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For every integer $n\ge 1$, the Riemann zeta function satisfies $\zeta (1-n)=(-1{)}^{n-1}{B}_n/n$. One proof applies meromorphic continuation of a Mellin transform from an asymptotic expansion to $(e^y-1{)}^{-1}$, whose transform is the Bose integral $\Gamma (s)\zeta (s)$, and divides residues by the residues of the Gamma function. The formula includes $\zeta (0)=-1/2$.

The fundamental strip of a Mellin transform is the interior of the set of real parts $\sigma$ for whichViewed in the complex plane, it is a vertical strip $\alpha <\mathrm{Re}s<\beta$; either bound may be infinite, and the interior may be empty. Both endpoints of the improper integral must be checked. Convergence on a boundary line, whether absolute convergence or conditional convergence, requires a separate test. The Mellin transform is holomorphic throughout a nonempty fundamental strip.^[ref]

A practical sufficient test uses Big O notation: if $f(x)=O(x^p)$ near zero and $f(x)=O(x^{-q})$ near infinity, with real $p,q$, then comparison with elementary improper integrals givesThus $-p<\sigma <q$ guarantees absolute convergence. These upper bounds on $|f|$ need not locate the maximal Mellin fundamental strip; faster actual decay can widen it. The strip property itself follows by bounding the weight at any intermediate $\sigma$ by the sum of two weights already known to be integrable. On a closed narrower strip, neighboring integrable weights also dominate the natural logarithm factors introduced by differentiation under the integral sign, making the transform a holomorphic function.

An exact two-endpoint example isHere both bounds are sharp: at $\sigma =-2$ the zero-end absolute convergence test contains ${\int }_0^{1}dx/x$, and at $\sigma =3$ the infinity-end test contains ${\int }_1^{\infty }dx/x$. The displayed rational function continues meromorphically outside the Mellin fundamental strip and has simple poles at $-2$ and $3$. Its other finite continued values outside the strip are values of a meromorphic continuation, not of a convergent defining integral. More general endpoint information leads to meromorphic continuation of a Mellin transform from an asymptotic expansion.

For a nonprincipal Dirichlet character modulo $N$, setThe numerator vanishes at $t=0$, so $H_{\chi }$ is analytic there and exponentially decreasing as $t\to \infty$. Initially for $\mathrm{Re}s>1$, $\Gamma (s)L(\chi ,s)={\int }_0^{\infty }{H}_{\chi }(t)t^{s-1}dt$. This integral already extends holomorphically to $\mathrm{Re}s>0$. Subtracting finite Taylor polynomials near zero, as in meromorphic continuation of a Mellin transform from an asymptotic expansion, continues it with possible simple poles only at nonpositive integers. Zeros of the reciprocal Gamma function cancel these, proving that $L(\chi ,s)$ is an entire function. This does not require a primitive Dirichlet character.

The Mellin transform iswhere 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 $\mathrm{Re}s>-{\sigma }_1$.

For $\mathrm{Re}s>1$, expand $(e^{\pi y}-1{)}^{-1}={\sum }_{r\ge 1}{e}^{-\pi ry}$. Absolute convergence justifies termwise integration, using the Gamma integral, and gives the scaled Bose integral

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:Continuity of $g_r$ at zero makes it bounded on $[0,1]$. The last integral is holomorphic on $\mathrm{Re}s>-{\sigma }_{r+1}$: 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 $r\ge j$ 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 isThis is meromorphic continuation of a Mellin transform from an asymptotic expansion.

To show why the correction matters, set ${\sigma }_j=1-1/j$, $c_j=2^{-j}$, andwhere $\eta$ is continuous, equals one for $0<y\le 1$ and vanishes for $y\ge 2$. 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 equalsThe series continues meromorphically on $\mathrm{Re}s>-1$ 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 givesAt $s=-j$ the numerator is $\Gamma (1)=1$ and the product of the nonzero denominator factors is $(-1{)}^{j}j!$. Therefore the residues of the Gamma function are

The generating function of the Bernoulli numbers gives the convergent expansion near zeroThe subtraction proof above applies to this expansion, including its zero coefficients, which give no pole. At $s=1-n$, for $n\ge 1$, 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:In particular the generating-series convention is $B_1=-1/2$, so the formula includes $\zeta (0)=-1/2$.