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For real $z>0$, defineThe integrand is positive and at most $e^{-t}$, so $0<F(z)<1$ and the improper integral converges. Substituting $s=z+t$ also gives $F(z)=z{e}^z{E}_1(z)$ in terms of the exponential integral; the original representation makes positivity and the large-$z$ behavior transparent.

Apply the exact finite geometric series identity inside the integral, without assuming that an infinite series can be integrated term by term:The Gamma integral gives ${\int }_0^{\infty }{e}^{-t}{t}^{k}dt=k!$. Therefore, with $N$ retained terms,This remainder formula proves the asymptotic expansion at every fixed order: $R_N=O(z^{-N})=o(z^{-N+1})$ as $z\to +\infty$. Its sign also gives alternating bounds: an odd $N$ overestimates $F(z)$, and an even $N$ underestimates it.

At $z=10$, an independent evaluation of the integral gives $F(10)\approx 0.9156333394$. Three retained terms giveFive retained terms give $S_5(10)=0.9164$ with error bounded by $5!/10^5=0.0012$. The actual errors are about $0.004367$ and $0.000767$, respectively. The error bound is conservative, but established without relying on the numerical integral evaluation.

Nevertheless, the formal infinite series diverges at every fixed $z>0$. The magnitudes $T_k=k!/z^k$ satisfy $T_{k+1}/T_k=(k+1)/z$, so they eventually increase without bound. In particular, the terms fail the necessary zero-term condition for a convergent series. The finite remainder identity remains valid after convergence has failed; this is precisely what makes the example useful.

For the planar heat equation with point injection of strength $p(s)$, zero initial data gives $\Psi (t,\mathbf{r})={\int }_0^{t}p(s)e^{-r^2/[4D(t-s)]}/[4\pi D(t-s)]ds$. If injection is $p_0$ until $t_0$ and zero later, then for $t>t_0$ this equals $p_0[\mathrm{Ei}(-b/(t-t_0))-\mathrm{Ei}(-b/t)]/(4\pi D)$, where $b=r^2/(4D)$. The exponential integral expansion gives $p_0\ln [t/(t-t_0)]/(4\pi D)$ when $b/(t-t_0)\ll 1$.

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.

Multiply the equation by $x^2$. Its left side is an exact derivative, yielding the first integralFor a solution continuous at the singular endpoint, the limiting value is $C=-2{ϵ}^3\gamma$. This can also be established from the exact integrating-factor formula below, without assuming uniform bounds on derivatives near zero.

Away from zero, the outer expansion has leading term $y\sim C/x^2$. The endpoint value at one requires $C\sim {ϵ}^3$ if an exponentially amplified homogeneous term is to be avoided. It follows already thatTo exhibit the matching and explain the required boundedness, first set $x=ϵX$, $y=ϵY(X)$ for a representative without an order-one homogeneous contribution. The leading inner expansion obeysIntegration giveswhere $\mathrm{Ei}$ is the exponential integral. For $X\to \infty$, $Y\sim X^{-2}$, matching the outer term ${ϵ}^3/x^2$. For $X\to 0$, $Y\sim -1/X$, so the intermediate solution behaves as $-{ϵ}^2/x$ and reaches order one on a second scale $x={ϵ}^2\xi$.

On this second scale, the first integral becomes, at leading order, ${\xi }^2{Y}_0^{\prime }-2Y_0=1$. Equivalently its differentiated equation is $Y_0^{\prime \prime }+(2/\xi -2/{\xi }^2)Y_0^{\prime }=0$, as in the hint. The matching solution that tends to zero as $\xi \to \infty$ isIt has $Y_0\sim -1/\xi$ at large $\xi$, matching the intermediate region, and $Y_0(0)=-1/2$. Thus the nested layers at an essential singular endpoint connect an order-${ϵ}^3$ outer solution to an order-one endpoint value without an unbounded overshoot.

The issue of uniqueness is clarified by exact integration. PutFor every fixed $ϵ>0$ and every prescribed exact value of $\gamma$, the solution satisfying the right endpoint isAs $x\to 0$, the integral is asymptotic to $\mu (x)/(2{ϵ}^2)$, proving $y(0)=\gamma$. Since $h(1)>0$, the two exact endpoint values fix the two constants: for a prescribed exact $\gamma$, this finite-$ϵ$ boundary-value problem is unique.

