epsilon\to0} \int_0^\infty x^\alpha\,e^{-ix\xi}\,e^{-\epsilon x}\,dx = \lim_{\epsilon\to0} \int_0^\infty x^\alpha\,e^{-x(i\xi+\epsilon)}\,dx . $$ We would now like to set $t=x(i\xi+\epsilon)$ and transform the integral into $$ \int_0^\infty \left(\frac{t}{i\xi+\epsilon}\right)^{\alpha} e^{-t}\,\frac{dt}{i\xi+\epsilon} = (i\xi+\epsilon)^{-\alpha-1} \int_0^\infty t^{\alpha} e^{-t}\,dt = (i\xi+\epsilon)^{-\alpha-1} \Gamma(\alpha+1). $$ We cannot do this directly but need to apply Cauchy's integral theorem over a wedge-like curve.

md2perpe
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  • Well, the part I cannot figure out is the relationship between its Fourier transform with Gamma function. – TaD Nov 29 '24 at 17:32
  • And it seems that I can get $\frac{1}{(i\xi+\varepsilon)^{1+\alpha}}=\left(\frac{\varepsilon-i\xi}{\xi^2+\varepsilon^2}\right)^{1+\alpha}\overset{\varepsilon\to0^+}{==}\left(\pi\delta_0-i\text{p.v.}\frac1\xi\right)^{1+\alpha}$ from $\lim\limits_{\varepsilon\to0}\frac{\varepsilon}{\pi(x^2+\varepsilon^2)}=\delta_0$. – TaD Nov 29 '24 at 17:33
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    $\left(\pi\delta_0-i\text{p.v.}\frac1\xi\right)^{1+\alpha}$ isn't a valid distributional expression unless $\alpha=0$. – md2perpe Nov 29 '24 at 18:09
  • Oh, I see. But for the case of $\alpha=0$, I got $\text{pv}\frac{1}{|\xi|}\exp(-\frac{\pi i}{2}\text{sign}(\xi))$. Then where will I get the $\pi\delta_0$? – TaD Nov 29 '24 at 19:03
  • When $\alpha=0$ you get $$ \frac{1}{i\xi+\epsilon} = -i \frac{d}{d\xi}\log(i\xi+\epsilon) = -i \frac{d}{d\xi} (\ln\sqrt{\xi^2+\epsilon^2}+i\arctan(\frac{\xi}{\epsilon})). $$ What do $\ln\sqrt{\xi^2+\epsilon^2}$ and $\arctan(\frac{\xi}{\epsilon})$ tend to as $\epsilon\to 0$? – md2perpe Nov 29 '24 at 19:27
  • Can I just use the conclusion of $\frac{1}{\xi-i\varepsilon}=\text{p.v.}\frac1\xi+i\pi\delta_0$ then multiply by $-i$? – TaD Nov 29 '24 at 19:37
  • Sure, if you consider that shown before. – md2perpe Nov 29 '24 at 19:49
  • That question above appears again. How can I get the term including $\delta$ by the polar form? Again for the case of $\alpha=0$, I only can get the term of $\frac{1}{i\xi}$. How can I get the $\pi\delta_0$? – TaD Nov 30 '24 at 04:30
  • Look at my earlier comment: what do $\ln\sqrt{\xi^2+\epsilon^2}$ and $\arctan(\frac\xi\epsilon)$ tend as $\epsilon\to0$? – md2perpe Nov 30 '24 at 07:30
  • Well, yes, I know that $\ln \sqrt{\xi^2+\varepsilon^2}\to \frac{1}{|\xi|}$ and $\arctan(\frac\xi\varepsilon)\to \frac{\pi}2\text{sgn}(\xi)$ so that I have the result. But I cannot get that from the form $|\xi|^{-1}\exp(\frac{i\pi}{2}\text{sign}(\xi))$ clearly. (I know it's pv$\frac{1}{i\xi}$ when $|\xi|\neq0$ but how can I know the value when $\xi\to0$? In other words, I cannot determine the coefficient of $\delta_0$) – TaD Nov 30 '24 at 08:07
