Frobenius solution to the hypergeometric equation

In the following we solve the second-order differential equation called the hypergeometric differential equation using Frobenius method, named after Ferdinand Georg Frobenius. This is a method that uses the series solution for a differential equation, where we assume the solution takes the form of a series. This is usually the method we use for complicated ordinary differential equations.

The solution of the hypergeometric differential equation is very important. For instance, Legendre's differential equation can be shown to be a special case of the hypergeometric differential equation. Hence, by solving the Hypergeometric differential equation, one may directly compare its solutions to get the solutions of Legendre's differential equation, after making the necessary substitutions. For more details, please check the hypergeometric differential equation

We shall prove that this equation has three singularities, namely at "x" = 0, "x" = 1 and around infinity. However, as these will turn out to be regular singular points, we will be able to assume a solution on the form of a series. Since this is a second-order differential equation, we must have two linearly independent solutions.

The problem however will be that our assumed solutions may or not be independent, or worse, may not even be defined (depending on the value of the parameters of the equation). This is why we shall study the different cases for the parameters and modify our assumed solution accordingly.

The equation

Solve the hypergeometric equation around all singularities:: $x(1-x)y"+left{ gamma -(1+alpha +eta )x
ight}y'-alpha eta y=0$

= Solution around "x" = 0 =

Let

: $egin{align} P_0(x) = -alpha eta, && P_1(x) = gamma - (1+alpha +eta )x, && P_2(x) = x(1-x)end{align}$

Then

: $P_2(0) = 0, P_2 (1)=0.,$

Hence, "x" = 0 and "x" = 1 are singular points. Let's start with "x" = 0. To see if it is regular, we study the following limits:

: $egin{align} lim_{x o a} frac{(x - a) P_1(x)}{P_2(x)} &= lim_{x o 0} frac{(x - 0)(gamma - (1 + alpha + eta)x)}{x(1 - x)} = lim_{x o 0} frac{x(gamma - (1 + alpha + eta)x)}{x(1 - x)} = gamma \ lim_{x o a} frac{(x - a)^2 P_0(x)}{P_2(x)} &= lim_{x o 0} frac{(x - 0)^2(-alpha eta)}{x(1 - x)} = lim_{x o 0} frac{x^2 (-alpha eta)}{x(1 - x)} = 0end{align}$

Hence, both limits exist and "x" = 0 is a regular singular point. Therefore, we assume the solution takes the form

: $y = sum_{r=0}^infty a_r x^{r + c}$

with "a"₀ ≠ 0. Hence,

: $egin{align} y' = sum_{r = 0}^infty a_r(r + c) x^{r + c - 1} && y"= sum_{r = 0}^infty a_r(r + c)(r + c - 1) x^{r + c - 2}.end{align}$

Substituting these into the hypergeometric equation, we get

: $egin{align} &x sum_{r = 0}^infty a_r(r + c)(r + c - 1) x^{r + c - 2} - x^2 sum_{r = 0}^infty a_r(r + c)(r + c - 1) x^{r + c - 2} + gamma sum_{r = 0}^infty a_r(r + c) x^{r + c - 1} \ &quad -(1 + alpha + eta) xsum_{r = 0}^infty a_r(r + c) x^{r + c - 1} -alpha eta sum_{r = 0}^infty a_r x^{r + c} = 0end{align}$ That is,: $egin{align} &sum_{r = 0}^infty a_r(r + c)(r + c - 1) x^{r + c - 1} -sum_{r = 0}^infty a_r(r + c)(r + c - 1) x^{r + c} +gamma sum_{r = 0}^infty a_r(r + c) x^{r + c - 1} \ &quad -(1 + alpha + eta) sum_{r = 0}^infty a_r(r + c) x^{r + c} -alpha eta sum_{r = 0}^infty a_r x^{r + c} =0end{align}$

In order to simplify this equation, we need all powers to be the same, equal to "r" + "c" - 1, the smallest power. Hence, we switch the indices as follows:

: $egin{align} &sum_{r = 0}^infty a_r(r + c)(r + c - 1)x^{r + c - 1} -sum_{r = 1}^infty a_{r - 1}(r + c - 1)(r + c - 2) x^{r + c - 1} +gamma sum_{r = 0}^infty a_r(r + c) x^{r + c - 1} \ &quad -(1 + alpha + eta) sum_{r = 1}^infty a_{r - 1}(r + c - 1) x^{r + c - 1} -alpha eta sum_{r = 1}^infty a_{r - 1} x^{r + c - 1} =0end{align}$