However, specifying only the leading value $\gamma \sim -1/2$ and uniform order-one size does not select a unique asymptotic family. To see this constructively, choose $x_m=\sqrt{2}{ϵ}^{3/2}$ and an arbitrary bounded real constant $A$. WriteThe right endpoint fixesEndpoint Laplace's method gives $I=ϵ\mu (1)[1+2ϵ+O({ϵ}^2)]$. Therefore every bounded choice of $A$ haswith changes in $A$ affecting $\gamma$ only by $O({ϵ}^{-3}{e}^{-1/ϵ})$. These solutions are uniformly order one: near zero their particular part has the finite limit $\gamma$, near one it follows the small outer solution, and the freely chosen value at $x_m$ is $Ah(x_m)=O(1)$. The inner-layer homogeneous contribution is also bounded. Indeed $h$ vanishes at zero, is order one near $x_m$, and is exponentially small at one.

Consequently the order-one selected value is $-1/2$, but algebraic asymptotic endpoint data leave an order-one interior freedom. This is beyond-all-orders boundary-value sensitivity. If “constant” were interpreted as fixing $\gamma =-1/2$ exactly and independently of $ϵ$, the omitted algebraic corrections would excite an exponentially large interior homogeneous term; a genuinely uniformly bounded exact family requires the higher-order tuning of $\gamma$. Constancy in $x$ does not preclude that parameter dependence.

Retaining the nonlinear factor $f$ determines the order-$ϵ$ forcing and the matching constants. Write $\delta =\sqrt{ϵ}$. The inner leading equation and its boundary condition at $r=1$ give $f_0=C(1-r^{-1/2})$. Matching to the far-field value fixes $C=1$.

The tail $-r^{-1/2}$ becomes $-\delta x^{-1/2}$ under $x=ϵr$, so a regular expansion in integer powers of $ϵ$ cannot suffice. This is a matched expansion with square-root and logarithmic corrections. DefineThe exponential integral here is normalized by the displayed definite integral. Integration by parts also gives $E(x)=2e^{-x}/\sqrt{x}-2\sqrt{\pi }\mathrm{erfc}\sqrt{x}$, using the complementary error function.

For the outer asymptotic expansion, put $F(x)=f(x/ϵ)$ andThe transformed equation gives $L{F}_1=0$ and $L{F}_2=-F_1{F}_1^{\prime }$, with both corrections vanishing at infinity. Matching the leading inner tail determines $F_1=-E/2$. Multiplying the equation for $F_2$ by its integrating factor yieldsFix the particular solution $G$ unambiguously byThen $(x^{3/2}{e}^x{G}^{\prime }{)}^{\prime }=E$, $G(\infty )=0$, and $G$ has no $x^{-1/2}$ term at zero. The general outer correction is $F_2=\beta E+G/4$, whereThe logarithm in this asymptotic expansion explains why the inner expansion needs an $ϵ\log ϵ$ term.

For the inner asymptotic expansion, all corrections larger than order $ϵ$ solve the homogeneous leading equation and vanish at $r=1$. At order $ϵ$, the forcing isIntegration gives the particular solution $1-\sqrt{r}+\log r$, which vanishes at $r=1$. Thus writeExpress this inner expansion in $x=ϵr$ and compare the terms that survive in the overlap $ϵ\ll x\ll 1$. The order-$\delta$ constant gives $A=\sqrt{\pi }$. At order $ϵ$, the term $C\log ϵ+\log r$ becomes $\log x+(C-1)\log ϵ$, so $C=1$. The singular term $-ϵA/\sqrt{x}$ matches $2ϵ\beta /\sqrt{x}$, giving $\beta =-\sqrt{\pi }/2$. Finally, the regular constants give $B+1=\pi +\mu /4$.

The requested expansions, including every correction through order $ϵ$, are thereforeThe inner formula satisfies the boundary condition at $r=1$ term by term; the outer formula satisfies the condition at infinity. Each formula has its own stated region of validity, and the matched asymptotic expansion relates them in their overlap.

