For instance, Mellin transform of function $f(x)=x$

$$\int_0^\infty f(x) x^{s-1} dx$$

returns the result $\delta(s)$ which is completely strange to me. Why only at $s=0$ the result is infinite? Why for instance for $s=2$ the result is $0$?

Moreover, for $f(x)=a^x$ the result is $\Gamma (s) (-\log (a))^{-s}$ which is again meaningless with $a\ge1$.

Why they just cannot use Delta Function or other infinite distributions?

What method do they use to compose these tables? Can I utilize the same method to obtain more sensible results?

Or maybe there is somewhere a more complete table?

  • 2
    $\begingroup$ $\delta(x) =\frac{1}{2\pi} \int_{-\infty}^\infty e^{i \omega x} d \omega = \lim_{N \to \infty} \frac{1}{2\pi} \int_{-N}^N e^{i \omega x} d \omega$ is true only in the sense of distributions, ie. for any Schwartz function $\varphi$ we have $$\lim_{N \to \infty} \int_{-\infty}^\infty \varphi(x) (\frac{1}{2\pi} \int_{-N}^N e^{i \omega x} d \omega) dx = \varphi(0)$$ (this is equivalent to the Fourier inversion theorem). But $\int_{-\infty}^\infty e^{i \omega x} d \omega$ diverges and isn't $=0$ for $x \ne 0$. $\endgroup$
    – reuns
    Jul 9 '17 at 1:51
  • $\begingroup$ @reuns I am not sure what you intended to say, but if to consider the integral from the last line in your comment, I definitely would like it to appear as DiracDelta, for any $x$. In general, I want to see the transforms in their "full form", but it seems unavailable anywhere. $\endgroup$
    – Anixx
    Jul 9 '17 at 7:34
  • $\begingroup$ If you have access to Mathematica (the full package, not Wolfram Alpha), you could add the option GenerateConditions -> True to find the range of validity of the Mellin transform. For example, MellinTransform[a^x, x, s, GenerateConditions -> True] returns ConditionalExpression[Gamma[s] (-Log[a])^-s, Re[Log[a]] < 0 && Re[s] > 0] $\endgroup$ Jul 9 '17 at 11:37

This question has not been well received, but I am intrigued by the delta function Mellin transform and would like to respond. I found this 2004 paper by Norbert Südland and Gerd Baumann instructive. The Mellin transform pair is $${\cal M}[f(x),s]=\int_0^\infty f(x)x^{s-1}\,dx$$ $${\cal M}^{-1}[g(s),x]=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}g(s)x^{-s}\,ds$$ If we take $g(s)=2\pi\,\delta(s)$ and set $a=0$ we find $${\cal M}^{-1}[g(s),x]=\frac{1}{i}\int_{-i\infty}^{i\infty}\delta(s)x^{-s}\,ds=1$$ This motivates the assignment ${\cal M}[1,s]=2\pi\,\delta(s)$ and, more generally, $${\cal M}[x^p,s]=2\pi\,\delta(s+p).$$ What is curious, is that Mathematica omits the $2\pi$ factor (for the same definition of Mellin transform). I think this is a mistake, and have inquired on the Mathematica SE site.
[Update: the missing factor of $2\pi$ has been fixed in Mathematica version 11.3.]

  • $\begingroup$ Sorry, but how this comes to be 1? $$\frac{1}{i}\int_{-i\infty}^{i\infty}\delta(s)x^{-s}\,ds$$ If you are assuming delta function to be zero on imaginable line, you are using grossly wrong definition of delta function (following integral definition it is infinite on the whole imaginable line). $\endgroup$
    – Anixx
    Jul 10 '17 at 7:35
  • $\begingroup$ This follows from equation 3 in the 2004 paper I linked to in my answer. $\endgroup$ Jul 10 '17 at 7:51
  • $\begingroup$ This is completely wrong definition, contradicting that via Fourier transform. $\endgroup$
    – Anixx
    Jul 10 '17 at 8:06
  • 1
    $\begingroup$ @CarloBeenakker The Mellin transform of $x^j$ is $2\pi\delta(i (s+j))$ which I illustrated at the following link: mathoverflow.net/q/294849. I submitted a problem report to Wolfram and Mathematica version is now reporting the correct result. $\endgroup$ Mar 19 '18 at 3:57
  • The integral $\int_{-\infty}^\infty e^{i \omega x} d \omega$ diverges for every $x$.

  • What is true is that for any Schwartz function $\varphi \in S(\mathbb{R})$ : $$\lim_{N \to \infty} \int_{-\infty}^\infty \varphi(x) (\frac{1}{2\pi} \int_{-N}^N e^{i \omega x} d \omega) dx = \varphi(0)$$

which is the definition of $\delta(x) = \lim_{N \to \infty} \frac{1}{2\pi} \int_{-N}^N e^{i \omega x} d \omega=\frac{1}{2\pi} \int_{-\infty}^\infty e^{i \omega x} d \omega $ in the sense of (tempered) distributions.

  • Then, given an interval $(a, b)$, we restrict to those Schwartz functions $\varphi$ such that $\varphi(x) e^{\sigma x}$ is Schwartz for every $\sigma \in (a,b)$. Let me denote this space of functions by $\{\varphi,\forall \sigma \in (a,b), \varphi(x) e^{\sigma x} \in S(\mathbb{R})\}$.

    In that case we can apply the Fourier inversion theorem to $\varphi(x)e^{\sigma x}$ to obtain the inverse bilateral Laplace transform formula

$$\varphi(0) = \lim_{N \to \infty} \int_{-\infty}^\infty \varphi(x)e^{\sigma x}(\frac{1}{2\pi} \int_{-N}^N e^{i \omega x} d \omega) dx = \lim_{N \to \infty} \int_{-\infty}^\infty \varphi(x)(\frac{1}{2i\pi} \int_{\sigma-iN}^{\sigma +iN} e^{s x} d s) dx$$

Which is the definition of $\displaystyle\delta(x) = \lim_{N \to \infty} \frac{1}{2i\pi} \int_{\sigma-iN}^{\sigma +iN} e^{s x} d s=\frac{1}{2i\pi} \int_{\sigma-i\infty}^{\sigma +i\infty} e^{s x} d s$ for $\sigma \in (-a,b)$ in the sense of distributions on $\{\varphi,\forall \sigma \in (a,b), \varphi(x) e^{\sigma x} \in S(\mathbb{R})\}$.

  • Finally the Mellin transform $\int_0^\infty x^{s-1} \varphi(x)dx$ is just the bilateral Laplace transform with a change of variable $x = e^{-y}$, so that (for $\sigma \in (a,b)$)

$\displaystyle\delta(x-1) =\frac{1}{2i\pi} \int_{\sigma-i\infty}^{\sigma +i\infty} x^{-s} d s$ in the sense of distributions on $\{\varphi,\forall \sigma \in (a,b), \varphi(e^x) e^{(\sigma-1) x} \in S(\mathbb{R})\}$.

From this, following the same lines (restricting to the correct function space) you can make sense to "the analytic functional" $\delta(s)$ with $s \in \mathbb{C}$ and apply the Laplace transform to it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.