# Questions on Mellin Transform of $x^j$ and Interpretation of Distributions with Complex Arguments

The Mellin transform/inverse transform pair are defined as follows:

(1) $\quad F(s)=\mathcal{M}_x[f(x)](s)=\int\limits_0^\infty f(x)\,x^{s-1}\,dx$

(2) $\quad f(x)=\mathcal{M}_s^{-1}[F(s)](x)=\frac{1}{2\,\pi\,i}\int\limits_{c-i\infty}^{c+i\infty}F(s)\,x^{-s}\,ds$

I've been struggling to understand Mathmatica evaluations such as the following. Formula (6) below is the Mellin transform of the contribution of zeta zero $k$ (i.e. $\rho_k$) in von Mangoldt's explicit formula for the second Chebyshev function which was my initial interest, but formulas (3) and (4) below are simpler examples, and formula (5) below is a more general example.

(3) $\quad\mathcal{M}_x[x](s)=\delta (s+1)$

(4) $\quad \mathcal{M}_x\left[x^{1+i}\right](s)=\delta (s+(1+i))$

(5) $\quad\mathcal{M}_x\left[x^j\right](s)=\delta(s+j)$

(6) $\quad\mathcal{M}_x\left[-\frac{x^{\rho_k}}{\rho_k}\right](s)=-\frac{\delta\left(s+\rho_k\right)}{\rho_k}$

The results illustrated above seem inconsistent with the definition of the inverse Mellin transform illustrated in (2) above. The Dirac delta function makes sense to me in the context of integration along the real axis, whereas the direction of integration in (2) above is orthogonal to the real axis. Consequently, inverse Mellin transforms such as the following don't seem to make sense to me.

(7) $\quad \mathcal{M}_s^{-1}[\delta(s+j)](x)=\frac{1}{2\,\pi\,i}\int\limits_{c-i\infty}^{c+i\infty}\delta(s+j)\,x^{-s}\,ds=x^j$

Question 1: Is the evaluation of the Mellin transform of $x^j$ in (5) above correct?

Question 2: Assuming the answer to Question 1 above is no, what is the correct Mellin transform of $x^j$?

Question 3: Assuming the answer to Question 1 above is yes, can anyone explain how the result of the Mellin transform of $x^j$ in (5) above makes sense in the context of the inverse Mellin transform illustrated in (7) above?

Assuming the answer to Question 1 above is yes and the answer to Question 3 above is no, it seems to me there's something wrong with the theory of Mellin transforms and/or Distributions which leads to the following two questions.

Question 4a: Can an alternate version of the Mellin transform/inverse transform pair be defined where the transform/inverse transform pair associated with $x^j$ makes more sense?

Question 4b: Can the distribution framework be extended to support a Dirac delta function evaluated along the imaginary axis in order to make more sense of the transform/inverse transform pair associated with $x^j$?

Question 4b above is explored in the answer I posted below.

I've also been trying to understand the evaluation of an integral associated with a Dirac delta function with a complex argument such as $\delta(s+(1+i))$ which was the result of evaluation (3) above. It seems to me integral (8) below should evaluate to 1, but I can't seem to get Mathematica to evaluate this integral. A simple substitution of variable $s=t-i$ in (8) below leads to (9) below which is obviously correct.

(8) $\quad\int\limits_{-\infty-i}^{\infty-i}\delta(s+(1+i))\,ds$

(9) $\quad\int\limits_{-\infty }^{\infty }\delta(t+1)\,dt=1$

Question (5): Shouldn't integral (8) above evaluate to 1? If not, why not and what is the proper interpretation of a Dirac delta function with a complex argument?