The constant can also be made explicit. Integration by parts in the definition of $G$ gives $G=-EI-{\int }_x^{\infty }E(t{)}^{2}dt$, and $E(x)I(x)\to 8$. Squaring the complementary error function expression for $E$ separates the integral into three elementary pieces:The first piece is $-4\log x-4\gamma -4\log 2+o(1)$, where $\gamma$ is the Euler--Mascheroni constant. Substitution $u=\sqrt{t}$ makes the second integral tend to $\sqrt{\pi }/2$, because $(\mathrm{erfc}u{)}^{\prime }=-2e^{-u^2}/\sqrt{\pi }$. Integration by parts in the third integral gives $1/2-1/\pi$. ConsequentlyEquivalently, defining $G$ by its convergent integral already determines the same matching constant without needing this closed form.

The Duhamel principle superposes the response to source impulses:This expression uses the source at the origin, is causal, and has zero initial data in the distributional sense. Its integral over the plane is ${\int }_0^{t}p(s)ds$, consistent with the injected total amount.

For the finite-duration source and $t>t_0$, set $b=(x^2+y^2)/(4D)$. The causal diffusion from a finite-duration planar point source becomesWith $\eta =b/\tau$, or by differentiating the supplied exponential integral, this isIf $b/(t-t_0)\ll 1$, both arguments are small. Their logarithmic constants and $\ln b$ cancel, leavingAt the origin this logarithmic expression is exact for $t>t_0$, obtained as the limit $b\to 0$. The leading spatial correction is $-p_{0}b{t}_0/[4\pi Dt(t-t_0)]$. If also $t\gg t_0$, the leading value is $p_0{t}_0/(4\pi Dt)$.

Set $z=x^2/\epsilon$. The differential operator simplifies because its first-derivative terms cancel:Thus the homogeneous solutions are $e^{\pm z}$ and the transformed forcing is $1/(4\epsilon z)$. For the decaying-sector variation of parameters solution choose the contour directionsThe first integrand decays at positive-real infinity and the second at positive-imaginary infinity. Join each endpoint to $x$ within the first quadrant, avoiding zero. This also fixes the exponentially small homogeneous contributions, which cannot be determined from an algebraic asymptotic expansion alone.

After $s=t^2/\epsilon$, write $A(z)={\int }_{+\infty }^z{e}^{-s}ds/s$ and $B(z)={\int }_{-\infty +i0}^z{e}^{s}ds/s$. Their derivatives are $A^{\prime }=e^{-z}/z$ and $B^{\prime }=e^z/z$. Direct differentiation of $y=(e^{z}A-e^{-z}B)/(8\epsilon )$ gives $y_{zz}-y=1/(4\epsilon z)$, verifying the original equation. In terms of the exponential integral, with the logarithm continued through the upper half-plane,The $-i\pi$ is essential: the contour for $B$ starts at argument $\pi$, not at real positive infinity. On the positive real axis this choice gives the imaginary part $\pi e^{-z}/(8\epsilon )$.

Repeated endpoint integration by parts gives $e^z{E}_1(z)\sim z^{-1}(1-z^{-1}+2!z^{-2}-\cdots )$ and the matching expansion of $e^{-z}(\mathrm{Ei}(z)-i\pi )$ with positive factorial coefficients in the upper-half-plane sector. Their odd inverse-power corrections cancel. Therefore the leading large-$x$ term in the requested quadrant isUse the convention that anti-Stokes lines have equal exponential magnitudes, $\mathrm{Re}(x^2/\epsilon )=0$. They are $\arg x=\pi /4+m\pi /2$. The Stokes lines are $\arg x=m\pi /2$. Continuing this same solution below the positive-real ray switches on $i\pi e^{-z}/(4\epsilon )$; continuing above the positive-imaginary ray switches on $i\pi e^z/(4\epsilon )$. These exponentials remain decaying up to the neighbouring anti-Stokes lines. Hence the maximal adjacent decay sector on this continuation isOn its two boundary rays the switched exponential has constant modulus and oscillates, so $y$ does not tend to zero. Within the sector the leading algebraic term above persists. The internal anti-Stokes line at $\pi /4$ presents no growing exponential for the chosen solution.

