We can create a forward transform $T$ which is a composition of the inverse Mellin $\mathcal{M}^{-1}$ and Laplace $\mathcal{L}$ transforms $$ T[\phi] = \mathcal{L}\mathcal{M}^{-1}[\phi]=\mathcal{L}\left[\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} x^{-s}\phi(s)\;ds \right] $$ $$ T[\phi] = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \mathcal{L}\left[x^{-s}\right]\phi(s)\;ds = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} q^{s-1}\Gamma(1-s)\phi(s)\;ds $$ by letting $s-1 \to -t$ we get $$ T[\phi] = \frac{-1}{2\pi i}\int_{c'-i\infty}^{c'+i\infty} q^{-t}\Gamma(t)\phi(1-t)\;dt = \frac{-1}{2\pi i}\int_{c'-i\infty}^{c'+i\infty} q^{-t}\phi^*(t)\;dt = \boxed{-\mathcal{M}^{-1}[\phi^*]} $$ where $\phi^*(t) = \Gamma(t)\phi(1-t)$. Then we have the backward transform $T^{-1}$ given by $$ T^{-1}[\psi]= \mathcal{M}\mathcal{L}^{-1}[\psi] = \mathcal{M}\left[\frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} e^{s t} \psi(s) \; ds\right] $$ $$ T^{-1}[\psi] = \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} \mathcal{M}\left[e^{s t}\right] \psi(s) \; ds = \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} (-s)^{-q} \Gamma(q) \psi(s) \; ds $$ with a variable change $-s\to t$ we get $$ T^{-1}[\psi] = \frac{-1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} t^{-q} \Gamma(q) \psi(-t) \; dt = \boxed{???} $$ but now I can't see how to simplify that down to one of the original transforms. Obviously all of these transforms only work when they exist and there are many ways this could break.

Question: Is there any way of rewriting this integral like the other one?

Question: Is there much meaning to this transform? Is this a valid way of generating inverse Mellin transform identities?

There seems to be a a few nice examples:

\begin{array}{|c|c|c|c|} \hline \phi(s) & T[\phi](q) & Result &Remark\\ \hline T[\Gamma(s)] & \frac{1}{1+q} & \mathcal{M}^{-1}[\Gamma(1-q)\Gamma(q)] = \frac{1}{1+s}& known^*\\ T[\Gamma(n+s)] & \frac{\Gamma(n+1)}{(1+q)^{n+1}} & \mathcal{M}^{-1}[\Gamma(1+n-q)\Gamma(q)] = T[\phi] & known^* \\ T[\Gamma(s)\sin(\frac{\pi s}{2})] & \frac{1}{1+q^2} & \mathcal{M}^{-1}[\Gamma(1-q)\Gamma(q)\sin(\pi(1-q)/2)] = T[\phi] & known^* \\ T[\Gamma(s)\zeta(s)] & -H_q & \mathcal{M}^{-1}[\Gamma(1-q)\Gamma(q)\zeta(1-q)]=-H_q & conjecture\\ T[n^{s-1}\Gamma(s)\zeta(s)] & -H_{nq} & \mathcal{M}^{-1}[n^{-q}\Gamma(1-q)\Gamma(q)\zeta(1-q)]=-H_{nq} & conjecture \\ T[\Gamma(s)\zeta(s,a)] & -H_{q+a-1} & \mathcal{M}^{-1}[\Gamma(1-q)\Gamma(q)\zeta(1-q,a)]=-H_{q+a-1} & conjecture\\ T[\Gamma(s)\psi(s)]&-\frac{\gamma+\log(1+q)}{1+q} & \mathcal{M}^{-1}[\Gamma(q)\Gamma(1-q)\psi(1-q)]=T[\psi] & conjecture \\ T[1/s] & \frac{1-\cosh(q)+\sinh(q)}{q} &\cdots& possibly\;breaks\\ T[1/s^n] & \;_{n+1}F_{n+1}(\mathbf{1};\mathbf{2};-q) &\cdots& ???\\ \hline \end{array} * known to Mathematica or in integral transform tables.

Above $H_q$ are the harmonic numbers. Something possibly more sketchy $$ T[\Gamma[s]\;_1F_0(s;1)]=\frac{1}{q} $$

Thanks in advance for any input!


2 Answers 2


With the biletaral Laplace transform it is much easier

$$\mathcal{M}[\phi(x)](s) = \int_0^\infty x^{s-1} \phi(x)dx = \int_{-\infty}^\infty e^{-su} \phi(e^{-u})du=\mathcal{L}[\phi(e^{-u})](s)$$ thus $$\mathcal{L}^{-1}[\mathcal{M}[\phi(x)](s)](u) = \phi(e^{-u}), \qquad \mathcal{M}^{-1}[\mathcal{L}[\phi(u)](s)](x) = \phi(-\log x)$$

Finally $$\mathcal{L}[\phi(u)](\sigma+2i\pi \xi)=\int_{-\infty}^\infty \phi(u) e^{-\sigma u} e^{-2i \pi \xi u}du=\mathcal{F}[\phi(u)e^{-\sigma u}](\xi)$$ and the Fourier transform is unitary $\mathcal{F}^{-1}[\Phi(\xi)](u) = \mathcal{F}[\Phi(\xi)](-u)=\mathcal{F}^*[\Phi(\xi)](-u)$

With the unilateral Laplace transform you'll obtain something similar to what you wrote.

  • $\begingroup$ This is a very helpful description, thank you. The top equation doesn't seem to fit the description of the bilateral Laplace transform. Shouldn't the limit on the second integral be from $-\infty$, or is there a hidden step function? Perhaps you could use a notation that distinguishes the unilateral and bilateral transforms just to make things clear? $\endgroup$ Nov 3, 2017 at 10:36
  • 1
    $\begingroup$ @BenedictWilliamJohnIrwin Oops sorry I meant $\int_{-\infty}^\infty$ and it was the most important point (ie. write everything in term of $\mathcal{F}$, multiplication by $1_{u >0}$, convolution and $x= e^{-u}$). $\int_0^\infty$ implies $\int_1^\infty$ on the Mellin side $\endgroup$
    – reuns
    Nov 3, 2017 at 10:39
  • $\begingroup$ That should do that trick, thank you for your help. $\endgroup$ Nov 3, 2017 at 10:45

Thanks to @reuns helping me see some new connections and manipulations, it seems we can write $$ T^{-1}[\psi] = \frac{-1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} t^{-q} \Gamma(q) \psi(-t) \; dt = \frac{-\Gamma(q)}{2 \pi i} \int_{c-i\infty}^{c+i\infty} e^{-q\log(t)} \psi(-t) \; dt $$ then with a substitution $-\log(t)=s$ $$ T^{-1}[\psi] = \frac{\Gamma(q)}{2 \pi i} \int_{c'-i\infty}^{c'+i\infty} e^{qs} \psi(-e^{-s})e^{-s} \; ds = \boxed{\Gamma(q)\mathcal{L}^{-1}[\psi^*]} $$ where $\psi^*(s)=\psi(-e^{-s})e^{-s}$. The main lesson here was that the connection between these two transforms is much more beautiful in terms of the bilateral Laplace transform, the reason being the unitary nature of the Fourier transform as illustrated by @reuns.


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