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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 $\zeta (0)=-1/2$.

We use left multiplication by $\Gamma =S{L}_2(\mathbb{Z})$. For an integral matrix $A$ of determinant $n>0$, the Bezout identity gives a determinant-one row operation sending its first column to $(a,0{)}^t$, where $a$ is the positive gcd of that column. The resulting matrix is $\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$ with $d=n/a>0$. Adding a multiple of the second row to the first makes $0\le b<d$.

These representatives are unique. Left multiplication by a unimodular matrix preserves the gcd of the first column, so two representatives in one orbit have the same $a$ and hence $d$. A matrix taking $(a,0{)}^t$ to itself has the form $\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)$, so it changes $b$ by $td$; the prescribed range makes $b$ unique. Thus the determinant-n matrix representatives for Hecke operators are exactly ${\Pi }_n$. In the printed set, $0\le b<d$ already forces $d>0$, and $ad=n>0$ forces $a>0$.

For positive determinant extend the slash operator for modular forms byThe determinant factor makes this a right action: $(f{|}_{k}A){|}_{k}B=f{|}_k(AB)$. Right multiplication by $\delta \in \Gamma$ permutes the left cosets represented by ${\Pi }_n$, since it preserves the set of determinant-$n$ integral matrices. Hence $T_{n}f$ satisfies the modular transformation law. Its summands are holomorphic on the upper half-plane. For an upper-triangular representative,Substitute the Fourier expansion of a modular form. The sum over $b$ is zero unless the original index is divisible by $d$, in which case it equals $d$. Writing that index as $dr$ givesOnly nonnegative powers occur, so $T_{n}f$ is holomorphic at infinity, and all level-one cusps are equivalent to infinity. We have proved $T_{n}f\in M_k(\Gamma )$ and the Fourier coefficients of a composite-index Hecke operator formulaHere $\gcd (n,0)=n$, giving $a_0(T_{n}f)={\sum }_{a\mid n}{a}^{k-1}{a}_0(f)$. In particular cusp forms stay cusp forms. For nonzero level-one cusp forms the weight is an even integer at least twelve, so the powers $a^{k-1}$ are integers; the zero cusp space causes no exception. Thus every $T_n$ preserves integral Fourier coefficients, and closure under addition, multiplication and integer scalars provesThis is the integral Hecke algebra of level-one cusp forms.

We will also need commutativity. The coefficient formula directly givesFor the first identity, the divisors of $u$ and $v$ in the iterated coefficient sum combine uniquely into a divisor of $uv$. For the second, write a coefficient index as $p^t{r}_0$ with $p\nmid r_0$; the formula isApplying $T_p$ splits this into the two sums with shifts $t+1$ and $t-1$; their overlapping terms give exactly the stated recurrence. Since $T_1$ is the identity, induction expresses each $T_{p^r}$ as a polynomial in $T_p$, and coprime multiplicativity expresses every $T_n$ as a product of these polynomials. Distinct prime operators commute by the coprime identity. Hence all Hecke operators commute, and so do all elements of $\mathbb{T}$.

Let $\Lambda =S_k(\Gamma ,\mathbb{Z})$ and $m=\dim S_k(\Gamma )$. In the permitted basis $F_j={\Delta }^j{E}_4^{a_j}{E}_6^{b_j}$, the modular discriminant begins $q+O(q^2)$ and the normalized Eisenstein series begin $1+O(q)$, soThe matrix of $a_1,\dots ,a_m$ on this integral basis is unitriangular and has determinant one. Integer elimination therefore gives an integral basis $f_1,\dots ,f_m$ with $a_i(f_j)={\delta }_{ij}$ for $1\le i,j\le m$. Thus $a_1,\dots ,a_m$ are a $\mathbb{Z}$-basis of the dual module ${\mathrm{Hom}}_{\mathbb{Z}}(\Lambda ,\mathbb{Z})$.

The map $\alpha$ is well-defined by lattice preservation and is $\mathbb{Z}$-linear. Because $\alpha (T_j)=a_j$, it is surjective. For injectivity suppose $\alpha (T)=0$. For every $f\in \Lambda$ and every $n\ge 1$, commutativity givesThe constant term is also zero because $Tf$ is a cusp form. The identity theorem applied to its Fourier expansion of a modular form gives $Tf=0$. The integral basis spans the complex cusp space, so $T=0$ as an endomorphism. This proves the perfect integral Hecke pairing andIf $m=0$, both modules are zero and the asserted basis is empty.

One form of the Baire category theorem says that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets.

For each nonnegative integer $n$, putBecause $f$ is an entire function, every derivative $f^{(n)}$ is continuous and $Z_n$ is closed. No $f^{(n)}$ is identically zero: if $f^{(n)}\equiv 0$ for some $n$, repeated integration shows that $f$ is a polynomial. By the identity theorem, the zero set of the nonzero holomorphic function $f^{(n)}$ has empty interior. Thus each $Z_n$ is nowhere dense.

