Deriving $ \frac{1}{n^s} = \frac{1}{\Gamma(s)} \int_0^{\infty}t^{s-1} e^{-nt} \, dt $ Backwards?

Question

Is it possible to start with $\dfrac{1}{n^s}$ and then, without knowing the Gamma function in advance, naturally (with reasons!) derive that

$$ \frac{1}{n^s} = \frac{1}{\Gamma(s)} \int_0^{\infty}t^{s-1} e^{-nt} \, dt $$

(which comes up when you analyze the Zeta function)?

I know it can be done by starting from

$$\Gamma(s) = \int_0^{\infty} t^{s-1}e^{-t} \, dt$$

and setting $t = nx$ so that

$$\Gamma(s) = n^s \int_0^\infty x^{s-1}e^{-nx} \, dx$$

but working backwards, maybe you can use

$$\frac{1}{n^s} = \int_n^\infty s \frac{dx}{x^{s+1}}$$

and somehow end up at it?

Answer 1

You first line is equivalent to: $$\mathcal{L}\left(x^{s-1}\right) = \Gamma(s)\,y^{-s} $$ (I used $y$ as a conjugate variable; that is not standard but there are obvious reasons for avoiding $s$) hence the identity "read backwards" deals with: $$ \mathcal{L}^{-1}(y^{-s})=\frac{1}{2\pi i}\int_{c-i\infty}^{c+\infty} e^{xy} y^{-s}\,dx $$ and the reciprocal $\Gamma$ function makes his appearance (assuming $s\in\mathbb{N}$) as the residue at the origin of $\frac{e^{xy}}{y^s}$.