• Before talking of analytic functionals (ie. things like $\delta(x-i)$) can you define the distributions $\delta(x)$ and $T_A(x) = \int_{-A}^A e^{ikx}dk$ rigorously ? Does $\lim_{A \to \infty} T_A$ converge ? – reuns Nov 28 '17 at 4:58
• @reuns An example of more interest to me is $U_\epsilon(z)=2\left(\frac{2\,\pi\,i\,z+e}{(2\,\pi)^2+(2\,\pi\,i\,z+e)^2}-\frac{2\,\pi\,i\,z-e}{(2\,\pi)^2+(2\,\pi\,i\,z-e)^2}\right)$. Does $U_\epsilon(z)$ converge to $\delta(x+1)+\delta(x-1)$ as $\epsilon\to 0^+$? – Steven Clark Nov 28 '17 at 18:20
• Answer my 1st comment. If in what you wrote you meant $z \in \mathbb{C}$, then it is a nonsense. – reuns Nov 28 '17 at 18:48
• @reuns I thought it fair to answer your question with a question since you answered my original post with a question. I meant $x\in \mathbb{R}$. If you meant $x\in \mathbb{C}$, then what you wrote is nonsense as well. I'd prefer you provide clear answers to the specific questions in my original posts versus providing related information and asking me related questions. I find this somewhat frustrating as it's a distraction from my specific interest at the time. Nevertheless, I generally find the information you provide useful in increasing my broader understanding over the longer term. – Steven Clark Nov 28 '17 at 20:44
• Do I really need to repeat what my 1st comment implied ? Read a course on the Fourier series and Fourier transform, then on the Fourier transform of distributions, only after you will be able to look at analytic functionals. – reuns Nov 28 '17 at 20:53

$\mathcal{M}_s[1](x)=2\,\pi\,\delta(i\,s)$, is a nonsense.

$\lim_{a \to \infty} \int_{-a}^a e^{-ik x }dx$ diverges for every $x$.

What mathematica says is incorrect because it doesn't mention "convergence in the sense of distributions".

With $\tau(x) =\int_{-\infty}^x \frac{2\sin(y)}{y}dy,\tau(-\infty) = 0,\tau(+\infty) = 2\pi$, we have a convergence in the sense of distributions

$$\int_{-\infty}^\infty e^{-ik x }dx = \lim_{a \to \infty} \int_{-a}^a e^{-ik x }dk = \lim_{a \to \infty} a\frac{2 \sin(ax)}{ax} \\ =\lim_{a \to \infty}\frac{d}{dx} \tau(ax) =\frac{d}{dx} 2\pi \ 1_{x > 0} = 2\pi \delta(x)$$ In the sense of distributions ? It means for any $\varphi \in C^\infty_c$ (more generally for any $\varphi \in L^1, \varphi' \in L^1$) $$\lim_{a \to \infty}\int_{-\infty}^\infty \varphi(x) (\int_{-a}^a e^{-ik x }dk) dx = 2\pi \varphi(0)$$ This is equivalent to the Fourier inversion theorem because it means $$\int_{-\infty}^\infty \hat{\varphi}(k)e^{iky}dk =\lim_{a \to \infty}\int_{-a}^a\hat{\varphi}(k)e^{iky}dk= \lim_{a \to \infty}\int_{-a}^a e^{iky}\int_{-\infty}^\infty \varphi(x)e^{-ikx}dx dk\\= \lim_{a \to \infty}\int_{-a}^a e^{iky}\int_{-\infty}^\infty \varphi(x+y)e^{-ik(x+y)}dx dk = \lim_{a \to \infty}\int_{-\infty}^\infty \varphi(x+y)\int_{-a}^a e^{-ik x} dkdx=2\pi \varphi(y)$$

What about $\lim_{a \to \infty} \int_{-a}^a e^{ik (z-x) }dx$ ? It converges to an analytic functional. For any $\phi : \mathbb{C} \to \mathbb{C}$ complex analytic such that for every $k,y$, $\int_{-\infty}^\infty |\phi(x+iy) x^k| dx < \infty$ then $$\lim_{a \to \infty}\int_{-\infty}^\infty \phi(x) \int_{-a}^a e^{ik(z- x) }dx = 2\pi \varphi(z)$$ For those $\phi$ complex analytic and Schwartz on every horizontal line (and only for those) it makes sense to talk for $z \in \mathbb{C}$ of the analytic functional $\delta_z$ such that $\langle \delta_z, \phi \rangle = \phi(z)$.

