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Mellin transform of $h(t)=f^{-1}(t)$ is:

$$M_s(h(t))=\int_0^\infty x^{s-1}h(t)dt\mathop=^{h(t)\to t}\int_0^\infty t f(t)^{s-1} df(t)=?$$

Could someone go through a step-by-step.

First part:

Using: $$h(t)=f^{-1}(t)$$ substituting: $u=f^{-1}(t)$ ->$f(u)=t$-> $df(u)=dt$

$$\int_0^\infty x^{s-1}h(t)dt=\\\int_0^\infty x^{s-1}f^{-1}(t)dt=\\\int_0^\infty f(u)^{s-1}f^{-1}(f(u))df(u)=\\\int_0^\infty f(u)^{s-1}u df(u)=\\\int_0^\infty tf(t)^{s-1}df(t)=?$$

Second part:

Assuming that $f$ is continuous and of local finite variation on $[0,\infty)$,

$$d_t(f(t)^{s-1})=(s-1)f(t)^{-1} f(t)^{s-1}df(t)$$

Integration by parts yields

$$\int_0^\infty tf(t)^{s-1}df(t)=\frac{1}{s-1}\int^\infty_0t f(t) d_t(f(t)^{s-1})=? $$

Similar question may find here,I can't integrate by parts.

Thanks.

Mariusz Iwaniuk
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    It appears you have too many variables in your Mellin transform ($x$ should be $t$). – Steven Clark Jul 18 '23 at 15:20
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    @TymaGaidash The Mellin transform only converges for $-1<\Re(s)<-\frac{1}{6}$, so I don't believe you can integrate along the line $s=0.2+i t$. – Steven Clark Jul 19 '23 at 02:13
  • @StevenClark Thanks. This clears up the convergence. The integral now works with $\int_{c-i\infty}^{c+i\infty},-1<c<-\frac16$ – Тyma Gaidash Jul 19 '23 at 02:16

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A function with points $(0,0)$ and $(\infty,\infty)$ will probably not work with the Mellin inversion theorem, but could with the Ramanujan master theorem, say if $s=-n,n\in\Bbb N$. If the function has these points and $\lim\limits_{t\to0}t\ f(t)^s=0$, then the DI method yields:

$$\begin{matrix}&\text D&\text I\\+&t&f(t)^{s-1}f’(t)\\-&1&\frac1s f(t)^s\end{matrix}$$

and since $\lim\limits_{t\to0}t\ f(t)^s=0$:

$$\begin{aligned}\int_0^\infty t\,f(t)^{s-1}f’(t)dt=\left.\left(\frac tsf(t)^s\right)\right|_0^\infty-\frac1s\int_0^\infty f(t)^sdt=\frac1s\int_0^\infty f(t)^sdt\end{aligned}$$

Тyma Gaidash
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