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For the Airy equation with opposite dispersive sign, $q_t-q_{xxx}=0$ on $x>0$, let $Q$ be the initial Half-range Fourier transform and $F_j(k,t)={\int }_0^t{e}^{i{k}^{3}s}{\partial }_x^{j}q(0,s)ds$. Three integrations by parts give the displayed global relation. The spatial transform is holomorphic for $\mathrm{Im}k<0$, whereas the finite-time spectral boundary transforms are entire functions. The symmetry $k\mapsto \alpha k$, ${\alpha }^3=1$, preserves their time exponent.

// Target: analysis.bigb

Keep the circular portion of a Hankel contour at fixed nonzero radius. Its numerator $I(z)={\int }_H{t}^{z-1}{e}^{t}dt$ is an entire function of $z$, because the contour avoids zero and its tails decay exponentially, uniformly on compact parameter sets. At a positive integer $n$, the integrand is single-valued and entire in $t$, so the bank integrals cancel and the circular integral is zero. Thus $I(n)=0$, canceling the simple zero of $2i\sin (\pi z)$ in the Hankel representation of the gamma function. At a nonpositive integer $-n$, the circular integral instead gives $2\pi i/n!$, yielding residue $(-1{)}^n/n!$.

A canonical product builds a prescribed discrete zero set from Weierstrass elementary factors. Choose the orders $p_n$ so that the logarithms of tail factors converge uniformly on each compact subset. Infinite product convergence from logarithmic tails then gives an entire function with exactly those zeros, counted with the prescribed multiplicities. When ${\sum }_n|a_n{|}^{-1}<\infty$, factors of order zero suffice.

For an entire function $f$, use the complex logarithm $L$ with argument in $(0,2\pi )$. The unit-circle integral in the title uses the branch's one-sided boundary values at its crossing with the cut. On a keyhole contour, the outward upper bank has $L=\log x$, and the inward lower bank has $L=\log x+2\pi i$. Their combined integral of $f(L-i\pi )$ is $-2\pi i{\int }_{\epsilon }^{1}f(x)dx$. The inner circle contributes $O(\epsilon |\log \epsilon |)$, which vanishes. The Cauchy integral theorem on the slit plane therefore gives the identity.

On $z=e^{i\theta }$, $|L-i\pi |=|\theta -\pi |\le \pi$. The estimation lemma consequently givesSubtracting the middle branch value $i\pi$ centres the weight and produces this bound. The contour crosses the cut only in the limiting boundary-value interpretation, so it is not a closed contour inside the logarithm's domain.

The displayed supremum norm is finite on every entire function, positive on every nonzero one by the identity theorem, and submultiplicative under pointwise multiplication. It is not complete: partial geometric sums ${\sum }_{j=0}^n(z/2{)}^j$ are Cauchy on the disk but their limit there extends meromorphically with a pole at two. More generally, no complete algebra norm exists on all entire functions because the coordinate function has algebraic spectrum of an element equal to the whole complex plane.

The complementary error function describes the Gaussian integral area remaining beyond the endpoint. For a real $x$,It is an entire function for complex arguments when defined by $1-\mathrm{erf}(z)$. On the real line it decreases from two to zero, with $\mathrm{erfc}(0)=1$ and $\mathrm{erfc}(-x)=2-\mathrm{erfc}(x)$.^[ref]

For example, $\mathrm{erfc}(1)\approx 0.157299$ is a sizeable tail, whereas $\mathrm{erfc}(8)\approx 1.12243\times 10^{-29}$ is extremely small. In the tested floating-point arithmetic, a direct evaluation of $1-\mathrm{erf}(8)$ gives zero because the computed error function has already rounded to one. Evaluating the complementary error function directly preserves this tail. It is the useful quantity in an upper-tail normal distribution probability and in the decaying side of an error-function diffusion front.

For large positive real $x$, the leading asymptotic expansion isThe first corrections are $1-1/(2x^2)+3/(4x^4)-\cdots$ multiplying the leading term. This is an asymptotic expansion, not an assertion that summing arbitrarily many terms converges at a fixed $x$.^[ref] The complementary error function tail bound provides finite-$x$ inequalities instead.

For a primitive nonprincipal Dirichlet character, the theta Mellin integral makes this an entire function satisfying $\Lambda (s,\chi )={\epsilon }_{\chi }\Lambda (1-s,\overline{\chi })$. The gamma poles cancel the trivial zeros of a Dirichlet L-function. For the principal Dirichlet character whose conductor of a Dirichlet character is one the completion is meromorphic, with poles at zero and one.

