I have came across a case where the standard formula for the Mellin transform of a convolution of two functions is not the product of the Mellin transforms and I don't understand exactly where the problem is coming from.

It is a bit of a particular case and I'm using a bit of a different convention for the Mellin transform, in particular because those computations are related to problems in physics — but I believe there might be a subtle mathematical problem here and that's why I post here and not on Physics SE.

Let me introduce all my conventions and then state precisely the problem. The Mellin transform of a function $f$ defined on $(0;1)$ and integrable on $[0;1]$ is defined as follows (we're only interested in integer values $n\in\mathbb{N}$ of the transform):

$$ M[f(x)](n):=\int_0^1\mathrm{d}x\,x^n f(x) $$

We can relate it quite easily to the standard definition of the Mellin transform that we'll denote with a curly $\mathcal{M}$:

$$ \mathcal{M}[f(x)](s):=\int_0^\infty \mathrm{d}x\ x^{s-1} f(x) \rightarrow M[f(x)](n)=\mathcal{M}[\Theta(1-x)f(x)](n+1) $$

Where $\Theta$ is the Heaviside function. That's why even though the definition looks a bit different, all the same properties should apply.

In the case the function $f$ is non integrable around $x=1$, i. we have $f(x)\underset{x\to 1}{\sim}\alpha\ (1-x)^\beta$ with $\alpha\in\mathbb{R}$, $1\leq\beta<2$, we define the Mellin transform in what we call the +-prescription, i.

$$ M[f_+(x)](n):=\int_0^1\mathrm{d}x\,(x^n-1) f(x) $$

In whole generality the definition of a +-prescription distribution $\tilde{\mathcal{T}^+_f}$ associated to a function $f$ acting on some test function $\phi\in\mathcal{D}(\mathbb{R}^+)$ would be

$$ <\tilde{\mathcal{T}^+_f},\phi>:=\int_0^1\mathrm{d}x\ \left[\phi (x)-\phi(1)\right]f(x) \left(= \int_0^{+\infty}\mathrm{d}x\ \left[\phi (x)-\phi(1)\right]\Theta(1-x)f(x)\right) $$

And since we can see the Mellin transform in the distribution sense for a given distribution $\mathcal{T}_f$ associated to a function $f$ ($\mathcal{T}_f$ is defined over $\mathcal{D}^\prime(\mathbb{R}^+)$) as $\mathcal{M}[f(x)](n)=<\mathcal{T}_f,t^{n-1}>$ we can adapt it to our case by defining

$$ M[f(x)](n)=<\tilde{\mathcal{T}_f},t^n>\text{ with }\tilde{\mathcal{T}_f}=\mathcal{T}_{\Theta(1-x)f(x)} $$

so that it fits both the standard and +-prescription definition. From now on we write $f_+$ as a shorthand for $\tilde{\mathcal{T}^+_f}$. Now we define the Mellin convolve of two functions $f$ and $g$ (and in a similar way above we can extend this to the distribution sense):

$$ f(x)\star g(x):=\int_0^1\int_0^1\mathrm{d}y\,\mathrm{d}z\,\delta(x-yz)f(y)g(z) $$

By integrating out one variable, we end up with the following integrals (see for example https://arxiv.org/abs/1407.1822 p.4), depending on whether $f$ is in the +-prescription or no:

$$ f(x)\star g(x)=\int_x^1\frac{\mathrm{d}y}{y}f\left(\frac{x}{y}\right)g(y) $$

\begin{equation}\label{PPConv} f_+(x)\star g(x)=\int_x^1\mathrm{d}y\ f\left(y\right)\left[\frac{1}{y}g\left(\frac{x}{y}\right)-g(x)\right]-g(x)\int_0^x\mathrm{d}y\ f(y) =\int_x^1\mathrm{d}y \ f\left(\frac{x}{y}\right)\left[\frac{g(y)}{y}-\frac{xg(x)}{y^2}\right]-g(x)\int_0^x\mathrm{d}y\ f(y) \end{equation}