  • The first limit is incorrect. – md2perpe Nov 30 '24 at 08:12
  • Ah sorry, it's $\ln|\xi|$, I want to say it's derivative is that. I mean I can know it's pv$\frac1{i\xi}+\pi\delta_0$ from the form of $\frac{1}{i \xi+\epsilon}=-i \frac{d}{d \xi} \log (i \xi+\epsilon)=-i \frac{d}{d \xi}\left(\ln \sqrt{\xi^2+\epsilon^2}+i \arctan \left(\frac{\xi}{\epsilon}\right)\right)$ but cannot get that from the form of polar. – TaD Nov 30 '24 at 08:14
  • In the case $\alpha=0$ I suggest that you don't use the polar form but just does what I suggested 12 hours ago. – md2perpe Nov 30 '24 at 08:25
  • OK, then do you mean there won't appear $\delta_0$ term when $\alpha\neq0$? – TaD Nov 30 '24 at 08:28
  • I don't know. I haven't studied it that deeply yet. – md2perpe Nov 30 '24 at 08:34
  • I have expanded my answer with ideas about how to do the Fourier transform. I leave to you to try to work out the details. – md2perpe Nov 30 '24 at 09:44
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#44f}{1 \over \pars{\ic\xi + 0^{+}}^{1+\alpha}} & = {1 \over \pars{\ic\xi}^{1 + \alpha}\,\,\bracks{1 - \ic\on{sgn}\pars{\xi}\,0^{+}}^{1+\alpha}} \\[5mm] & = {1 \over \pars{\ic\xi}^{1 + \alpha}\,\, \braces{1 - \ic\on{sgn}\pars{\xi\bracks{1 + \alpha}}\,0^{+}}} \\[5mm] & = {1 \over \pars{\ic\xi}^{1 + \alpha}} \bracks{{\rm P.V.}\,1 + \ic\pi\on{sgn}\pars{\xi\bracks{1 + \alpha}}\,\delta\pars{1}} \\[5mm] & = \bbx{\color{#44f}{{\rm P.V.}{1 \over \pars{\ic\xi}^{1 + \alpha}}}} \\ & \end{align}

Felix Marin
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  • I don't completely agree. For the first "=", you don't have $i\xi+0^+=(i\xi)(1-i \text{sgn}(\xi)0^+)$ but $(i\xi)(1-i\text{p.v.}\frac1\xi 0^+)$. – TaD Nov 27 '24 at 09:26
  • I got $\displaystyle\frac{1}{(i\xi+\varepsilon)^{1+\alpha}}=\left(\frac{\varepsilon-i\xi}{\xi^2+\varepsilon^2}\right)^{1+\alpha}\overset{\varepsilon\to0^+}{==}\left(\pi\delta_0-i\text{p.v.}\frac1\xi\right)^{1+\alpha}$ from lemma $\displaystyle\lim_{\varepsilon\to0}\frac{\epsilon}{\pi(x^2+\epsilon^2)}=\delta_0$, whose proof is here. – TaD Nov 27 '24 at 09:33
  • For the case of $\alpha=0$, it seems that your answer doesn't hold. – TaD Nov 29 '24 at 18:58
  • @TaD In checking my answer, I guess it is valid for $\displaystyle \xi \not= 0$. Otherwise, the result will be $\displaystyle {\rm P.V.}\left(1 \over {\rm i}\xi\right) + \pi,\delta\left(\xi\right)$. We should keep it until the whole expression appears inside an integration. – Felix Marin Nov 29 '24 at 19:10
  • OK, but I still have some question (maybe they're basic). For the first equation, why it hold? We don't have $(i\xi+0^+)=(i\xi)(1-i\text{sgn}(\xi)0^+)$. Maybe I misunderstood some definition? – TaD Nov 29 '24 at 19:40