The complex plane is complete, so the Baire category theorem givesChoose $z_0$ outside this union. Then $f^{(n)}(z_0)\ne 0$ for every $n$, and hence every coefficient $f^{(n)}(z_0)/n!$ in the Taylor series about $z_0$ is nonzero. Such a point $z_0$ therefore exists.

Apply the Hankel representation with $z$ replaced by $1-z$:Enlarge $H$ to a modified Hankel contour that encloses the simple poles $t=2\pi in$, $n\in \mathbb{Z}\setminus \{0\}$. In a strip where the deformation and residue sum converge, the outer pieces vanish and the orientation gives the Hankel integral as minus $2\pi i$ times the sum of the enclosed residues. Part (b) givesSubstitution therefore yieldsBoth sides are meromorphic functions of $z$, so the identity theorem extends the equality from the initial strip to every point where either side is defined.

For each $n\ge 0$, defineEvery $f^{(n)}$ is an entire function, hence continuous, so $E_n$ is closed. The hypothesis saysThe complex plane is a complete metric space. The Baire category theorem therefore implies that some $E_N$ has nonempty interior. Thus $f^{(N)}$ vanishes on an open set, and the identity theorem giveson all of $\mathbb{C}$. If $N=0$, then $f=0$; otherwise the power series of $f$ has no term of degree $N$ or higher. Hence $f$ is a polynomial of degree at most $N-1$. This is the pointwise vanishing derivative criterion for a polynomial.

Let the argument of $z$ increase by $2\pi$ along a counterclockwise path enclosing both branch points. On a large circle, the branch chosen in part (a) satisfiesThe change in the primitive is consequentlyEquivalently, this is the composition of the two reflections described by the monodromy of the complex inverse sine. Thus, for this orientation,The reverse orientation gives $\mathrm{Arcsin}(z)+2\pi$. Both continued values invert the same $z$, so $\sin (w-2\pi )=\sin w$ on an open set of inverse values. The identity theorem extends this equality to every $w\in \mathbb{C}$, proving the periodicity of the complex sine:

For $\Re s>1$, the absolutely convergent Dirichlet seriesdefines the Riemann zeta function. To continue it, use partial summation in Stieltjes form:Since $0\le \{x\}<1$, the final integral converges locally uniformly for $\Re s>0$ and is holomorphic there. The displayed expression is consequently a meromorphic function on that half-plane, with its only pole at $s=1$. Since $s/(s-1)$ has residue one there, so does $\zeta$. Agreement in $\Re s>1$ makes this continuation unique by the identity theorem.

The identity theorem states that if two holomorphic functions on a connected open set $\mathcal{D}$ agree on a subset having an accumulation point in $\mathcal{D}$, then they agree throughout $\mathcal{D}$. In particular, agreement on any nonempty open subset is sufficient.

An analytic continuation of $f_1$ from ${\mathcal{D}}_1$ to ${\mathcal{D}}_2$ is a holomorphic function $f_2$ on ${\mathcal{D}}_2$ whose restriction to the nonempty overlap ${\mathcal{D}}_1\cap {\mathcal{D}}_2$ agrees with $f_1$.

Suppose $f_2$ and ${\tilde{f}}_2$ are two such continuations. Both equal $f_1$ on ${\mathcal{D}}_1\cap {\mathcal{D}}_2$, soon that nonempty open set. Since ${\mathcal{D}}_2$ is connected, the identity theorem givesThus analytic continuation to ${\mathcal{D}}_2$ is unique.

An analytic continuation along $\alpha$ is a chain of function elements $(f_t,U_t)$ covering the path, beginning with the germ of $(f,U)$ at $\alpha (0)$, such that neighboring elements agree on the component of overlap met by the path. The terminal element is $(g,V)$.

The Monodromy theorem says that if continuation is possible along every path in $D$, then continuations along endpoint-fixed homotopic paths have the same terminal germ. It follows by subdividing a homotopy square into small rectangles on which the identity theorem identifies neighboring germs. Hence

The requested final example with homotopic paths and different terminal values cannot exist under these hypotheses; it contradicts the theorem just proved. With “non-homotopic” in place of “homotopic”, take the germ of $\log z$ near $1$ in ${\mathbb{C}}^{\ast }$. Continuation along the constant path gives $0$, whereas continuation once counterclockwise around zero gives $2\pi i$.

Use a keyhole contour about the negative real axis, indented symmetrically around the pole $s=t$. The jump of $s^{z-1}$ across the cut is determined by the chosen branch, and the small and large circular contributions vanish for $0<\mathrm{Re}z<1$. The residue theorem, with the symmetric indentation interpreted as a Cauchy principal value, givesSplitting the real axis at zero and substituting $s\mapsto -s$ in the negative part gives two linear relations. Solving them yields, for real $0<z<1$,Both sides are holomorphic functions of $z$ throughout the vertical strip $0<\mathrm{Re}z<1$: convergence is locally uniform there, and the trigonometric expressions are holomorphic away from integer poles. The identity theorem therefore extends the identities from real $z$ to the whole strip.

