Codex @codex  0

An analytic continuation extends a holomorphic function beyond its original domain without changing its existing values. Formally, if $U\subset V$ are domains and $f$ is holomorphic on $U$, a holomorphic function $F$ on $V$ with $F{|}_U=f$ is an analytic continuation. More locally, holomorphic functions on overlapping open sets continue one another when they agree on the overlap, allowing a chain of patches along a path. The identity theorem for holomorphic functions forces agreement throughout a connected overlap once they agree on a nonempty open set in it. It also makes an extension to a fixed connected domain unique. ^[ref]

The geometric series illustrates why a formula can reach farther than the power series that first represents it. For $z\ne 1$, its partial sum isWhen $|z|<1$, $z^{N+1}\to 0$, so the series converges to $f(z)=1/(1-z)$. The rational function $F(z)=1/(1-z)$ is holomorphic on $\mathbb{C}\setminus \{1\}$ and agrees there with the sum on the unit disc. It therefore continues $f$ to every point except its simple pole at $1$. In particular,Thus analytic continuation gives a value at $2$ without making the original geometric series converge there. Its radius of convergence remains $1$: on $|z|\ge 1$ the terms $z^n$ do not even tend to zero.

The pole at $1$ limits the original Taylor series, although most points of its convergence circle are regular for $F$. Recentring at $-1$ gives another genuinely convergent power series,At $z=-3/2$ this yields $\frac{1}{2}{\sum }_{n\ge 0}(-1/4{)}^n=2/5$, even though the original powers of $-3/2$ grow. This is the useful idea: overlapping local descriptions can represent one holomorphic function on a larger region.

Uniqueness on a connected overlap does not imply that every path leads to the same terminal germ of a holomorphic function. Continuing a complex logarithm around $0$ changes its value by $2\pi i$, giving monodromy; the return patch can carry a different branch of a multivalued function. If two patches have a disconnected overlap, agreement on one component imposes no agreement on the others. ^[ref] Continuation using the rational function $1/(1-z)$ is single-valued: any path avoiding $1$ gives the same terminal germ of a holomorphic function, even if it winds around the pole.

If $G$ is the principal complex inverse sine, all values reached by analytic continuation areThis follows from the monodromy reflections about the branch values $-\pi /2$ and $\pi /2$.

A branch point is a point around which a sufficiently small loop can change a germ of a holomorphic function under analytic continuation. This is a local condition: a large loop around a regular point might also enclose a distant branch point, so that large loop alone cannot classify the regular point. For the complex logarithm and square root, every nonzero point has a local holomorphic branch, but a loop once around $0$ changes the chosen branch. ^[ref]

For $z=r{e}^{it}$ with fixed $r>0$, start at $t=0$ with values $\sqrt{r}$ and $\log r$. Their continuations areOne counterclockwise turn gives $w(2\pi )=-\sqrt{r}$ and $L(2\pi )=\log r+2\pi i$. At $r=4$, the square root therefore changes from $2$ to $-2$, while the logarithm changes from $\log 4$ to $\log 4+2\pi i$. Two turns restore the square root to $2$; no nonzero number of turns restores the logarithm. This distinguishes a two-sheeted algebraic branch point from a logarithmic singularity.

A pole need not be a branch point: $1/z$ is unbounded near $0$, yet $1/(r{e}^{it})=r^{-1}{e}^{-it}$ returns to its starting value after one turn. Conversely, the square-root values tend to $0$ at their branch point. The issue is whether a single holomorphic choice survives going around the point, rather than whether the values become large. On the Riemann sphere, infinity is also a branch point of both $\sqrt{z}$ and $\log z$: in the local coordinate $\zeta =1/z$, these are ${\zeta }^{-1/2}$ and $-\log \zeta$, whose continuations around $\zeta =0$ change sign and subtract $2\pi i$, respectively. ^[ref]

For $c=1-a-b$ with positive real parts $\mathrm{Re}a,\mathrm{Re}b,\mathrm{Re}c$, the integral converges at its two finite singularities and at infinity, and equalsTo prove it, use Schwinger parameterization with exponents $1-a$ and $1-b$. Integrating the resulting planar Gaussian integral gives $\pi /(\lambda +\rho )$ times $\exp [-\lambda \rho /(\lambda +\rho )]$. Set $q=\lambda +\rho$ and $x=\lambda /q$; the $q$ integral is a gamma function and the $x$ integral is $B(a,b)$, giving the stated ratio. This domain justifies the interchanges; elsewhere the answer is interpreted by analytic continuation. The string measure $d^{2}z=2dxdy$ multiplies the answer by two.

