An integral for the difference of zeta functions $\zeta (s-1)-\zeta(s)$

Question

Starting with:

$$\zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }{\frac {x^{s-1}}{e^{x}-1}}\,\mathrm {d} x$$

How can we prove for $s > 2$ the following conjecture:

$$\zeta (s-1)-\zeta(s)={\frac {2}{\Gamma (s+1)}}\int _{0}^{\infty }{\frac {x^s e^x}{(e^{x}-1)^3}}\,\mathrm {d} x$$

Integration by parts may be a way to go with:

$$u=x^s \qquad du=s x^{s-1} \\ dv=\frac {e^x dx}{(e^{x}-1)^3} \qquad v=- \frac{1}{2} \frac {1}{(e^{x}-1)^2}$$

Which gives us:

$$\zeta (s-1)-\zeta(s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }{\frac {x^{s-1} }{(e^{x}-1)^2}}\,\mathrm {d} x$$

It's a little nicer, but I'm not sure how to prove it either.

---

It just came to me: we should probably do another integration by parts, that's it.

Answer 1

The (inverse) Laplace transform gives $$ \zeta(s)=\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x-1}\,dx \stackrel{\text{IBP}}{=} \frac{1}{\Gamma(s+1)}\int_{0}^{+\infty}\frac{x^s e^x}{(e^x-1)^2}\,dx$$ hence, from the last expression, $$ \zeta(s-1) = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}e^x}{(e^x-1)^2}\,dx $$ and $$\zeta(s)-\zeta(s-1)=\frac{1}{\Gamma(s)}\int_{0}^{+\infty}x^{s-1}\left[\frac{1}{e^x-1}-\frac{e^x}{(e^x-1)^2}\right]\,dx = -\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{(e^x-1)^2}\,dx.$$ Integrate by parts once again and you are done.