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An M/M/1 queue has a Poisson arrival process of rate $\lambda$, independent exponential service times of rate $\mu$, and one server, with an unlimited waiting room and the usual first-come-first-served discipline. Its number-in-system process is a continuous-time birth-death chain with $q_{j,j+1}=\lambda$ and $q_{j,j-1}=\mu$ for $j\ge 1$. For $\rho =\lambda /\mu <1$, detailed balance givesThis sums to one and satisfies ${\pi }_j\lambda ={\pi }_{j+1}\mu$. The nonexplosive irreducible chain therefore has an invariant probability and is positive recurrent. Equivalently its negative drift above zero gives finite mean return cycles. Here queue length includes the customer in service.

In equilibrium, PASTA gives an arriving customer's number of predecessors the distribution $\pi$. Conditional on $j$ predecessors, memoryless property makes its total waiting-plus-service time a sum of $j+1$ independent $\mathrm{Exp}(\mu )$ variables. Its Laplace transform isThereforeThis is the sojourn time including service, not just the time until service begins.

The Laplace transform isUsing the supplied Gaussian integral after $t=x^2$ gives

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

Taking the Laplace transform in $t$ and using the initial condition $u(r,0)=u_0$, the radial heat equation in three dimensions becomesPut $W=rU$ and $q=\sqrt{s}$. ThensoFiniteness at the centre forces $B(s)=0$.

The transformed boundary condition isSincesubstitution and cancellation of the two $u_0/s$ terms givesHenceand the required explicit transform is

A frailty random variable is an unobserved positive multiplicative risk factor. A proportional frailty model hasThe scale of $U$ is not separately identifiable from $h_0$: multiplying $U$ by a constant and dividing $h_0$ by it leaves the model unchanged. We may therefore normalize $\mathbb{E}U=1$, which lets $h_0$ represent the mean initial hazard multiplier and makes relative frailty interpretable.

If $U\sim \mathrm{Exponential}(1)$ and $h_0(t)=\lambda$, then the Laplace transform of $U$ givesThus $h(0)=\lambda$, while $h(t)\to 0$ as $t\to \infty$. Survivors become increasingly enriched for low-frailty individuals.

Apply the time Laplace transform to the Airy equation. Extend $g$ past $T$ in any suitable way and writeFor example, setting $g=0$ after $T$ makes $G(p)={\int }_0^T{e}^{-pt}g(t)dt$; the final solution for $t<T$ is independent of this extension. The Laplace transform of a derivative givesTake the principal cube root $\lambda =p^{1/3}$ for $\mathrm{Re}p>0$. The three characteristic roots of the spatial ordinary differential equation areSince $|\arg \lambda |<\pi /6$, only $-\lambda$ has negative real part. Thus spatial decay leaves just one homogeneous exponential, which the single prescribed Neumann boundary condition determines.

The Airy resolvent kernel on the whole real line isIt decays at both ends, is continuous together with its first derivative, and satisfies $R_p^{\prime \prime }(0+)-R_p^{\prime \prime }(0-)=1$. Consequently $({\partial }_s^3+p)R_p={\delta }_0$ as a distribution. A particular solution is the Green-function representationAdding the decaying homogeneous mode to impose the boundary derivative givesEvery term is known. If ${\tilde{u}}_0(r)={\int }_0^{\infty }{e}^{-ry}{u}_0(y)dy$ denotes the spatial Laplace transform, thenThe Bromwich inversion formula now gives the required integral representation:Here $\gamma >0$ is to the right of any singularities required by the growth of the data. The usual decay or growth hypotheses are understood for this Laplace transform construction; when absolute inversion is unavailable, the vertical integral is interpreted as the limit of truncated Bromwich contours. The transformed ordinary differential equation verifies the partial differential equation and initial condition, while differentiating at $x=0$ gives exactly $G(p)$ and hence $u_x(0,t)=g(t)$. The derivative compatibility $u_0^{\prime }(0)=g(0)$ makes the two data agree at the corner.

IfthenA nonnegative random variable with finite expectation is finite almost surely, so the process explodes.

