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For $1<x<2$ the specified value is $g(x)=i\sqrt{(2-x)(x^2-1)}$. Continue towards $0$ around the branch point $1$. Along the upper bank, the argument of $(z-2)(z^2-1)$ increases from $\pi$ to $2\pi$, so its square root reaches the negative real value; along the lower bank it decreases from $\pi$ to $0$, giving the positive value. Since the radicand at $0$ is $2$,

Choose the square root with $-\pi /2<\arg z<3\pi /2$, so the branch cut is the negative imaginary axis. The change of variable $u=e^{-i\pi /4}\sqrt{z}$ maps this sheet onto $\mathrm{Re}u>0$ and gives $z=i{u}^2$. Writing the exponential function as $e^{k\varphi (z)}$, its phase becomes exactly quadratic:The saddle point is therefore $u=1$, or $z_s=i$. On the descending line $u=1+iv$, $v\in \mathbb{R}$, the phase is $-1-v^2$. Its image is the parabolaThe contour runs from left to right, corresponding to $v$ decreasing from $+\infty$ to $-\infty$. A contour deformation in the right half of the $u$-plane moves the original indented contour to this line. The connecting tails vanish in the descending sectors, and no branch point or pole is crossed. Since $dz=-2(1+iv)dv$, the Gaussian integral and the odd function $v{e}^{-k{v}^2}$ giveHere the method of steepest descent actually gives an exact answer for $k>0$, because the quadratic phase and linear transformed amplitude have no further even correction.

Because $\mathrm{Re}a>0$, the endpoint factor $(1+t^2{)}^a$ vanishes at the branch points $t=\pm i$. A finite contour may therefore run from $-i$ to $i$ on either side of a chosen branch cut; the diagram shows contours bowing through the left and right half-planes.

For $a=1$, $f(t)=1$ is entire, so path independence makes either contour give, up to orientation,Hence a nonzero constant multiple gives the solution

A branch point of a multivalued analytic function is a point around which analytic continuation along a closed loop can return a different value. A branch cut is a curve or union of curves removed from the domain so that loops causing this monodromy are excluded and one single-valued branch can be chosen.

On taking $z$ once anticlockwise around zero, $\theta$ increases by $2\pi$ and $w$ changes by $2\pi i$; hence zero is a branch point. Under the coordinate $\zeta =1/z$, a loop around $\zeta =0$ corresponds to a loop around infinity and produces the opposite increment, so infinity is also a branch point. Removing any curve joining the two branch points prevents such loops. With $0<\arg z<2\pi$, one possible branch cut is therefore

PutThis slit complex plane is a simply connected domain. On $D$ there is one branch of $\sqrt{1-z^2}$ that equals $1$ at zero, and its reciprocal is holomorphic. The primitive of a holomorphic function on a simply connected domain therefore makes the integral from zero path-independent, so $G$ is single-valued.

For unrestricted paths, going once around either branch point changes the sign of the square root. The resulting analytic continuation therefore need not return to the original germ, and the value of $g(z)$ can depend on the path.

A branch point is a point around which analytic continuation changes a value; a branch cut removes curves so continuation becomes single-valued, and a branch is one such single-valued analytic choice. Defineon $\mathbb{C}\setminus ((-\infty ,-1]\cup [1,\infty ))$, choosing the square root equal to $1$ at zero. Monodromy around $\pm 1$ explains multivaluedness.

On the upper lip $0+$ of the cut convention, the branch remains positive on $(0,1)$, so ${\int }_{0+}^{1}dt/\sqrt{1-t^2}=\pi /2$. Define $\mathrm{Arccos}z=\pi /2-\mathrm{Arcsin}z$ on the same slit domain.

AlsoChanging logarithm branches changes the value by multiples of $\pi$ (already visible at $z=1$). Principal logarithms give a single branch on the plane cut from $\pm i$ outward. Differentiating both sides and checking the value at zero proveswhere the compatible principal branches are defined.

The laplace transform is $\hat{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t),dt$, and inversion iswhere the vertical Bromwich contour lies to the right of all singularities. For $t>0$ it is closed leftwards, and for $t<0$ rightwards, where the exponential decays.

Transforming the PDE and imposing boundedness as $x\to \infty$ giveswith the square root cut conventionally along $(-\infty ,0]$. ThereforeThere are poles at $p=0,-1$ (with cancellation assessed in the complete expression) and a branch point at $0$; close left for $t>0$ and wrap the cut while including the relevant residues.