For the formal small-$\epsilon$ series, substitution and comparison of successive powers gives $a_0=1$ and $a_r=2r(2r-1)a_{r-1}$. Thus all coefficients areThis is a factorially divergent asymptotic expansion. The ratio of consecutive term magnitudes is $(2r+2)(2r+1)/|z{|}^2$. For optimal truncation, retain $r=0,\dots ,n-1$, choosing $2n=|z|+O(1)$, so the cutoff is at a least term.

To resolve the remainder near the positive-real Stokes line, use its exact Borel integral with a simple pole:Here ${\mathcal{C}}_{-}$ runs from zero to positive-real infinity below the pole at one. Partial fractions reproduce the exponential-integral solution, including its positive imaginary part. The finite geometric identity for $(1-t^2{)}^{-1}$ gives the displayed exact remainder after integrating the first $n$ monomials.

Write $z=2n+i\mu \sqrt{n}+\nu$, with real bounded $\mu$ and bounded $\nu$. For a Gaussian saddle coalescing with a simple pole, put $t=1+u/\sqrt{n}$. The exponent becomes $-u^2-i\mu u+O(n^{-1/2})$ and $dt/(1-t^2)=-du/(2u)+O(n^{-1/2})du$. ThereforeThe integral is its Cauchy principal value plus $i\pi$. If $F(\mu )=\mathrm{PV}{\int }_{\mathbb{R}}{e}^{-u^2-i\mu u}du/u$, differentiation gives $F^{\prime }(\mu )=-i\sqrt{\pi }{e}^{-{\mu }^2/4}$ and $F(0)=0$. Consequently $F(\mu )=-i\pi \mathrm{erf}(\mu /2)$, and $J_n=-i\pi \mathrm{erfc}(\mu /2)/2+O(n^{-1/2})$.

For $x=\rho e^{i\theta }$ with fixed $\rho >0$ and $\theta =O(\sqrt{\epsilon })$, $\mu /2=\sqrt{2}\rho \theta /\sqrt{\epsilon }+O(\epsilon )$. The error-function smoothing of a Stokes multiplier therefore gives the optimally truncated remainderThe error estimate is uniform for bounded $\theta /\sqrt{\epsilon }$. On the Stokes line the multiplier is half its switched-on value; above it the exponentially small term switches off, and below it switches on. This term is invisible at every algebraic order. Near the line the remainder can exceed the least retained term by a square-root factor because a pole meets the saddle point.

The printed auxiliary Gaussian hint has inconsistent constants. Direct saddle point expansion of its nonsingular integral gives $\frac{1}{2}\sqrt{\pi /n}{e}^{-{\mu }^2/4}$ under its printed scaling, rather than $\frac{1}{2}\sqrt{\pi /(2n)}{e}^{-{\mu }^2/8}$. The derivation above uses the correct saddle point normalization and independently fixes both the smoothing width and the contour sign.

For the finite-mass similarity solution above, the total disk mass isSubstitute $\xi =1/(\tau \sqrt{x})$, so $x={\tau }^{-2}{\xi }^{-2}$ and $x^{-3/2}|dx|=2\tau d\xi$. This givesThe requested linear increase of mass is therefore correct. It is supplied by the infinite-radius boundary, rather than by a source-free isolated finite disk. The inward mass accretion rate iswhich tends to zero at the origin and to $3\pi C{\bar{\nu }}_0{\Sigma }_0$ at infinity.

For Keplerian rotation, the specific angular momentum is $\ell =\sqrt{G{M}_{\star }r}$. The total disk angular momentum would beThis integral diverges at $\xi =0$, corresponding to infinite radius. There is no finite constant total angular momentum for the nonzero printed similarity solution. The seemingly time-independent last expression is an infinite quantity; it cannot prove a finite conservation law.

The problem remains visible under a cutoff. With $r\le r_{0}R$, write $J_R=J_{\ast }{E}_1(1/(\tau \sqrt{R}))$, where $J_{\ast }=4\pi C{\Sigma }_0\sqrt{G{M}_{\star }}{r}_0^{5/2}$ and $E_1$ is the exponential integral. ThenEven the finite change after subtracting a common logarithmic divergence is not zero. Direct integration of the disk diffusion equation gives the boundary identityfor this profile the outer expression has a nonzero limit, giving the same finite change rate. An inner zero torque alone does not impose zero outer angular-momentum flux.