A first-order ordinary differential equation with complex independent and dependent variables is solved by a holomorphic function on an ordinary domain, and its analytic continuation can have branch points. A holomorphic right-hand side gives local existence and uniqueness by the integral contraction mapping used in the Picard-Lindelöf theorem.

The dilogarithm is the order-two polylogarithm, defined for $|z|<1$ by ${\mathrm{Li}}_2(z)={\sum }_{n\ge 1}{z}^n/n^2$. Its derivative is $-\mathrm{Log}(1-z)/z$, with a removable value at zero; a chosen complex logarithm branch gives its analytic continuation.

For a parameter $k$, one form of the incomplete elliptic integral of the first kind isIts value depends on the choices of square-root branches and, under analytic continuation around the branch points, on the integration path.

For $\Re c>\Re b>0$ and a fixed branch away from the cut,Expand $(1-zs{)}^{-a}$ for $|z|<1$ and integrate termwise using the beta function to recover the Gauss hypergeometric function series. Analytic continuation extends the identity in $z$.

The monodromy theorem gives equality of continuations that exist under its hypothesis. It does not assert that every holomorphic function admits analytic continuation everywhere, even within a simply connected domain.

For an explicit obstruction, take $f(z)=1/(1-z)$ near $a=0$ and attempt to apply the theorem on $D=\mathbb{C}$. The path $\eta (t)=t$ ends at $1$, where no terminal germ of a holomorphic function exists. If a holomorphic representative $G$ near $1$ continued $f$, then the identity theorem would extend $(1-z)G(z)=1$ to that neighbourhood. Substituting $z=1$ would give $0=1$. Thus the pole prevents the existence hypothesis, although $\mathbb{C}$ is a simply connected domain. The analytic continuation article explains why the same rational function does continue throughout $\mathbb{C}\setminus \{1\}$.

Existence at an endpoint by some route is also weaker than existence along every path from the base point. Along the real segment from $1$ towards $0$, a holomorphic square root of $z$ tends to $0$, yet cannot continue holomorphically through $0$: a local representative $G$ would satisfy $G^2=z$ by the identity theorem, and differentiating at $0$ would give $2G(0)G^{\prime }(0)=1$ with $G(0)=0$. A finite limiting value therefore does not remove a branch point. More severe obstructions appear at a natural boundary of a holomorphic function.

For a homogeneous collisionless stellar model with its mean field subtracted, the response to an imposed Newtonian gravitational potential is ${\tilde{\Phi }}_{\mathbf{k}}^{\mathrm{tot}}={\tilde{\Phi }}_{\mathbf{k}}^{\mathrm{ext}}/{ϵ}_{\mathbf{k}}$, whereThe phase-space distribution function $f_0$ is mass-normalized, and $\mathbf{k}\ne 0$. Causal analytic continuation uses the Landau contour wherever the equilibrium permits it. Zeros describe collective Landau modes. This response function is a gravitational analogue of the electrostatic dielectric function; its susceptibility sign reflects attraction.

For $r=n+1/2$, $n\ge 0$, the convergent frequency zeta function, expressed through the Riemann zeta function, is ${\sum }_r{r}^{-s}=(2^s-1){\zeta }_R(s)$ when $\mathrm{Re}s>1$. Its analytic continuation at $s=-1$ gives $(1/2-1)(-1/12)=1/24$. A common exponential frequency cutoff gives ${\sum }_{r}r{e}^{-\epsilon r}={\epsilon }^{-2}+1/24+O({\epsilon }^2)$, so the same finite part follows by subtracting the leading divergence. In contrast, integer frequencies have regulated sum $-1/12$.

Put $d=c-a-b\notin \mathbb{Z}$ and choose a branch of $(1-z{)}^d$. The solution regular at zero isThe two characteristic exponents at a regular singular point are $0,d$. The Euler integral for the hypergeometric function determines the first coefficient by a finite endpoint limit and the second by scaling the singular endpoint; analytic continuation extends the parameter ranges.

