# analytic continuation

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##### 1: 6.4 Analytic Continuation
###### §6.4 AnalyticContinuation
Analytic continuation of the principal value of $E_{1}\left(z\right)$ yields a multi-valued function with branch points at $z=0$ and $z=\infty$. …
6.4.7 $\mathrm{g}\left(ze^{\pm\pi i}\right)=\mp\pi ie^{\mp iz}+\mathrm{g}\left(z% \right).$
##### 2: 28.7 Analytic Continuation of Eigenvalues
###### §28.7 AnalyticContinuation of Eigenvalues
28.7.4 $\sum_{n=0}^{\infty}\left(b_{2n+2}\left(q\right)-(2n+2)^{2}\right)=0.$
##### 6: 15.17 Mathematical Applications
By considering, as a group, all analytic transformations of a basis of solutions under analytic continuation around all paths on the Riemann sheet, we obtain the monodromy group. …
##### 7: 16.15 Integral Representations and Integrals
These representations can be used to derive analytic continuations of the Appell functions, including convergent series expansions for large $x$, large $y$, or both. …
##### 9: 1.10 Functions of a Complex Variable
###### §1.10(ii) AnalyticContinuation
If $f_{2}(z)$, analytic in $D_{2}$, equals $f_{1}(z)$ on an arc in $D=D_{1}\cap D_{2}$, or on just an infinite number of points with a limit point in $D$, then they are equal throughout $D$ and $f_{2}(z)$ is called an analytic continuation of $f_{1}(z)$. … Analytic continuation is a powerful aid in establishing transformations or functional equations for complex variables, because it enables the problem to be reduced to: (a) deriving the transformation (or functional equation) with real variables; followed by (b) finding the domain on which the transformed function is analytic.
###### Schwarz Reflection Principle
Then the value of $F(z)$ at any other point is obtained by analytic continuation. …
##### 10: 8.15 Sums
8.15.2 $a\sum_{k=1}^{\infty}\left(\frac{{\mathrm{e}}^{2\pi\mathrm{i}k(z+h)}}{\left(2% \pi\mathrm{i}k\right)^{a+1}}\Gamma\left(a,2\pi\mathrm{i}kz\right)+\frac{{% \mathrm{e}}^{-2\pi\mathrm{i}k(z+h)}}{\left(-2\pi\mathrm{i}k\right)^{a+1}}% \Gamma\left(a,-2\pi\mathrm{i}kz\right)\right)=\zeta\left(-a,z+h\right)+\frac{z% ^{a+1}}{a+1}+\left(h-\tfrac{1}{2}\right)z^{a},$ $h\in[0,1]$.