• Mathematica has knowledge of many relationships based on convergence in the distributional sense, but it only reports the result, not the convergence type. Furthermore, the proof of the inverse Mellin transform $\frac{1}{2\,\pi\,i}\int\limits_{-i\,\infty}^{i\,\infty}(2\,\pi\,\delta(i\,s))\,x^{-s}\,ds=1$ doesn't require any Mathematica evaluations. It only requires one to understand $\int\limits_{-i\,\infty}^{i\,\infty}f(i\,s)\,ds=i\int\limits_{-\infty}^{\infty}f(t)\,dt$ and $\int\limits_{-\infty}^{\infty}\delta(t)\,g(t)\,dt=g(0)$ when $g(t)$ is continuous at $t=0$. It's really not complicated. – Steven Clark Dec 26 '17 at 15:01
• Do you dispute $\int\limits_{-i\,\infty}^{i\,\infty}\frac{\sin(A\,i\,s)}{\pi\,i\,s}\,ds=i$? – Steven Clark Dec 26 '17 at 15:14
• I think perhaps you're making the same mistake I originally did. I originally thought this result was inconsistent with existing theory. In my case I thought a paradigm shift was required along with some new theory, but in your case you believe there is only the existing theory. I eventually realized with the simple translation $s=-i\ t$ the integral of $\delta(i\, s)$ along the imaginary axis is really an integral of $\delta(t)$ along the real axis which is well understood and consistent with existing theory such as that you provided in your answer. – Steven Clark Dec 26 '17 at 22:52
• The reason I asked you about the integral $\int\limits_{-i\,\infty}^{i\,\infty}\frac{\sin(A\,i\,s)}{\pi\,i\,s}\,ds=i$ is because the integral $i\int\limits_{-\infty}^{\infty}\frac{\sin(A\,t)}{\pi\,t}\,dt=i$ produces the same result, and I thought this insight might help you get past the mental block you seem to be having with respect to the change of variables. – Steven Clark Dec 27 '17 at 1:13
• Don't you understand it's not $\delta(i\,s)$ when evaluated along the imaginary axis (i.e. along the line $s=-i\,t$)? My advice is to forget worrying about evaluating the integral of $\delta(i\,s)$ along the imaginary axis. Just make the variable substitution $s=-i\,t$, evaluate the integral of $\delta(t)$ along the real axis where it belongs, and everything works itself out. I think we've beat this horse to death, and it's not worth spending any further time debating this topic. – Steven Clark Dec 27 '17 at 18:39

Mathematica evaluates the Mellin transforms for 1 and $x^j$ as follows.

(a) $\quad\mathcal{M}_x[1](s)=\delta(s)$

(b) $\quad\mathcal{M}_x\left[x^j\right](s)=\delta(s+j)$

The following Mathematica inverse Mellin transform evaluations are consistent with (a) and (b) above.

(c) $\quad\mathcal{M}_s^{-1}[\delta(s)](x)=1$

(d) $\quad\mathcal{M}_s^{-1}[\delta(s+j)](x)=x^j$

The inverse Mellin inverse transform is defined as follows.

(e) $\quad\mathcal{M}_s^{-1}[F(s)](x)=\frac{1}{2\,\pi\,i}\int\limits_{c-i\,\infty}^{c+i\,\infty}F(s)\,x^{-s}\,ds$

The inverse Mellin transform evaluations (c) and (d) above which are illustrated in integral form below don't make sense because the $\delta(s)$ function is intended to be integrated along the real axis and the inverse Mellin transform is integrated along a line perpendicular to the real axis.

(f) $\quad\mathcal{M}_s^{-1}[\delta (s)](x)=\frac{1}{2\,\pi\,i}\int\limits_{c-i\,\infty}^{c+i\,\infty}\delta(s)\,x^{-s}\,ds=1$

(g) $\quad\mathcal{M}_s^{-1}[\delta(j+s)](x)=\frac{1}{2\,\pi\,i}\int\limits_{c-i\,\infty}^{c+i\,\infty}\delta(s+j)\,x^{-s}\,ds=x^j$

The following Mathematica evaluation illustrates the Mellin transform of 1 should be $2\,\pi\,\delta(i\,s)$ instead of $\delta(s)$.

(h) $\quad\frac{1}{2\,\pi\,i}\int\limits_{-i\,\infty}^{i\,\infty}(2\,\pi\,\delta(i\,s))\,x^{-s}\,ds=1$

Integral (h) above is along the imaginary axis, but substituting $s=-i\,t$ into integral (h) above results in the following integral along the real axis.

(i) $\quad\frac{1}{2\,\pi\,i}\int\limits_{-\infty }^{\infty}(2\,\pi\,\delta\,(t))\,x^{i\,t}\,i\,dt=x^0=1$

The result above implies the Mellin transform of $x^j$ should be $2\,\pi\,\delta(i\,(s+j))$ instead of $\delta(s+j)$, and the corresponding inverse Mellin transform can be evaluated as follows.