Conversely, suppose the series diverges. For $s>0$, independence and the Laplace transform of an exponential variable giveThe sum in the exponent diverges. If infinitely many ${\lambda }_n\le s$, infinitely many terms are at least $\log 2$; otherwise, eventually $s/{\lambda }_n\le 1$ andThus $\mathbb{E}{e}^{-s{T}_{\infty }}=0$. Since $e^{-s{T}_{\infty }}$ is strictly positive on $\{T_{\infty }<\infty \}$, this forces $T_{\infty }=\infty$ almost surely. Hence the nonexplosion criterion for a pure birth process is

The string is initially at rest, so $y(x,0)=y_t(x,0)=0$. Taking the Laplace transform in $t$ converts the wave equation intoFor $\mathrm{Re}p>0$, the solution that decays as $x\to \infty$ is $\hat{y}=A(p)e^{-px/c}$. The transformed boundary condition $\hat{y}(0,p)=\hat{f}(p)$ fixes $A=\hat{f}$, and therefore

Let $N$ be the number of visits to the server made by one customer. Then $N$ has the geometric distribution with success parameter $\delta$, and its service periods $E_1,E_2,\dots$ are independent exponential random variables of rate $\mu$. For $s\ge 0$, the Laplace transform of the total service time $T=E_1+\cdots +E_N$ isThus

At each service completion, a successful departure reduces the queue length by one and occurs at rate $\delta \mu$ while the queue is nonempty; feedback completions leave its length unchanged. The queue-length process is therefore the birth-death process of an ordinary $M/M/1$ queue with service rate $\delta \mu$. It is positive recurrent exactly when $\lambda <\delta \mu$, and then, with $\rho =\lambda /(\delta \mu )$,

The Laplace transform of a function on $t\ge 0$ isfor values of $s$ for which the integral converges. Integration by parts gives the Laplace transform of a derivative

Write ${\hat{N}}_i(s)=\mathcal{L}\{N_i\}(s)$. Transforming the first two equations and using the initial data givesThe third equation then givesand henceA partial fraction decomposition therefore yields the sequential radioactive decay formula

Now let ${\lambda }_1={\lambda }_2=\lambda$ and put $\Delta ={\lambda }_3-\lambda >0$. The transformed contribution originating from the initial $N_1$ population becomesEquivalently, solve the middle equation first to obtainand use an integrating factor in the final equation:Evaluating the elementary integral gives

The zero initial condition turns the time Laplace transform of the heat equation intoDecay as $x\to \infty$ selectsThe boundary value givesso

A proportional frailty model specifiesIts population survival function is the Laplace transform of the frailty density,If $m=\mathbb{E}U<\infty$, replacing $U$ by $U/m$ and $h_0$ by $m{h}_0$ leaves their product, and hence the model, unchanged. The normalization $\mathbb{E}U=1$ identifies this otherwise arbitrary scale and makes $h_0(0)$ the initial population hazard when $H_0(0)=0$.

Taking the Laplace transform in $t$ and using the zero initial condition givesThe solution bounded as $x\to \infty$ isThe transformed boundary condition isso this is the sinusoidally forced heat equation on a half-line. Integrating in $x$ gives

Use the principal square root, with a branch cut along the negative real axis. The Bromwich contour is a vertical line to the right of $p=\pm i\omega$ and the branch point $p=0$. For $t>0$, close it in the left half-plane, indent around the cut, and let the large semicircle recede. The simple poles at $p=\pm i\omega$ contributeHence

On the upper and lower sides of the cut, put $p=-s$; the two values of $\sqrt{p}$ are $i\sqrt{s}$ and $-i\sqrt{s}$. Their jump and the opposite contour orientations combine to giveThis is precisely Bromwich inversion with a square-root branch cut: the residues produce the permanent periodic response, while the cut integral is the transient that decays with time.

Taking the Laplace transform of the differential equation and usinggivesHenceWhenever $s$ is not an eigenvalue of $A$, the matrix $sI-A$ is invertible, and thusThis is the Laplace-transform solution of a constant-coefficient vector ODE.

Set $g=0$. The given homogeneous solution is $y(t)=e^{tA}{y}_0$, so the preceding formula saysfor every initial vector $y_0$. Equality on every vector gives the matrix identityon the common domain of convergence and in particular away from the eigenvalues of $A$. This is the Laplace transform of a matrix exponential.

