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A positive, exponentially increasing hazard function $h(t)=q{e}^{\beta t}$ gives the survivor functionIts mean is ${\beta }^{-1}{e}^{q/\beta }{E}_1(q/\beta )$, where $E_1$ is the exponential integral. Unlike the exponential distribution, its future mean waiting time is not the reciprocal of its current hazard function.

Under the frozen-age exponential distribution convention, the mean waiting time is the reciprocal of the outgoing transition intensity. The beliefs therefore specifyBoth rates double over ten years. ConsequentlyThese calibrate the age-specific constant-rate means; they are not exact mean ages of future events when age-dependent rates continue increasing throughout the waiting time.

For a literal continuously ageing interpretation, a rate $q{e}^{\beta u}$ gives a Gompertz distribution for the waiting time. For $\beta >0$, its mean isHere $E_1$ is the exponential integral. Solving $m(a,{\beta }_{12})=10$ and $m(a{e}^{10{\beta }_{12}},{\beta }_{12})=5$, and the analogous pair with means 2 and 1, instead giveswith rates and slopes in inverse years. These are an alternative continuous-age calibration, not the frozen-age answer boxed above.

At fixed $r$, the leading equation is $(r^2{f}_0^{\prime }{)}^{\prime }=0$. Its solution compatible with the eventual decaying far field and the boundary value at one is $f_0=1/r$. This decay implies $f=O(ϵ)$ at distances $r=O({ϵ}^{-1})$. Balancing radial derivatives against the linear screening term identifiesThe distant region is governed by the modified Helmholtz equation; its decaying homogeneous profile is $e^{-x}/x$.

For a matched asymptotic expansion, write the fixed-$r$ approximation as $f=f_0+ϵf_1+{ϵ}^2{f}_2+\cdots$, allowing logarithms of $ϵ$ in the coefficients. Successive equations areThe boundary condition imposes $f_0(1)=1$ and $f_1(1)=f_2(1)=0$. Before matching, their integrated forms can be writtenIn particular a pure power series with parameter-independent coefficients will be insufficient: the logarithmic overlap creates a switchback term.

In the distant region set $f=ϵF_0(x)+{ϵ}^2{F}_1(x)+\cdots$. The scaled equation is $f_{xx}+2f_x/x-f=x{f}^3/ϵ$, soDecay and leading matching give $F_0=e^{-x}/x$. The radial modified Helmholtz equation gives the supplied particular integral in terms of the exponential integral:As $x\downarrow 0$, the small-argument expansion of the exponential integral yieldswhere $\gamma$ is the Euler--Mascheroni constant. Substitute $x=ϵr$ to compare the two expansions in $1\ll r\ll {ϵ}^{-1}$:The constant at order $ϵ$ fixes $C_1=-1$. Its $1/r$ coefficient then fixes $\alpha +\frac{1}{2}\log 2=1$, and the constant at order ${ϵ}^2$ fixes $D$. HenceThe required inner expansion at fixed $r$ isAt fixed positive $x=ϵr$, the outer expansion isBoth display every term through the requested order, including the ${ϵ}^2\log ϵ$ switchback term in the fixed-$r$ region.

To form an additive composite expansion, subtract the common overlap from the sum of the inner and outer expressions. Their retained common part isThus one composite is $ϵF_0(ϵr)+{ϵ}^2{F}_1(ϵr)-{ϵ}^2(D+1/2)/r$. The last term can be screened by multiplying it by $e^{-ϵr}$ without changing either retained expansion. This gives a useful exponentially decaying version:Its boundary value is $1+o({ϵ}^2)$. If exact satisfaction of the boundary value is desired, use insteadThe small-argument expansion of the exponential integral shows that this normalized composite has the same two retained expansions; it equals one at $r=1$ and tends to zero at infinity.

As $x\downarrow 0$, the exponential integral haswhere $\gamma$ is the Euler--Mascheroni constant. Integrating the Taylor series of $E_1^{\prime }(x)=-e^{-x}/x$ gives every nonconstant coefficient; the constant follows from the limiting definition of $\gamma$. A logarithm of a stretched variable can create a switchback term in a matched asymptotic expansion.