Composing a hypothetical conformal equivalence $\mathbb{C}\to \mathbb{H}$ with a Cayley transform $\mathbb{H}\to \mathbb{D}$ would give a nonconstant bounded entire function, contradicting the Liouville theorem.

The sine function extends to an entire function on the complex plane by the displayed exponential formula. Substituting $z=x+iy$ and grouping the real and imaginary terms gives $\sin (x+iy)=\sin x\cosh y+i\cos x\sinh y$, for real $x,y$. Thus its values grow exponentially in imaginary directions, even though the real sine is bounded.

If $f$ is holomorphic on an open set $U$, its conjugate reflection $f^{\ast }$ is holomorphic on $U^{\ast }=\{z:\overline{z}\in U\}$, with $(f^{\ast }{)}^{\prime }(z)=\overline{f^{\prime }(\overline{z})}$. The identity follows by conjugating the difference quotient with increment $\overline{h}$; the two conjugations restore complex differentiability.

If $U$ is invariant under complex conjugation and is connected, and $f$ is real at a set of real points accumulating inside $U$, then $f=f^{\ast }$ by the identity theorem. In particular, an entire function real at every $1/n$ has this symmetry. This is a symmetry statement on an already defined domain; Schwarz reflection also constructs extensions across a boundary.

The saddle and endpoint asymptotic expansions of this entire function have algebraic coefficients $C{\omega }^j/z$ in sectors $(2j-1)\pi /3<\arg z<(2j+1)\pi /3$, where $C=\Gamma (1/3)/3$ and $\omega =e^{2\pi i/3}$. Deforming the contour onto the decay ray $\arg (zt)=2j\pi /3$ gives its complete integral $C{\omega }^j$. The endpoint contribution is $-e^{-z^3}(3z^3{)}^{-1}{\sum }_{r\ge 0}(-1{)}^r(2/3{)}_r/z^{3r}$. The algebraic coefficient switches at $\arg z=\pi /3,\pi ,5\pi /3$; the equal-exponential-magnitude anti-Stokes lines are the intervening rays $\pi /6+j\pi /3$.

A distribution $u$ on a plane open set with ${\partial }_{\bar{z}}u=0$ is represented by a holomorphic function. Indeed $\Delta u=4{\partial }_z{\partial }_{\bar{z}}u=0$, and elliptic regularity makes $u$ smooth. For a direct proof, mollify locally with a radial mollifier ${\eta }_{\epsilon }$. The resulting smooth functions are harmonic. Their mean value property gives $(u\ast {\eta }_{\epsilon })\ast {\eta }_r=u\ast {\eta }_{\epsilon }$ on a smaller ball for fixed small $r$. Passing to the weak convergence of distributions limit gives $u\ast {\eta }_r=u$ there. This proves smoothness without assuming local integrability beforehand. The ordinary Cauchy-Riemann equations now prove holomorphy.

The equation ${\partial }_{z}u=0$ similarly characterizes antiholomorphic functions. On the whole plane, the homogeneous solutions are respectively arbitrary entire functions $h(z)$ and the functions $h(\bar{z})$ with $h$ entire. Consequently any two fundamental solutions for one of these Wirtinger derivatives differ by the corresponding homogeneous solution.

For a nonconstant entire function, its derivative is nonzero somewhere and therefore bounded away from zero on a small disc. Recurrence of planar Brownian motion and the Strong Markov property imply infinite total occupation time in that disc: repeat entrances into a smaller concentric disc after exits from the larger one. On each entrance there is a uniformly positive chance of remaining inside the larger disc for a fixed positive duration. Iterating the conditional failure bound gives infinitely many such sojourns almost surely.

The displayed integral is bounded below by a positive constant times this occupation time, so it diverges. The same conclusion follows by comparing with the divergence of a positive planar Brownian occupation integral. At every finite time the clock is finite, because the Brownian path has compact range and $f^{\prime }$ is continuous. Thus the conformal Brownian clock covers every nonnegative image time; conformal invariance of planar Brownian motion then yields a Brownian image with no missing finite or infinite lifetime.