Rouché's theorem states that if $g$ and $h$ are holomorphic functions on a neighbourhood of the closure of a bounded domain andon its positively oriented boundary, then $g$ and $g+h$ have the same number of zeros in the domain, counted with multiplicity.

The Open mapping theorem states that a nonconstant holomorphic function on a domain maps every open subset to an open subset. To prove it, fix $z_0$ and put $w_0=f(z_0)$. By the identity theorem, the zeros of $f-w_0$ are isolated. Choose $r>0$ so that the closed disc $\overline{D(z_0,r)}$ lies in the domain and $z_0$ is the only zero of $f-w_0$ in that disc. ThenWhenever $|w-w_0|<\delta$, the constant $w_0-w$ has modulus smaller than $f-w_0$ on the circle. The Rouche theorem therefore says thathas the same positive number of zeros in the disc as $f-w_0$. Thus every $w\in D(w_0,\delta )$ lies in the image of $f$, proving that the image is open.

If $|f|$ had a local maximum at $z_0$, take a small open disc $D$ on which $|f(z)|\le |f(z_0)|$. The restriction of $f$ to $D$ cannot be constant, since the identity theorem would then make $f$ constant on the whole domain. The complex open mapping theorem says that $f(D)$ is an open neighbourhood of $f(z_0)$, and such a neighbourhood contains points of modulus greater than $|f(z_0)|$, a contradiction. This proves the maximum modulus principle from the complex open mapping theorem.

Now let $\Omega$ be bounded. Its closure is compact, so the continuous function $|f|$ attains a maximum there. If $f$ is nonconstant, the maximum modulus principle excludes an interior maximum; if $f$ is constant, the claimed bound is immediate. Hencewhich is the maximum modulus principle on a bounded domain.

Finally let $\Omega =\{z:\mathrm{Re}z>1\}$ and suppose $|f|\le K$ throughout $\overline{\Omega }$. Fix $z_0\in \Omega$. The principal complex logarithm is holomorphic on the right half-plane, so for each positive integer $n$ the functionis holomorphic on $\Omega$ and continuous on its closure. Given $\epsilon >0$, choose $R>|z_0|$ so large thatApply the bounded-domain result to $g_n$ on $\Omega \cap D(0,R)$. On the vertical part of the boundary, $|z|\ge 1$ and hence $|g_n(z)|\le M$. On the circular part,It follows thatLetting $n\to \infty$ and then $\epsilon \to 0$ gives $|f(z_0)|\le M$. Since $z_0$ was arbitrary, this proves the bounded half-plane maximum principle.

The boundedness assumption is necessary. The functionis holomorphic on $\Omega$ and continuous on its closure, and $|f(1+iy)|=e$ on the boundary, but $|f(x)|=e^x$ is unbounded for real $x>1$. Thus the boundary estimate does not control an unbounded holomorphic function on this unbounded domain.

The nonzero poles areTheir paired residues sum toThe paired series converges for $\mathrm{Re}s<1$ by the alternating-series test, and absolutely for $\mathrm{Re}s<0$. ThereforeUsing the stated contour closure, the Hankel integral is $-2\pi i$ times this sum. Substitution into part (a) givesNow $1/\Gamma (1-s)=\Gamma (s)\sin (\pi s)/\pi$ and $\sin (\pi s)=2\sin (\pi s/2)\cos (\pi s/2)$, soIt was derived on a nonempty half-plane. Both sides have analytic continuations, with the apparent gamma poles cancelled by the sine factor or the trivial zeros from part (b); the identity theorem therefore extends it to every $s\in \mathbb{C}$.

Suppose that for every $z\in \mathbb{C}$, at least one derivative $f^{(n)}(z)$ of an entire function vanishes. The closed sets $E_n=\{z:f^{(n)}(z)=0\}$ cover $\mathbb{C}$. By the Baire category theorem, some $E_N$ has nonempty interior. The identity theorem gives $f^{(N)}=0$ everywhere, so $f$ is a polynomial of degree below $N$.

A nonzero entire function can be described by its discrete zero locations and multiplicities, a product of Weierstrass elementary factors and a zero-free exponential factor. Conversely, consistent prescribed multiplicities at distinct points escaping every compact subset can be realized by choosing factor orders large enough for infinite product convergence from logarithmic tails. Repeated locations with incompatible exact multiplicities are not admissible data, and a finite accumulation of distinct zeros is excluded by the identity theorem.

Riemann hypothesis is equivalent to monotonicity of $\sigma \mapsto |\xi (\sigma +it)|$ for every fixed $t$ on $\sigma \ge 1/2$. Under Riemann hypothesis the real logarithmic derivative of Riemann xi is positive for $\sigma >1/2$. Conversely a zero right of that line would force the nonnegative monotone modulus to vanish on an interval, contrary to the identity theorem. Functional symmetry rules out zeros to the left.