How do you prove that $\int_{0}^{\infty} \frac{t^{z-1}}{e^t - 1} \text{ dt}$ converges for $\operatorname{Re}(z) > 1$?

Question

The integral comes from the Mellin Transform representation of the Riemann zeta function, and is such that:

$$ \zeta{(z)}=\frac{1}{\Gamma(z)} \int_{0}^{\infty} \frac{t^{z-1}}{e^t - 1} \text{ dt} $$

Since this is a nonelementary integral, you cant prove convergence by evaluating it with the fundamental theorem of calculus and taking a limit (I think, maybe with polylogarithms but I don't know much about those). I do know that the function can be split up into $\int_{0}^{1}$ and $\int_{1}^{\infty}$ to deal with the problematic region separately.

As far as bounding functions and a direct comparison test, I found that:

$e^t -1 \geq t $

And so,

$\frac{t^{z-1}}{e^t - 1} \leq \frac{t^{z-1}}{t} = t^{z-2}$

For which the integral $\int_{0}^{1}t^{z-2} \text{ dt} $ converges for $\operatorname{Re}(z) > 1$ and therefore so does $\int_0^1 \frac{t^{z-1}}{e^t - 1} \text{ dt}$

I am not sure how to handle the other part of the integral though, what can I use to compare $\int_{1}^{\infty} \frac{t^{z-1}}{e^t - 1} \text{ dt}$ to in order to show that it converges? There might just be something really simple that I am missing.

Answer 1

As $t \to 0$, the integrand behaves as $t^{z -2} \implies \Re\left(z - 2\right) > -1$.

Answer 2

The complex number $z$ in the integrand is fixed, so we can ignore the factor $\Gamma(z)$. To prove the integral converges, it suffices to show it is absolutely convergent. Since $|t^{z-1}| = t^{a - 1}$ when $z = a+bi$, show $$ \int_0^\infty \frac{t^{a-1}}{e^t-1}\,dt $$ converges when $a > 1$. We might as well write $a-1$ as a single expression, say $c$, and show $$ \int_0^\infty \frac{t^{c}}{e^t-1}\,dt $$ converges when $c > 0$. Note $e^t > 1$ for $t > 0$, so the integrand is positive on $(0,\infty)$.

The function $t^c/(e^t-1)$ is continuous on $(0,\infty)$ and near $t=0$ the integrand behaves like $t^c/t = t{c-1}$, which is integrable on $(0,1)$ since $\int_{\varepsilon}^1 t^{c-1}\,dt = (t^c/c)|_{\varepsilon}^1 = 1/c-\varepsilon^c/c \to 1/c$ as $\varepsilon \to 0^+$. We used $c>0$ in the last limit calculation.

Thus we can split the integral into two parts, say $[0,1]$ and $[1,\infty)$. There is no trouble on $[0,1]$ (because the integrand is a bounded continuous function there), and on $[1,\infty)$ we have the lower bound $e^t - 1 > (1/2)e^t$, so $$ \int_1^\infty \frac{t^{c}}{e^t-1}\,dt < \int_1^\infty \frac{2t^{c}}{e^t}\,dt. $$ Split up $e^t$ on the right into $e^{t/2}e^{t/2}$ so we can get $t^c/e^{t/2}$, which has a positive upper bound on $[1,\infty)$, say $t^c/e^{t/2} \leq B_c$. Then $$ \int_1^\infty \frac{2t^{c}}{e^t}\,dt \leq 2B_c\int_1^\infty e^{-t/2}\,dt < \infty. $$

Answer 3

The integral

$$\zeta(s)=\frac{1}{\Gamma(s)} \int\limits_{0}^{\infty} \frac{t^{s-1}}{e^t-1}\,dt\tag{1}$$

can be evaluated using the relationship

$$\frac{1}{e^t-1}=\sum_{n=1}^\infty e^{-n t}.\tag{2}$$

which is valid for $t>0$.

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Since

$$\int\limits_0^\infty e^{-n t} t^{s-1}\,dt=\Gamma(s)\, n^{-s},\quad Re(s)>0$$

term-wise integration of formula (2) above leads to

$$\zeta(s)=\sum_{n=1}^\infty n^{-s}$$

which is only valid for $\Re(s)>1$.