Finally, the Mellin exchange formula tells you that:

$$ M[f(x)](n)\cdot M[g(x)](n)= M[f(x)\star g(x)](n) $$

Now let's examine the problematic example where the formula above doesn't hold. Let's take the following two functions:

$$ f(x)=\frac{x}{(x-1)\sqrt{x(1-x)}},\quad g(x)=\frac{1}{\sqrt{x(1-x)}} $$

We immediately see that since $f(x)\underset{x\to 1}{\sim}-(1-x)^{-\frac{3}{2}}$, we have to compute the Mellin transform of $f$ in the +-prescription. Using a geometric series expansion and then Mathematica, we can analytically compute Mellin transforms of $f$ and $g$:

$$ M[f_+(x)](n)=\int_0^1\mathrm{d}x\ (x^n-1)\frac{x}{(x-1)\sqrt{x(1-x)}}=\sum_{i=0}^{n-1}\int_0^1\mathrm{d}x\ x^i\sqrt{\frac{x}{1-x}} = \frac{2 \sqrt{\pi}\ \Gamma \left(n+\frac{3}{2}\right)}{\Gamma (n+1)}-\pi $$ $$ M[g(x)](n)=\int_0^1\mathrm{d}x\ x^n \frac{1}{\sqrt{x(1-x)}}=\frac{\sqrt{\pi}\ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)} $$

Therefore we have:

$$ \text{LHS}(n):=M[f_+(x)](n)\cdot M[g(x)](n)= \frac{\pi\ \Gamma \left(n+\frac{1}{2}\right) \left(2 \Gamma \left(n+\frac{3}{2}\right)-\sqrt{\pi}\ \Gamma (n+1)\right)}{\Gamma (n+1)^2} $$

Now if we compute the $f_+(x)\star g(x)$ convolution, Mathematica gives us the following result:

$$ f_+(x)\star g(x)=\int_x^1\mathrm{d}y\ f\left(y\right)\left[\frac{1}{y}g\left(\frac{x}{y}\right)-g(x)\right]-g(x)\int_0^x\mathrm{d}y\ f(y) = \frac{2 \left(K\left(\frac{x-1}{x}\right)+\sqrt{\frac{1}{x}-1} \arccos \left(\sqrt{x}\right)\left(\frac{x-1}{x}\right)+1\right)}{x-1}-\frac{2 \left(\arcsin \left(\sqrt{x}\right)-\frac{1}{\sqrt{\frac{1}{x}-1}}\right)}{\sqrt{(1-x) x}} $$

where $E$ is the complete elliptic integral and $K$ the complete elliptic integral of the first kind. We have in particular that

$$ f_+(x)\star g(x)\underset{x\to 1}{=}-\frac{\pi}{\sqrt{1-x}}+\frac{\pi }{2}+O\left(\sqrt{1-x}\right) $$ $$ f_+(x)\star g(x)\underset{x\to 0}{=}\frac{2-\pi }{\sqrt{x}}+O\left(\sqrt{x}\right) $$

The convolution is therefore integrable at $x=1$ (and of course at $x=0$) and therefore its Mellin convolution is computed as:

$$ \text{RHS}(n):=M[f_+(x)\star g(x)](n)=\int_0^1 \mathrm{d}x\ x^n f_+(x)\star g(x) $$

We cannot compute analytically $\text{RHS}(n)$, but we can do numerical checks against $\text{LHS}(n)$ and we immediately realise that there is a mismatch between $\text{LHS}(n)$ and $\text{RHS}(n)$. We take the first values, i. for $n=0,1,2,3,4,5$, for both functions:

$\text{LHS}(n)$ : $\{0.,2.4674,3.23846,3.66255,3.94266,4.14612\}$

$\text{RHS}_1(n)$ : $\{-6.28319,-3.81578,-3.04472,-2.62064,-2.34052,-2.13707\}$

Problem: We find now that the formula Mellin exchange formula doesn't hold, i.e. we have $M[f_+(x)](n)\cdot M[g(x)](n)\neq M[f_+(x)\star g(x)](n)$ - why is that the case?