For $X={\mathbb{P}}^1\times \mathbb{C}$, the two standard affine charts and their intersection form an acyclic cover for the sheaves of holomorphic forms. Their Čech cohomology gives $H_{\bar{\partial }}^{0,0}=\mathcal{O}(\mathbb{C})$ and $H_{\bar{\partial }}^{2,1}=\mathcal{O}(\mathbb{C})$, with all other groups for $p\in \{0,2\}$ zero. For the top-form sheaf, the transition $d(1/z)=-z^{-2}dz$ removes every Laurent series power except $z^{-1}$ from the first cohomology quotient; its coefficient may be an arbitrary entire function on the affine factor.

An entire function is a holomorphic function whose domain of a function is the whole complex plane. Its derivative exists at every $z\in \mathbb{C}$. Every polynomial, the complex exponential function $e^z$, the product $z{e}^z$, and the complex sine function $\sin z$ are entire functions. The Taylor series of an entire function about any finite centre converges throughout the complex plane.

The complex exponential function is a particularly revealing entire function. Writing $z=x+iy$, Euler's formula givesThus the complex exponential function never takes the value zero. Conversely, for any $w\ne 0$, choose an argument $\theta$ of $w$. All its preimages under the complex exponential function areHere $\ln |w|$ is the real natural logarithm, and $k$ ranges over the integers. Substitution gives $e^{z_k}=|w|e^{i\theta }=w$; equality of moduli and arguments shows these are all the solutions. The complex exponential function therefore has image exactly $\mathbb{C}\setminus \{0\}$, with infinitely many preimages for each nonzero value.

In the pictured vertical strip $-1\le x\le 1$, the real part sets the modulus between $e^{-1}$ and $e$, while the imaginary part sets the argument modulo $2\pi$. The two drawn bands each make one full turn around the same annulus; their corresponding points differ by $2\pi i$ and have identical images. Allowing all real $x$ gives every positive modulus, so the whole complex plane covers every nonzero output.

An entire function has exceptionally strong global constraints. The Liouville theorem says that a bounded function which is also an entire function must be a constant function. Accordingly, the nonconstant entire functions above are unbounded somewhere in the complex plane, even when their restrictions to the real axis are bounded: the complex sine function satisfies $\sin (iy)=i\sinh y$, growing through the hyperbolic sine. The Little Picard theorem adds a different constraint: an unbounded nonconstant entire function still cannot leave two distinct complex values out of its image. These results make entire functions useful when global bounds or restrictions on the image can establish rigidity without solving for the function explicitly.

In the opening of his 1880 memoir, Émile Picard credits Karl Weierstrass with the terminology of entire functions and describes them by everywhere-convergent power series. ^[ref]

The displayed formula is a submultiplicative norm on the algebra of entire functions: definiteness follows from the identity theorem. No submultiplicative norm on the whole algebra can be complete. The coordinate function $Z(z)=z$ has every complex number in its algebraic spectrum of an element, because $Z-\lambda$ vanishes somewhere and has no entire inverse. This contradicts compactness of spectra in a Banach algebra.

If an entire function $g$ satisfies $\mathrm{Re}(cg)\ge b$ for constants $c\ne 0$ and real $b$, then $e^{-cg}$ is entire and has modulus at most $e^{-b}$. The Liouville theorem makes it constant, and differentiating gives $g^{\prime }=0$ since the exponential never vanishes. The conclusion also applies after rotating and translating any containing half-plane; separate bounds on the real and imaginary parts are unnecessary.

Let $f,g$ be entire functions, with $g$ not identically zero. If $|f(z)|\le C|g(z)|$ everywhere, then $f/g$ is bounded and holomorphic away from the isolated zeros of $g$. The removable singularity extends it across each zero, giving a bounded entire function. Liouville's theorem makes it constant, so $f=cg$ with $|c|\le C$. If $g$ is identically zero the inequality forces $f=0$. For example, domination by $|\cos z|$ forces a constant multiple of $\cos z$.

An entire function is of exponential type if constants $A,B>0$ give the displayed bound everywhere. A genus-zero canonical product with ${\sum }_n|a_n{|}^{-1}<\infty$ obeys $|{\prod }_n(1-z/a_n)|\le \exp (|z|{\sum }_n|a_n{|}^{-1})$, since $1+u\le e^u$ for $u\ge 0$.

If two nonzero entire functions have identical zeros and orders, their quotient extends to a nowhere-zero entire function. Its logarithmic derivative has a primitive on the simply connected plane, producing a holomorphic logarithm $h$ and the displayed relation. Any two choices differ by one constant integer multiple of $2\pi i$. On a multiply connected domain the logarithm can fail to exist globally.