= Solution
The <Mellin transform> is
$$
\boxed{M(f,s)=\int_0^\infty f(y)y^{s-1}\,dy,}
$$
where this integral converges. Rapid decay at infinity makes the integral over $[1,\infty)$ an entire function of $s$, but alone gives no control near zero. With the additional expansion, the integral initially converges absolutely for $\operatorname{Re}s>-\sigma_1$.
For $\operatorname{Re}s>1$, expand $(e^{\pi y}-1)^{-1}=\sum_{r\geq1}e^{-\pi ry}$. Absolute convergence justifies termwise integration, using the <Gamma integral>, and gives the scaled <Bose integral>
$$
\boxed{M\left(\frac1{e^{\pi y}-1},s\right)
=\sum_{r\geq1}(\pi r)^{-s}\Gamma(s)
=\pi^{-s}\Gamma(s)\zeta(s).}
$$
There is a missing hypothesis in the general continuation claim: one needs $\sigma_j\to+\infty$. The original PDF, like the TeX, only says that the sequence increases. Under the intended additional hypothesis, split the integral at one and subtract $r$ terms of the <asymptotic expansion>:
$$
M(f,s)=\int_1^\infty f(y)y^{s-1}\,dy
+\sum_{j=1}^r\frac{c_j}{s+\sigma_j}
+\int_0^1 y^{s+\sigma_{r+1}-1}g_r(y)\,dy.
$$
Continuity of $g_r$ at zero makes it bounded on $[0,1]$. The last integral is <holomorphic> on $\operatorname{Re}s>-\sigma_{r+1}$: on every compact subset, its integrand and all its $s$ derivatives are dominated by an integrable power of $y$ times a power of $|\log y|$. The expressions for successive $r$ agree on their common initial domain and hence everywhere they overlap by the <identity theorem>. As $\sigma_{r+1}\to\infty$, these half-planes cover $\mathbb C$. Taking $r\geq j$ with $\sigma_{r+1}>\sigma_j$ isolates the term $c_j/(s+\sigma_j)$, while all other terms are <holomorphic> near $-\sigma_j$. Thus the corrected claim is
$$
\boxed{\operatorname{Res}_{s=-\sigma_j}M(f,s)=c_j,
\quad\text{with no other poles, provided }\sigma_j\to\infty.}
$$
This is <meromorphic continuation of a Mellin transform from an asymptotic expansion>.
To show why the correction matters, set $\sigma_j=1-1/j$, $c_j=2^{-j}$, and
$$
f(y)=\eta(y)\sum_{j\geq1}2^{-j}y^{1-1/j},
$$
where $\eta$ is continuous, equals one for $0<y\leq1$ and vanishes for $y\geq2$. On $[0,1]$, the remainder after division by $y^{\sigma_{r+1}}$ is a uniformly convergent series of nonnegative powers of $y$, so it extends continuously to zero with value $c_{r+1}$. Away from zero, define $g_r$ by the required remainder quotient; it is continuous on the rest of $[0,\infty)$ as well. Thus all the printed hypotheses hold. Its <Mellin transform> equals
$$
\sum_{j\geq1}\frac{2^{-j}}{s+1-1/j}
+\int_1^2 f(y)y^{s-1}\,dy.
$$
The series continues meromorphically on $\operatorname{Re}s>-1$ and has genuine poles at $-1+1/j$, accumulating at $-1$. A <meromorphic function> on $\mathbb C$ cannot have such an accumulation of poles. This is the <accumulating asymptotic exponents obstruct Mellin continuation> counterexample.
For the <Gamma function>, the <Gamma function recurrence> gives
$$
\Gamma(s)=\frac{\Gamma(s+j+1)}{s(s+1)\cdots(s+j)}.
$$
At $s=-j$ the numerator is $\Gamma(1)=1$ and the product of the nonzero denominator factors is $(-1)^jj!$. Therefore the <residues of the Gamma function> are
$$
\boxed{\operatorname{Res}_{s=-j}\Gamma(s)=\frac{(-1)^j}{j!}\quad(j\geq0).}
$$
The generating function of the <Bernoulli numbers> gives the convergent expansion near zero
$$
\frac1{e^{\pi y}-1}=\sum_{n\geq0}\frac{B_n\pi^{n-1}}{n!}y^{n-1}.
$$
The subtraction proof above applies to this expansion, including its zero coefficients, which give no pole. At $s=1-n$, for $n\geq1$, the residue of its transform is $B_n\pi^{n-1}/n!$, while the residue of $\pi^{-s}\Gamma(s)$ is $\pi^{n-1}(-1)^{n-1}/(n-1)!$. Their quotient defines a <holomorphic> continuation of the <Riemann zeta function> at this point, including when the first residue is zero. Dividing gives the <Bernoulli formula for zeta values at nonpositive integers>:
$$
\boxed{\zeta(1-n)=\frac{(-1)^{n-1}B_n}{n}\quad(n\geq1).}
$$
In particular the generating-series convention is $B_1=-1/2$, so the formula includes $\zeta(0)=-1/2$.
Back to article page