Furthemore, if we take the Mellin transform of $f_+(x)\star g(x)$ in the +-prescription, i.

$$ \text{RHS}_+(n)=\int_0^1\mathrm{d}x\ (x^n-1)f_+(x)\star g(x) $$

Then the values fit those of $\text{LHS}(n)$:

$\text{RHS}_+(n)$ : $\{0.,2.4674,3.23846,3.66255,3.94266,4.14612\}$

And omitting the first term $n=0$, the relative error gives us for $n=1,\ldots,5$:

$\left\lvert 1-\frac{\text{RHS}_+(n)}{\text{LHS}(n)}\right\rvert$ : $\{1.40998\times 10^{-14},2.31688\times 10^{-9},6.31408\times 10^{-11},4.45677\times 10^{-12},2.30926\times 10^{-13}\}$

I have no clue why in this particular case I have to actually manually "correct" the Mellin exchange formula by forcing the computation to be in the +-prescription. It's disturbing as the +-prescription convolution formula gets rid as expected of the non-integrable singularity at $x=1$, as we can see looking at its Laurent series around $x=1$, and the standard Mellin transform is properly defined for $n\geq 0$. The fact that there are no non-integrable singularity that tells us we need to use the +-prescription is making this problem non-trivial (I'm working on automatisation of such computations for Mellin inverse purposes, so it's paramount for me to be able to tell what's the correct Mellin transform of the convolution).

General question: How do I know exactly in which situations I should use the +-prescription for Mellin transform computations when dealing with convolutions of functions defined themselves with +-prescription?

Remark: It's really only an empirical/hand-waving statement, but I realised in thus particular example that the sum of the powers of the dominant terms of $f$ and $g$ Laurent series expansion is basically $-2$, given $f(x)\underset{x\to 1}{\sim}-(1-x)^{-\frac{3}{2}}$ and $g(x)\underset{x\to 1}{\sim}(1-x)^{-\frac{1}{2}}$, and therefore I have applied a similar testing process for functions of the form

$$ f_\alpha(x)=\frac{x}{(x-1)\left[x(1-x)\right]^{1-\alpha}},\quad g_\alpha(x)=\frac{1}{\left[x(1-x)\right]^\alpha}\text{ with }0<\alpha<1 $$

And basically we get exactly the same problem. On the contrary, if I take now

$$ \tilde{f}(x)=\frac{x}{(x-1)\sqrt{x}},\quad g(x)=\frac{1}{\sqrt{x(1-x)}} $$

So I just change $f(x)\to \tilde{f}(x)=\sqrt{1-x}f(x)$, the Mellin exchange formula holds, i. I don't need to add any "manual" +-prescription at the end. I have however no idea how to formalise this, or where I could find any material to read about those exact issues on the Mellin exchange formula (usually textbooks/articles concentrate more on the inversion process itself), so I would be very thankful for any help.

  • 1
    $\begingroup$ Forgive my apparently short attention span, but... what's the question? $\endgroup$ Apr 2 at 19:17
  • $\begingroup$ So, you compute the same integral with 2 different numerical schemes and get $10^{-14}$ difference? $\endgroup$
    – user619894
    Apr 2 at 19:40
  • $\begingroup$ Your first formula is not the standard definition of the Mellin transform, so please state the exact relationship that you're trying to validate with respect to the Mellin convolution theorem. For $f(x)$ the Mellin transform integral $\int\limits_0^1 \frac{x}{(x-1) \sqrt{x (1-x)}}\, x^{s-1} \, dx$ doesn't even converge in the usual sense. $\endgroup$ Apr 2 at 20:09
  • $\begingroup$ @JakobStreipel I'm sorry for not stating this more explicitely, that was my fault : my question is that I don't understand why the Mellin transform of the convolution of my two functions $f$ and $g$ doesn't give the same result as the product of the Mellin transforms of those functions. I edited the post and stated the "problem" and "general question" in bold in the post. $\endgroup$
    – Hakanaou
    Apr 3 at 0:22
  • $\begingroup$ @user619894 I'm sorry if it ended up being confusing, I compute two different integrals, one is $\int_0^1\mathrm{d}x\ x^n f_+(x)\star g(x)$ (and then when I compare it to the product of Mellin transforms I get a discrepancy) and the other is $\int_0^1\mathrm{d}x\ (x^n-1) f_+(x)\star g(x)$ and then I get the right result (at least numerically-wise) $\endgroup$
    – Hakanaou
    Apr 3 at 0:25


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