Today I Learned

February 20th

Today I learned about the Mellin transformation, from here mostly. We take the following definition.

Terrence Tao mentions this is like a "multiplicative Fourier transform,'' which we can already see in the $\RR^+.$ Quickly, we'll be a bit more explicit with this. Rewrite the definition as\[2\{\mathcal Mf(t)\}(s)=\int_{\RR^\times}f(t)|t|^s\,\frac{dt}{|t|}.\]So we claim $2\{\mathcal Mf(t)\}$ is the Fourier transform of $f(t):\RR^\times\to\CC.$ Note $dt/|t|$ is a multiplicative Haar measure on $\RR^\times,$ for $t\mapsto|c|t$ gives $dt/|t|\mapsto dt/|t|.$ Additionally, our character group $\RR^\times\to\CC$ is $|t|\mapsto |t|^s$; we don't show this here, but we do remark that it makes our desired (abstract) Fourier transform as above.

There are also some concrete connections to the Fourier transform. For clarity, we define the Fourier transform by\[\{\mathcal Ff(t)\}(s):=\int_\RR f(t)e^{-2\pi its}\,dt.\]To translate from $\mathcal M$ to $\mathcal F,$ we begin by applying the variable change $t\mapsto e^{-u}$ with $dt\mapsto-e^{-u}\,du.$ This gives\[\{\mathcal Mf(t)\}(s)=\int_\infty^{-\infty}f\left(e^{-u}\right)e^{-us}e^u\cdot-e^{-u}\,du=\int_{-\infty}^\infty f\left(e^{-u}\right)e^{-us}\,dt.\]This is almost the Fourier transform (actually this is the "two-sided Laplace transform''). To finish, we fix $s=a+2\pi ib,$ which implies\[\{\mathcal Mf(t)\}(s)=\int_{-\infty}^\infty\left(f\left(e^{-u}\right)e^{-ua}\right)e^{-2\pi iub}\,du=\left\{\mathcal Ff\left(e^{-u}\right)e^{-ua}\right\}(b).\]I agree—this is a bit ugly. Regardless, I think we can see the multiplicativity hiding in the $e^{-u}.$

However, this relationship is somewhat useful. For example, we get an inversion formula for free from Fourier inversion. In particular, we recall\[f(t)=\int_\RR\{\mathcal Ff(t)\}(s)e^{2\pi its}\,dt,\]which when applied to $\{\mathcal Mf(t)\}$ gives\[f\left(e^{-u}\right)e^{-ua}=\int_{-\infty}^\infty\{\mathcal Mf(t)\}(s)e^{2\pi iub}\,db.\]Rearranging, this is\[f\left(e^{-u}\right)=\int_{-\infty}^\infty\{\mathcal Mf(t)\}(s)e^{u(a+2\pi ib)}\,db.\]We would like to integrate over $s=a+2\pi ib,$ but some care is required because we are integrating $db.$ Interpreting this as a variable change, we see\[f\left(e^{-u}\right)=\frac1{2\pi i}\int_{a-i\infty}^{a+i\infty}\{\mathcal Mf(t)\}(s)e^{us}\,ds.\]At this point, the $t=e^{-u}$ substitution is no longer helpful, so we rewrite this as\[f(t)=\frac1{2\pi i}\int_{a-i\infty}^{a+i\infty}\{\mathcal Mf(t)\}(s)t^{-s}\,ds.\]This is our inversion formula. Note the introduction of complex numbers. Also note that we are ignoring convergence issues, as usual.

I guess I should actually do an example. I think the example I'm supposed to care about is\[\left\{\mathcal Me^{-t}\right\}(s)=\int_0^\infty e^{-t}t^{s-1}\,dt.\]However, this is exactly $\Gamma(s),$ which is nice. In general, if we do $\left\{\mathcal Me^{-pt}\right\}$ for $p \gt 0,$ then we get\[\left\{\mathcal Me^{-pt}\right\}(s)=\int_0^\infty t^{s-1}e^{-pt}\,dt.\]To make $\Gamma$ appear now, we have to do a variable transformation $pt\mapsto t.$ This gives\[\left\{\mathcal Me^{-pt}\right\}(s)=\int_0^\infty\frac{t^{s-1}}{p^{s-1}}e^{-t}\,\frac{dt}p=\frac{\Gamma(s)}{p^s}.\]I guess this feels multiplicative, in that taking exponents amounts to a multiplication factor.

In other news, happy birthday to me I guess.