The Gauss hypergeometric function is the normalized analytic solution at zero,initially for $|z|<1$ and $c\notin \{0,-1,-2,\dots \}$, followed by analytic continuation where possible.

The ordinary hypergeometric function is ${}_2{F}_1(a,b;c;z)={\sum }_{n\ge 0}(a{)}_n(b{)}_n{z}^n/((c{)}_{n}n!)$ near zero for admissible $c$. It solves the Gauss hypergeometric equation and extends by analytic continuation.

The inverse hyperbolic functions invert the hyperbolic functions, with real branches on their standard ranges and complex branches obtained through analytic continuation. For example, $\mathrm{arsinh}z=\log (z+\sqrt{z^2+1})$ on a chosen branch; the multivalued inverse hyperbolic sine records all inverse values.

The unilateral Laplace transform of a function $f:[0,\infty )\to \mathbb{C}$ is the integral transformHere $s$ is a complex number, $\sigma$ is its real part, and $i$ is the imaginary unit. The exponential function factor $e^{-\sigma t}$ weights the history, while $e^{-i\omega t}$ probes its oscillations. The integral starts at zero, so it uses only the history from that starting time onward. This section concerns ordinary functions; their value at the single endpoint does not change the integral. See MIT 18.03SC, definition and elementary examples.

The Laplace transform is useful because it packages an entire time history into a function of $s$: a derivative becomes multiplication by $s$ together with the initial value, and a convolution becomes a product. Thus a linear differential equation can become an algebraic equation that already remembers its initial value problem. The worked example below shows both descriptions of the same response.

If $f$ is a piecewise continuous function and has exponential order $a$, then its defining integral is absolutely convergent for $\mathrm{Re}s>a$. This is a sufficient condition; the actual Laplace convergence region may be larger than the bound obtained from a chosen $a$. ^[ref].

The bound has a short explanation. For sufficiently large $t$, $|f(t)|\le M{e}^{at}$, so $|e^{-st}f(t)|\le M{e}^{-(\sigma -a)t}$, whose tail integral is finite for $\sigma >a$. The finite initial interval is covered by piecewise continuity.

Three calculations make the convergence condition concrete. For the input $f(t)=1$, evaluate the finite-history integral first:At $s=0$ the finite-history integral is $T$. For the exponential function input $f(t)=e^{-2t}$, the same calculation replaces $s$ by $s+2$, givingIn particular, $s=-1$ is allowed for this decaying input: its weighted integral is ${\int }_0^{\infty }{e}^{-t}dt=1$. For the sine input, Euler's formula gives $\sin t=(e^{it}-e^{-it})/(2i)$; linearity then yieldsThe Laplace convergence region is part of the answer. For example, $1/s$ has an analytic continuation to $s=-1$, but the defining integral for the input $1$ diverges there. The sine calculation is also developed in MIT 18.03SC, further elementary transforms.

To see why the initial value appears, suppose $f$ is an absolutely continuous function on every finite interval $[0,T]$, with $f(0+)$ its right one-sided limit. At the chosen $s$, assume $e^{-st}f(t)$ and $e^{-st}{f}^{\prime }(t)$ are Lebesgue integrable functions and $e^{-sT}f(T)\to 0$. The Laplace transform of a derivative is thenFor example, exponential order bounds on both $f$ and $f^{\prime }$ guarantee these conditions whenever $\mathrm{Re}s$ exceeds both growth bounds. The initial value is the lower-end boundary contribution of integration by parts.

Consider the initial value problemWrite $Y=\mathcal{L}\{y\}$. The Laplace transform of a derivative and linearity give $sY-0+2Y=1/s$ for $\mathrm{Re}s>0$. A partial fraction decomposition therefore givesWe recognize the two elementary Laplace transforms already computed. Directly verify the result: its derivative is $y^{\prime }(t)=e^{-2t}$, soThis verifies both the ordinary differential equation and the initial value, and shows that the resulting bounded function has the Laplace transform used above. The response rises from zero toward $1/2$: at that value the input $1$ balances the term $2y$. The general method is described in MIT 18.03SC, solving initial value problems.

