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For $b$ not a nonpositive integer, the regular solution normalized by $M(a,b,0)=1$ has the power seriesHere $(a{)}_n$ is the rising factorial. Substitution into the Kummer differential equation proves the coefficient recurrence. This normalization agrees with NIST's definitions.

The falling factorial is the product of $j$ successive descending factors, with $(t{)}_0=1$. Another notation is $t^{\underline{j}}$; this article uses $(t{)}_j$ for descending factors, whereas the rising factorial uses the same notation for ascending factors. For a nonnegative integer $t$, it counts ordered selections of $j$ distinct objects and vanishes when $j>t$. Powers expand as $t^j={\sum }_{\ell =0}^{j}S(j,\ell )(t{)}_{\ell }$, where $S(j,\ell )$ is a Stirling number of the second kind.

First take $z>0$. Set $s=t^3-1$ in the tail integral. This puts it in the endpoint form needed for Watson's lemma:The binomial coefficients are $(-1{)}^r(2/3{)}_r/r!$, where $(a{)}_r=\Gamma (a+r)/\Gamma (a)$ is the rising factorial. Integrating each endpoint term givesAlso, scaling the complete positive-ray integral gives ${\int }_0^{\infty }{e}^{-z^3{t}^3}dt=\Gamma (1/3)/(3z)$. HenceThe endpoint term is exponentially small on this positive ray but must be retained for continuation to other sectors.

For complex $z$, use $s=zt$ so $G(z)=z^{-1}{\int }_0^z{e}^{-s^3}ds$. Put $C=\Gamma (1/3)/3$ and $\omega =e^{2\pi i/3}$. The complete steepest descent contours from the cubic saddle are the rays $\arg s=2\pi j/3$, whose integrals are $C{\omega }^j$. Deform the finite contour onto one of these rays followed by the endpoint contour. The cubic exponential integral Stokes sectors giveFor $0\le \arg z\le 2\pi$, take $j=0$ below $\pi /3$, $j=1$ between $\pi /3$ and $\pi$, $j=2$ between $\pi$ and $5\pi /3$, and $j=3$ above $5\pi /3$. The corresponding algebraic coefficients are $C,C\omega ,C{\omega }^2,C$. A negative-real $z$ provides a useful sign check: $G(z)=G(|z|e^{i\pi })$ is real and positive, so its two lateral algebraic coefficients must be complex conjugates.

The subdominant algebraic term switches across the Stokes linesThere $z^3$ is negative real and the endpoint exponential is dominant. The intervening anti-Stokes lines are $\arg z=\pi /6+j\pi /3$, where $\mathrm{Re}{z}^3=0$ and the exponential changes from growth to decay. The naming convention here distinguishes switching from equal exponential magnitude.

To resolve the jump near ${\theta }_s=\pi /3$, optimally truncate the endpoint series at $N$ with $N+2/3=R+O(1)$, where $R=|z{|}^3$. Let $P_N$ denote that truncated endpoint contribution. The Borel remainder for a cubic exponential integral derived in solution, together with the Gaussian pole transition for a large gamma parameter in solution, givesThis estimate is on the transition scale $\theta -\pi /3=O(R^{-1/2})$ after optimal truncation. The error-function smoothing of a Stokes multiplier has width $O(|z{|}^{-3/2})$. Before the line $S\to 0$ and the coefficient is $C$; after it $S\to 1$ and the coefficient is $C\omega$, agreeing with the sector expansions. On the line the leading coefficient is their average. The original $G$ is an entire function; only its asymptotic decomposition has these sector-dependent coefficients.

The origin is a regular singular point of this Kummer differential equation. Apply the Frobenius method, writing $y={\sum }_{n=0}^{\infty }{a}_n{x}^{n+r}$ with $a_0\ne 0$. The lowest power gives the indicial equationFor $n\ge 1$, comparison of the coefficient of $x^{n+r-1}$ givesSince $0<c<1$, both roots give valid independent branches. Normalize $a_0=1$:Here $(c{)}_n$ is the rising factorial, and the first series is the confluent hypergeometric function of the first kind $M(1,c,x)$. The two leading powers are distinct, proving linear independence. We describe real solutions on $x>0$, with their one-sided behavior at zero.

For the inhomogeneous equation a particular solution is $-(x+c)/2$, verified by direct substitution. Thus $y=A{y}_1+B{y}_2-(x+c)/2$. Since $y_2^{\prime }\sim (1-c)x^{-c}$, a finite derivative at zero forces $B=0$. Then $y(0)=0$ forces $A=c/2$, and the required solution isIn particular $y^{\prime }(0)=0$. The singular homogeneous branch cannot be added without violating the derivative condition.

Near zero, write the amplitude as $t^{-1/4}(1+t{)}^{-1/4}={\sum }_{m\ge 0}(\genfrac{}{}{0ex}{}{-1/4}{m})t^{m-1/4}$. The large parameter localizes the integral at $t=0$. Termwise integration of the local expansion givesConsequentlyHere $(1/4{)}_m$ is the rising factorial. This is an asymptotic series, not an assertion that the full integrated Taylor series converges.

The needed result is Watson's lemma: if a locally integrable amplitude has an expansion $g(s)\sim \sum c_m{s}^{{\rho }_m-1}$ as $s\downarrow 0$, with ${\rho }_m>0$ increasing and suitable growth at infinity so its Laplace integral converges, then ${\int }_0^{\infty }{e}^{-ks}g(s)ds\sim \sum c_m\Gamma ({\rho }_m)k^{-{\rho }_m}$, with the corresponding finite-order remainder bounds. Apply it after $s=t^2$, when $g(s)=\frac{1}{2}{s}^{-5/8}(1+\sqrt{s}{)}^{-1/4}$ and ${\rho }_m=3/8+m/2$. The tail is exponentially small after any fixed positive cutoff.

Insert the Frobenius solution and its differentiated series into the equation. Collecting the terms before shifting an index givesThe coefficient of the lowest power $x^{\sigma -1}$ is $a_0\sigma (\sigma +\gamma -1)$. Because $a_0\ne 0$, the indicial equation givesFor the coefficient of $x^{n+\sigma -1}$ with $n\ge 1$, shift the index in the second sum. The hypergeometric Frobenius recurrence isWhenever the left coefficient is nonzero, division yieldsFor noninteger $\gamma$, the two independent local solutions can therefore be identified aswith a consistent local branch for the fractional power. The hypergeometric function coefficients follow by iterating the recurrence using rising factorials.

There is a parameter qualification to the PDF's statement for all real constants: a zero denominator cannot be divided through. At an undetermined coefficient at Frobenius resonance, use the undivided coefficient equation to test compatibility. For example, $\gamma =0,\sigma =0,n=1$ requires $0=\alpha \beta a_0$, so a pure exponent-zero series with $a_0\ne 0$ is impossible if $\alpha \beta \ne 0$. When $\gamma =1$ the two indicial roots coincide, and a second independent solution generally needs a logarithmic term. The indicial roots and undivided relation remain the valid coefficient conclusions in these exceptional cases.