(j) $\quad\frac{1}{2\,\pi\,i}\int\limits_{-\Re(j)-i\,\infty}^{-\Re(j)+i\,\infty}(2\,\pi\,\delta(i\,(s+j)))\,x^{-s}\,ds=x^j$

Mathematica isn't capable of evaluating integral (j) above, but Mathematica is capable of evaluating the following integral along the real axis which is derived from (j) above by substituting $s=-i\,t-\Re(j)$.

(k) $\quad\frac{1}{2\,\pi\,i}\int\limits_{-\infty}^{\infty}(2\,\pi\,\delta(t-\Im(j)))\,x^{i\,t+\Re(j)}\,i\,dt=x^j$

For further illustration of the correctness of the Mellin transform $\mathcal{M}_s[1](x)=2\,\pi\,\delta(i\,s)$, consider the limit representation of $\delta(x)$ illustrated in (l) below and the two associated Mathematica evaluations illustrated in (m) and (n) below both of which corroborate $\mathcal{M}_s[1](x)=2\,\pi\,\delta(i\,s)$. The result illustrated in (m) below was obtained using the Mathematica $InverseMellinTransform$ function. Note for $A=\infty$ the result illustrated in (m) below evaluates to $1$ for $x>0$. The result illustrated in (n) below was obtained using the Mathematica $Integrate$ function using the assumption $x<A$.

(l) $\quad T_A(x)=\frac{1}{2\,\pi}\int\limits_{-A}^A\,e^{i\,t\,x}\, dt=\frac{\sin(A\,x)}{\pi\,x},\quad A\to\infty$

(m) $\quad\mathcal{M}_s^{-1}[2\,\pi\,T_A(i\,s)](x)=\theta(A-\log(x))-\theta(-A-\log (x)),\quad 0<x<1$

(n)$\quad\frac{1}{2\,\pi\,i}\int_{-i\,\infty}^{i\,\infty}2\,\pi\,T_A(i\,s)\,x^{-s}\,ds=1\,,\quad 1\leq x<A$

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I originally posted the following exploration into defining a $\delta(s)$ function along the imaginary axis. This approach was misguided but provided some insight which eventually helped me to realize the correct answer I posted above and also provided the context for some of the comments below, so I decided to retain the original exploration I posted below rather than delete it.

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I believe I've made some progress in making sense of a delta function integrated along the imaginary axis, but this seems to require an extension of the distributional framework (at least as I understand it). In the remainder of this answer, I use $\delta^*(s)$ to refer to a delta function intended to be integrated along the imaginary axis to distinguish it from the traditional delta function $\delta(s)$ intended to be integrated along the real axis. Likewise I use $\theta^*(s)$ to refer to a Heaviside step function intended to be evaluated along the imaginary axis to distinguish it from the traditional Heaviside step function $\theta(s)$ intended to be evaluated along the real axis.

The Mellin transform of $1$ and the associated inverse Mellin transform are as follows. Mathematica evaluates the Mellin transform of 1 as $\delta(s)$, but I believe the result should be $\delta^*(s)$ versus the traditional definition of $\delta(s)$.

(1) $\quad\mathcal{M}_x[1](s)=\int\limits_0^\infty 1\,x^{s-1}\,dx=\delta^*(s)$

(2) $\quad \mathcal{M}_s^{-1}[\delta^*(s)](x)=\frac{1}{2\,\pi\,i}\int\limits_{-i\infty}^{i\infty}\delta^*(s)\,x^{-s}\,ds=1$

The Mellin transform pair illustrated in (1) and (2) above seems to imply the following.

(3) $\quad\int\limits_{-i\infty}^{i\infty}\delta^*(s)\,ds=2\,\pi\,i$

Consider the following approximation to $\delta^*(s)$ which is an approximation to the Mellin transform of 1. Mathematica indicates the integral below converges for $Re(s)<0$, but evaluation at $Re(s)=0$ along the imaginary axis produces some encouraging results which are illustrated below.