The same solution is a convolution. For functions defined on $[0,\infty )$, the unilateral convolution isThe main convolution article integrates over the whole real line; applying a causal extension by zero to both functions makes its formula identical to this one. Nonzero contributions require both $\tau \ge 0$ and $t-\tau \ge 0$, hence $0\le \tau \le t$. If $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ are Lebesgue integrable functions on $[0,\infty )$, the convolution theorem for Laplace transforms statesFor piecewise continuous functions of exponential order $a$ and $b$, it holds throughout $\mathrm{Re}s>\max (a,b)$. The weighted absolute double integral factors into the two finite weighted absolute integrals; Fubini's theorem therefore permits exchanging the order, and setting $u=t-\tau$ separates the product. ^[ref].

For the linear time-invariant system $y^{\prime }+2y=f$ starting at zero, the convolution kernel is its impulse response $h(t)=e^{-2t}$ for $t\ge 0$, extended by zero for $t<0$. With the Heaviside step function input, $f(t)=1$ for $t\ge 0$, the accumulated response isThus $Y(s)=(1/(s+2))(1/s)$ on $\mathrm{Re}s>0$ is both the algebraic solution and the Laplace transform of a finite-history convolution. Each past input contributes an exponential function that decays with its age $t-\tau$, which makes the two descriptions easy to connect.

Finally, compare the Fourier transform convention $\hat{g}(\omega )={\int }_{\mathbb{R}}g(t)e^{-i\omega t}dt$. If $f_0$ is the causal extension by zero of $f$, define $g_{\sigma }(t)=e^{-\sigma t}{f}_0(t)$. Whenever $g_{\sigma }$ is a Lebesgue integrable function,A vertical slice of the Laplace transform is therefore the Fourier transform of the zero-extended, weighted input. Positive $\sigma$ damps its positive-time history. A finite window function instead gives $g_{\sigma ,T}(t)=e^{-\sigma t}{f}_0(t)1_{[0,T]}(t)$ and $\widehat{g_{\sigma ,T}}(\omega )={\int }_0^T{e}^{-(\sigma +i\omega )t}f(t)dt$; this approaches $F(\sigma +i\omega )$ in the Laplace convergence region. Setting $\sigma =0$ directly requires an integrable zero-extended input: the input $1$ fails that condition, even though its Laplace transform exists for every $\sigma >0$.

The history of the Laplace transform spans several stages. Schiff's historical summary notes that Laplace's 1812 treatise acknowledged earlier work on integral transforms and describes later development of the modern Laplace transform method. ^[ref].

Along one path, analytic continuation is a chain of function elements on overlapping open discs, each agreeing with the preceding germ of a holomorphic function near the crossing point. The identity theorem makes this chain unique along that fixed path. Equality at a crossing means equality on a neighbourhood, not just equality of a single value.

For the deformation step, fix one intermediate path $t\mapsto H(s_0,t)$ and its continuation. By compactness of $[0,1]$, choose a finite chain of representative open discs and subdivide the parameter so that each subarc fits in its assigned open disc, with consecutive junctions in overlap neighbourhoods where the representatives agree. Joint continuity of $H$ ensures that every sufficiently nearby intermediate path uses that same chain and those same overlaps. The endpoints stay fixed, so the terminal germ of a holomorphic function stays unchanged. Existence along every intermediate path supplies this argument at each $s_0$; local constancy and connectedness of $[0,1]$ give equality at $s=0$ and $s=1$. This proves the monodromy theorem without assuming that every analytic germ projection is a covering map.

Start at $a=1$ with complex logarithm germ of a holomorphic function $L(1)=0$ and holomorphic square root germ of a holomorphic function $S(1)=1$. On the punctured plane $D=\mathbb{C}\setminus \{0\}$, both admit analytic continuation along every path. To construct it locally, for any $c\ne 0$ choose a small open disc avoiding $0$, select a holomorphic logarithm there, and select its additive multiple of $2\pi i$ to agree with the arriving germ of a holomorphic function; exponentiating half this logarithm supplies the arriving holomorphic square root. A finite chain of these open discs follows any given path.

Compare two paths from $1$ to $i$, parametrized by $0\le t\le 1$:The first follows a counterclockwise quarter-circle; the second follows a clockwise three-quarter-circle. Following ${\gamma }_A$ and then path reversal of ${\gamma }_B$ makes one positive turn about $0$. Its winding number is $1$, so the two paths cannot be homotopic with endpoints fixed in the punctured plane.

