Vtyjkyuk

This equation:
$$int_0^infty fracx^s-1e^x-1 dx=Pi(s-1)zeta(s) $$
was used by Riemann in his famous paper from 1859.
Seemingly it follows from:
$$ int_0^infty e^-nxx^s-1dx= fracPi(s-1)n^s$$

a result achieved by repeated integration by parts. How does the first equation follows from the second?

First we have that $$int_0^infty fracx^s-1e^x-1dx = int_0^infty x^s-1sum _n=0^inftye^-nxdx$$ Beliveing in uniform convergence we swap the sum and the integral to get $$sum _n=0^inftyint_0^infty x^s-1e^-nxdx$$ with a simple change of variable, we set $nx = t$ so $dt = ndx$, then $$sum _n=0^inftyint_0^infty x^s-1e^-nxdx = sum _n=0^inftyint_0^infty frac1nleft(fractnright)^s-1e^-tdt = underbracesum_n=0^infty n^-s_textDef. of zeta(s)underbraceint_0^infty t^s-1e^-tdt_textDef. of Gamma(s) = zeta(s)Gamma(s)$$ Just remember now the relation between the pi function and the Euler gamma function $$Pi(z) = Gamma(z+1)$$

A faster way to find the second equation would be to simply make the substitution $u = nx$ which leads to :

$$ int_0^infty e^-u left( fracunright)^s-1 fracdun = fracPi(s-1)n^s quad quad (*)$$

You can also write :

$$ int_0^infty fracx^s-1e^x - 1dx = int_0^infty e^-x x^s-1frac11 - e^-xdx$$

And, you can write (geometric series) :

$$ forall x >0, frac11 - e^-x = sum_n=0^infty e^-nx$$

which leads to

$$ int_0^infty fracx^s-1e^x - 1dx = int_0^infty e^-x x^s-1sum_n=0^infty e^-nx dx $$

We can easily switch sum and integral, indeed :

$$ forall n in mathbb N, sum_k=0^n e^-(n+1)x x^s-1 leqslant e^-x x^s-1 = varphi(x) $$

so we can apply the dominated convergence theorem with $varphi in L^1(mathbb R_+^*)$. Using $(*)$ :

$$ int_0^infty fracx^s-1e^x - 1dx = sum_n=0^infty fracPi(s-1)(n+1)^s $$

Which is exactly :

$$ int_0^infty fracx^s-1e^x - 1dx = Pi(s-1)zeta(s) $$