(4) $\quad\delta^*(s)\approx T_a(s)=\int\limits_a^\infty 1\,x^{s-1}\, dx=-\frac{a^s}{s},\quad a\to 0^+$

The following three plots illustrate the real part, imaginary part, and absolute value of $T_a(s)$ evaluated along the imaginary axis for $a=0.1$ (green), $a=0.01$ (orange), and $a=0.001$ (blue). In the third plot below for the absolute value of $T_a(s)$, all three evaluation limits for $a$ result in the same evaluation and hence the green, orange, and blue curves overlap each other.

Figure (1): $\Re(T_a(s))$ Evaluated Along the Imaginary Axis

Figure (2): $\Im(T_a(s))$ Evaluated Along the Imaginary Axis

Figure (3): $\left|T_a(s)\right|$ Evaluated Along the Imaginary Axis

The following plot illustrates $T_a(s)$ evaluated along the real axis for $a=0.1$ (green), $a=0.01$ (orange), and $a=0.001$ (blue). Assuming $a\to 0^+$, $T_a(s)$ diverges to $+\infty$ as $s\to-\infty$, and $T_a(s)$ converges to $0$ as $s\to+\infty$.

Figure (4): $T_a(s)$ Evaluated Along the Real Axis

Now consider the following integral which is an approximation to and consistent with integral (3) above. Note that $i\,(\pi-2\,\text{Si}\,(-\infty))=2\,\pi\,i$.

(5) $\quad\int\limits_{-i\,N}^{i\,N}T_a(s)\,ds=i\,(\pi-2\,\text{Si}\,(N\,\log(a)))\approx 2\,\pi\,i\,,\quad a\to0^+\land N\to\infty^-$

Now consider the following integral which is an approximation to and consistent with the inverse Mellin transform of $\delta^*(s)$ in (2) above. Mathematica indicates the integral below converges for $\Re(\log(x))>0$, but it actually seems to converge for $x>0$.

(6) $\quad\frac{1}{2\,\pi\,i}\int\limits_{-i\,N}^{i\,N}T_a(s)\,x^{-s}\,ds=\frac{(\text{Ei}(-i\,N\,(\log(a)-\log (x)))-\text{Ei}(i\,N\,(\log(a)-\log(x))))}{2\,\pi\,i}\approx 1\,,\qquad x>0\land a\to 0^+\land N\to\infty^-$

The following plot illustrates the evaluation of formula (6) above evaluated at $a=0.001$ and $N=1,000$. The blue and orange curves represent the real and imaginary parts of formula (6) respectively. I'll note formula (6) only converges to 1 for $x>0$.

Figure (5): Formula (6) illustrated at $a=0.001$ and $N=1,000$

Next consider the following approximation to $\theta^*(s)$.

(7) $\quad\theta^*(s)\approx\int\limits_{-i\,\infty}^s T_a(z)\,dz=-\text{Ei}(s\,\log(a))+i\,\pi\,,\quad \Re(s)=0\land a\to 0^+$

The following three plots illustrate the real part, imaginary part, and absolute value of formula (7) above evaluated along the imaginary axis for $a=0.1$ (green), $a=0.01$ (orange), and $a=0.001$ (blue).

Figure (6): Real part of formula (7) Evaluated Along the Imaginary Axis

Figure (7): Imaginary part of formula (7) Evaluated Along the Imaginary Axis

Figure (8): Absolute Value of formula (7) Evaluated Along the Imaginary Axis

The following plot illustrates an expanded view of the real part of formula (7) defined above.

Figure (9): Expanded View of Real part of formula (7) Evaluated Along the Imaginary Axis

Finally consider the following function which represents $T_a(s)$ evaluated at $s=i\,t$ and $a=e^{-A}$. The evaluation of $T_A(t)$ along the real axis as $A\to\infty^-$ is analogous to the evaluation of $T_a(s)$ along the imaginary axis as $a\to 0^+$, but I can't seem to derive the integrals for $T_A(t)$ which are analogous to the integrals associated with $T_a(s)$ defined and illustrated above.

(8) $\quad T_A(t)=\frac{\sin(A\,t)}{t}+\frac{i\,\cos(A\,t)}{t},\quad t\in \mathbb{R}\land A\to\infty-$

12/5/2017 Update: I wasn't happy with the following aspects of the limit representation $Ta(s)=-\frac{a^s}{s}$ for $\delta^*(s)$ defined in formula (4) above, so this update explores an alternate limit representation for $\delta^*(s)$ which eliminates these problems and therefore I believe is more correct.