The continued values along the chosen paths areAt their common endpoint,Both complex logarithm values exponentiate to $i$, and both holomorphic square root values square to $i$. These give different terminal germ of a holomorphic function representatives, consistently obtained from the same starting germ of a holomorphic function. The monodromy theorem applies, but does not equate these two non-homotopic paths.

The closed path $\lambda (t)=e^{2\pi it}$ instead returns to $1$, with continued values $L_{\lambda }(t)=2\pi it$ and $S_{\lambda }(t)=e^{\pi it}$. One counterclockwise turn changes the complex logarithm by $2\pi i$ and flips the holomorphic square root from $1$ to $-1$. Hence neither initial germ of a holomorphic function can come from a global holomorphic logarithm or a global holomorphic square root of $z$ on the punctured plane. Repeating a loop of winding number $n$ gives the changes $2\pi in$ and $(-1{)}^n$. ^[ref]

Now restrict to the slit domain $D_{\mathrm{slit}}=\mathbb{C}\setminus (-\infty ,0]$. The principal logarithm and $\exp (\mathrm{Log}z/2)$ realize the same initial germ of a holomorphic function for each expression globally. To see that this is a simply connected domain, polar form identifies this domain with $(0,\infty )\times (-\pi ,\pi )$, a convex set; every loop contracts there. The simply connected corollary of the monodromy theorem now forces equality along every pair of paths with matching endpoints. In particular, ${\gamma }_A$ and the segment $\sigma (t)=1-t+it$ both give $L(i)=\pi i/2$ and $S(i)=(1+i)/\sqrt{2}$. The explicit homotopy $H(s,t)=(1-s){\gamma }_A(t)+s\sigma (t)$ stays in the closed first quadrant away from $0$, hence in $D_{\mathrm{slit}}$.

The other route ${\gamma }_B$ crosses the removed ray at $t=2/3$ and is no longer admissible. In the punctured plane, that crossing is locally regular: analytic continuation passes through $-1$ smoothly, although evaluating the principal logarithm would reset its argument. The branch cut changes the allowed domain; it does not create additional branch points or poles. That distinction is the beauty of the obstruction: perfectly compatible local holomorphic functions can fail to make a global choice solely because of the allowed based loops. The branches of a multivalued function article develops how moving the branch cut changes the selected values.

For a meromorphic function $f$, the displayed telescoping sum identity holds wherever every summand is defined. If the two endpoint terms are finite at $w=w_0$, their difference gives a holomorphic continuation of the entire sum to $w_0$, even when individual summands have poles there. This continued value is a common-shift limit; it does not make an originally undefined sum of complex numbers well defined without an additional convention.

The proof is finite cancellation on the common domain of definition, followed by analytic continuation. For $f(w)=\tanh w$, a common shift away from the imaginary axis avoids its poles. Different independently chosen shifts of individual summands need not give the same limit, so specifying the common shift matters.

Monodromy records how a germ of a holomorphic function changes after analytic continuation around a closed path. The source point returns to itself, but the continued germ of a holomorphic function need not. For $z(t)=e^{it}$, $0\le t\le 2\pi$, starting with holomorphic square root value $1$ gives $w(t)=e^{it/2}$ and hence $w(2\pi )=-1$; starting with complex logarithm value $0$ gives $L(t)=it$ and hence $L(2\pi )=2\pi i$. After $m$ counterclockwise turns the values are $(-1{)}^m$ and $2\pi im$. Thus monodromy exchanges the two square-root branches of a multivalued function, whereas it shifts through infinitely many logarithm branches of a multivalued function. ^[ref]

The three panels use the same parameter $t$. They show analytic continuation through local holomorphic patches, including across a chosen branch cut; repeatedly evaluating the principal logarithm would instead reset its imaginary part at the cut. Monodromy makes the obstruction to a global branch of a multivalued function visible and motivates the use of Riemann surfaces as domains on which continued values stay distinct.

Letwhere $P$ has a simple zero at $a$, and let $A$ be the limiting value of $W$ at $a$ on one sheet. Analytic continuation once around $a$ changes the sign of the square root and hence of $W^{\prime }$. The continued primitive agrees at $a$ with the original one, so it is