More generally, the printed family $A+B{e}^{-\xi }$ can have finite angular momentum at both ends only if $A=0$ and $A+B=0$, hence it is trivial. Thus the request for a nonzero finite constant $J_d$ is incompatible with the stated viscosity exponent, similarity ansatz and infinite radial domain. A physical finite-domain or altered similarity problem would need different boundary data or assumptions; silently assigning a finite value to the divergent integral does not repair it.

The expansion is singular near the moving lower limit, so expanding $e^{-\epsilon /x}$ throughout the original integral would lose its constant term. Introduce the symmetric exponential integralUnder $x=\epsilon /t$, the discarded piece becomesThis is a uniform Taylor expansion because $t\ge 1$. To derive the logarithmic expansion of a symmetric exponential integral without needing another special function, first exploit the same substitution to writeThe estimate follows from $0\le 1-e^{-\epsilon /x}\le \epsilon /x$. Here $\gamma$ is Euler's constant and the small-argument expansion of the exponential integral fixes the constant as well as the logarithm.

Differentiate under the integral and integrate the derivative of $e^{-x-\epsilon /x}/x$ over $(0,\infty )$; the exponentially small endpoint terms vanish. This gives the ordinary differential equationFor the Frobenius method at its regular singular point, write $J=A\log \epsilon +B+\epsilon (C_1\log \epsilon +D)+\cdots$. Substitution gives $C_1=A$ and $D=B-2A$. The preceding leading calculation sets $A=-1$, $B=-2\gamma$, henceSubtracting the moving-cutoff correction to a symmetric exponential integral yields the requested answer, including the logarithmically enhanced first-order term:Thus the $\epsilon \log \epsilon$ switchback term must be retained when asking for all contributions through order $\epsilon$.

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.

For real positive weight $w$, the weighted inner product isThe functions must have finite weighted norm; the conjugate is omitted for real functions. Multiplying the differential equation by $x{e}^{-x}$ givesThus the Sturm-Liouville form has $p=x{e}^{-x}$, $q=0$ and $w=e^{-x}$. With the positive-sign spectral convention defineThe existing Laguerre differential operator on polynomials uses the opposite operator $L=-\mathcal{A}$, with eigenvalues $-n$; keeping this sign convention explicit avoids confusing the two descriptions.

For admissible $f,g$, integration by parts givesThe boundary form must vanish for every pair in the self-adjoint domain. In the usual regular-at-zero Laguerre realization one imposes ${\lim }_{x\downarrow 0}x{e}^{-x}{y}^{\prime }=0$, excluding the logarithmic branch; at infinity one uses the weighted square-integrable domain with vanishing boundary form. A condition $y(\infty )=0$ would be inappropriate for polynomial eigenfunctions. For any pair of polynomials, the factor $x$ makes the boundary form zero at zero, while the exponential beats all polynomial growth at infinity. These functions also have finite weighted norms.

Self-adjointness yields $(n-m)⟨y_m,y_n{⟩}_w=0$, giving orthogonality of Sturm-Liouville eigenfunctions:For the specified polynomials, direct differentiation gives $\mathcal{A}u=u$ and $\mathcal{A}v=3v$. Hence their eigenvalues in the displayed convention are 1 and 3, respectively, or −1 and −3 for $L=-\mathcal{A}$.

To find the quadratic, put $h=a{x}^2+bx+c$. The equation $x{h}^{\prime \prime }+(1-x)h^{\prime }+2h=0$ reduces to $(4a+b)x+(b+2c)=0$. Thus $b=-4a$, $c=2a$. Since $h(1)=-a=1$,These are scalar multiples of the Laguerre polynomials of degrees one, three and two.

For $n=0$, the first integral is $x{e}^{-x}{y}^{\prime }=C$, so all solutions with the prescribed interior normalization areTwo independent choices areThey both equal one at $x_0$, and their Wronskian is $e^x/x\ne 0$. For sufficiently large $x$, the last unit interval contributes at least $e^{x-1}/x$, proving that $y_2\to \infty$. These are the zero-eigenvalue solutions of the Laguerre differential equation, with the second expressible through the exponential integral. It also has a logarithmic singularity at zero and is not an admissible polynomial or weighted square-integrable eigenfunction. An interior value alone does not enforce the endpoint conditions of the spectral problem.