Under the Cameron-Martin theorem for a linear drift, the probability that Brownian motion with drift $-\mu$ reaches $x$ isUsing the supplied Laplace transform with $\lambda ={\mu }^2/2$ givesand henceThis is the survival function of the exponential distribution with rate $2\mu$.

For $x,y>0$, the Laplace transform identityexhibits $k_1$ as an inner product of the feature functions $t\mapsto e^{-tx}$ in $L^2(0,\infty )$. More explicitly, for any real $c_i$ and positive $x_i$,Therefore $k_1$ is a positive-semidefinite kernel.

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.

Laplace identities. The one-sided Laplace transform isfor $\mathrm{Re}p$ sufficiently large that the integral converges. For a piecewise smooth function of exponential order, integration by parts givesDifferentiating under the integral givesIteration also yields $\mathcal{L}[f^{\prime \prime }]=p^{2}F-pf(0)-f^{\prime }(0)$. For real $k$ and $p>0$, taking the imaginary part of $(p-ik{)}^{-1}$ gives

Transformed differential equation. Let $G=\mathcal{L}[g]$, using the initial values $g(0)=1,g^{\prime }(0)=0$. ThenSumming the three contributions givesFor $\alpha \ne 0$, the condition $G(p)\to 0$ as $p\to \infty$ fixes the integration constant:It is simpler to invert the derivative: $\mathcal{L}[tg]=-G^{\prime }=1/(p^2+{\alpha }^2)=\mathcal{L}[\sin (\alpha t)/\alpha ]$. ThusThis is the sinc function. Its Taylor series begins $1-{\alpha }^2{t}^2/6+O(t^4)$, verifying both initial values. For a direct differential check set $h=tg=\sin (\alpha t)/\alpha$. Then $h^{\prime \prime }+{\alpha }^{2}h=t{g}^{\prime \prime }+2g^{\prime }+{\alpha }^{2}tg=0$. For $\alpha =0$, take the continuous limit $\boxed{g(t)=1}$, which also follows from $G=1/p$.

For a piecewise continuous function of exponential order on $t\ge 0$, its unilateral Laplace transform isThe Bromwich inversion formula, at a point of continuity, isThe Bromwich contour is a straight vertical line, traversed upwards, with $\gamma$ to the right of the abscissa of convergence and hence of all singularities of the transform. Under the usual piecewise smoothness assumptions, at a jump the symmetric inversion limit gives the mean of the two one-sided values.

In the transform in the question, put $\lambda =e^{-t}$. Then $d\lambda =-e^{-t}dt$ andThus the restricted Mellin transform on the unit interval is the Laplace transform of $f(t)=h(e^{-t})$. Apply the Bromwich inversion formula with $t=-\log \lambda$, where $0<\lambda <1$, to obtainHere $\gamma$ lies in the convergence half-plane of $H$. At a discontinuity this formula has the same mean-value interpretation. The real natural logarithm defines ${\lambda }^{-s}$ unambiguously.

The Laplace transform is $\overline{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t)dt$ wherever it converges. For sufficiently large $\mathrm{Re}p$, suitable exponential bounds justify integration by parts and differentiation under the integral sign, givingFor a nonnegative integer $n$ and real $p>0$, substitute $u=pt$ in the gamma integral. The result extends holomorphically to $\mathrm{Re}p>0$, givingThis also follows recursively from the second identity, starting with $\mathcal{L}[1]=1/p$.

Put $G=\overline{g}$, $g_0=g(0^{+})$ and $g_1=g^{\prime }(0^{+})$. For a solution with finite initial values and the required transform bounds, the transformed equation isThe initial-value terms cancel, leaving $(p-p^2)G^{\prime }+(3-p)G=0$. On the real interval $p>1$, a nonzero solution satisfiesIntegration gives $G=C(p-1{)}^2/p^3=C(1/p-2/p^2+1/p^3)$. Inverting term by term yieldsTaking $C=1$ gives a nonzero solution, verified by direct substitution. It is $L_2(t)$, and the rational transform agrees with the general Laplace transform of a Laguerre polynomial. No claim about solutions with singular initial behavior is needed to produce this regular solution.