1. Inability to evaluate formulas (5) and (6) above at infinite integration limits.
2. Non-zero value of $\Im(T_a(s))$ illustrated in figure (2) above.
3. Non-zero value of real part of formula (7) illustrated in figures (6) and (9) above.

The alternate limit representation for $\delta^*(s)$ is defined as follows.

(9) $\quad\delta^*(s)=T_1(s)=\frac{2\,\sinh(A\,s)}{s},\quad \Re(s)=0\land A\to\infty$

The following two plots illustrate the real part and imaginary part of $T_1(s)$ evaluated along the imaginary axis for $A=2$ (green), $A=4$ (orange), and $A=8$ (blue).

Figure (10): $\Re(T_1(s))$ Evaluated Along the Imaginary Axis

Figure (11): $\Im(T_1(s))$ Evaluated Along the Imaginary Axis

Now consider the following integral which is consistent with integral (3) above.

(10) $\quad\int\limits_{-i\,\infty}^{i\,\infty}T_1(s)\,ds=2\,\pi\,i$

Now consider the following integral which evaluates to $1$ for $x>0$ and $A>\log(x)$ and therefore is consistent with the inverse Mellin transform of $\delta^*(s)$ in (2) above.

(11) $\quad\frac{1}{2\,\pi\,i}\int\limits_{-i\,\infty}^{i\,\infty}T_1(s)\,x^{-s}\,ds=\frac{1}{2}\left(1+\frac{1}{\text{sgn}(A-\log(x))}\right)=1\,,\qquad x>0\land A\to\infty$

Next consider the following limit representation of $\theta^*(s)$.

(12) $\quad\theta^*(s)=\int\limits_{-i\,\infty}^s T_1(z)\,dz=2\,\text{Shi}(A\,s)+i\,\pi\,,\quad \Re(s)=0\land A\to\infty$

The following two plots illustrate the real part and imaginary part of formula (12) above evaluated along the imaginary axis for $A=2$ (green), $A=4$ (orange), and $A=8$ (blue).

Figure (12): Real part of formula (12) Evaluated Along the Imaginary Axis

Figure (13): Imaginary part of formula (12) Evaluated Along the Imaginary Axis

• Not quite. Plotting $\frac{\sin(A t)}{t}$ is of course not a proof of the convergence in the sense of distributions. Why don't you answer to my comments above ? – reuns Dec 4 '17 at 10:31
• @reuns You still don't seem to get it. The function $\frac{\sin(A\,t)}{t}$ is not the right test function. If you feel you have a better answer than the one I posted, then you're certainly free to post your own answer. – Steven Clark Dec 4 '17 at 14:47
• ? I know quite well this subject... What do you mean exactly with your last comment ? You start with "I use $\delta^*$ meant to be integrated along the imaginary axis" which doesn't make any sense. In the inverse Laplace/Mellin transform we integrate $e^{-st} F(s)$ on a vertical line, because $F(\sigma+i \omega)$ is the Fourier transform of $f(t) e^{-\sigma t}$. – reuns Dec 4 '17 at 14:53
• I did write an answer but I didn't post it because I'm quite sure you won't study it. The first step is to prove $\lim_{A \to \infty} \langle T_A, \varphi \rangle = \varphi(0)$ for any $\varphi$ Schwartz or $C^\infty_c$ and $T_A(x) = \frac{1}{2\pi} \int_{-A}^A e^{i \omega x} d\omega = \frac{\sin(A x)}{\pi x}$, which is really the Fourier inversion theorem. The next step is to look at $T_{A,z}(x) = \frac{1}{2\pi} \int_{-A}^A e^{i \omega (x-z)} d\omega$ for $z \in \mathbb{C}$ and to find for which analytic functions we have $\lim_{A \to \infty} \langle T_A, \varphi \rangle = \varphi(z)$. – reuns Dec 4 '17 at 14:56
• I meant $\lim_{A \to \infty} \langle T_{A,z},\varphi \rangle = \varphi(z)$. Do you see why this is what we need to prove $\frac{1}{2\pi}\int_0^\infty y^{ix-1}dy=\frac{1}{2\pi}\int_{-\infty}^\infty e^{i \omega x}d \omega = \lim_{A \to \infty} T_A(x) = \delta(x)$ in the sense of distributions ? – reuns Dec 4 '17 at 15:14