Thus, isolating the first term of the sums starting from 0 we get

: $egin{align} &a_0 (c(c-1) + gamma c) x^{c - 1} + sum_{r = 1}^infty a_r(r + c)(r + c - 1) x^{r + c - 1} -sum_{r = 1}^infty a_{r - 1}(r + c - 1)(r + c - 2) x^{r + c - 1} \ &quad + gamma sum_{r = 1}^infty a_r(r + c) x^{r + c - 1} -(1 + alpha + eta) sum_{r = 1}^infty a_{r - 1}(r + c - 1) x^{r + c - 1} -alpha eta sum_{r = 1}^infty a_{r - 1} x^{r + c - 1} = 0end{align}$

Now, from the linear independence of all powers of "x", that is, of the functions 1, "x", "x"², etc., the coefficients of "x"^k vanish for all "k". Hence, from the first term, we have

: $a_{0} (c(c - 1) + gamma c) = 0,$

which is the indicial equation. Since "a"₀ ≠ 0, we have

: $c(c - 1 + gamma) = 0.,$

Hence,

: $egin{align} c_1 = 0 && c_2 = 1 - gammaend{align}$

Also, from the rest of the terms, we have

: $egin{align} &((r + c)(r + c - 1) + gamma(r+c)) a_r \ &quad + (-(r + c - 1)(r + c - 2) - (1 + alpha + eta)(r + c - 1) - alphaeta) a_{r - 1} = 0end{align}$

Hence,

: $egin{align} a_r &= frac{(r + c - 1)(r + c - 2) + (1 + alpha + eta)(r + c - 1) + alphaeta} {(r + c)(r + c - 1) + gamma(r + c)} a_{r - 1} \ &= frac{(r + c -1)(r + c + alpha + eta - 1) + alphaeta} {(r + c)(r + c + gamma - 1)} a_{r - 1}end{align}$

But

: $egin{align} &(r + c - 1)(r + c + alpha + eta - 1) + alphaeta \ &quad = (r + c - 1)(r + c + alpha - 1) + (r + c - 1)eta + alphaeta \ &quad = (r + c - 1)(r + c + alpha - 1) + eta(r + c + alpha - 1)end{align}$ Hence, we get the recurrence relation: $a_r = frac{(r + c + alpha - 1)(r + c + eta - 1)}{(r + c)(r + c + gamma - 1)} a_{r - 1}, ext{ for } r geq 1.$

Let's now simplify this relation by giving "a"_"r" in terms of "a"₀ instead of "a"_{"r" − 1}. From the recurrence relation (note: below, expressions of the form ("u")_"r" refer to the Pochhammer symbol).: $egin{align} a_1 &= frac{(c + alpha)(c + eta)}{(c + 1)(c + gamma)} a_0 \ a_2 &= frac{(c + alpha + 1)(c + eta + 1)}{(c + 2)(c + gamma + 1)} a_1 = frac{(c + alpha + 1)(c + alpha)(c + eta)(c + eta + 1)} {(c + 2)(c + 1)(c + gamma)(c + gamma + 1)} a_0 = frac{(c + alpha)_2 (c + eta)_2}{(c + 1)_2 (c + gamma)_2} a_0 \ a_3 &= frac{(c + alpha + 2)(c + eta + 2)}{(c + 3)(c + gamma + 2)} a_2 = frac{(c + alpha)_2 (c + alpha + 2)(c + eta )_2 (c + eta + 2} {(c + 1)_2 (c + 3)(c + gamma)_2 (c + gamma + 2)} a_0 \ &= frac{(c + alpha)_3 (c + eta)_3} {(c + 1)_3 (c + gamma)_3} a_0end{align}$

As we can see,

: $a_r =frac{(c + alpha)_r (c + eta)_r}{(c + 1)_r(c + gamma)_r} a_0, ext{ for } r geq 0$

Hence, our assumed solution takes the form

: $y = a_0 sum_{r = 0}^infty frac{(c + alpha)_r (c + eta)_r}{(c + 1)_r (c + gamma)_r} x^{r + c}.$

We are now ready to study the solutions corresponding to the different cases for "c"₁ − "c"₂ = γ − 1 (it should be noted that this reduces to study the nature of the parameter γ: whether it is an an integer or not).

Analysis of the solution in terms of the difference γ − 1 of the two roots

γ not an integer

Then "y"₁ = "y"|_{"c" = 0} and "y"₂ = "y"|_{"c" = 1 − γ}. Since: $y = a_0 sum_{r = 0}^infty frac{(c + alpha )_r (c + eta)_r}{(c + 1)_r (c + gamma)_r} x^{r + c},$ we have: $egin{align} y_1 &= a_0 sum_{r = 0}^infty frac{(alpha)_r (eta)_r}{(1)_r (gamma)_r} x^r \ &= a_0 cdot$ }_2 F_1}(alpha, eta; gamma; x) \ y_2 &= a_0 sum_{r = 0}^infty frac{(alpha + 1 - gamma)_r (eta + 1 - gamma)_r} {(1 - gamma + 1)_r (1 - gamma + gamma)_r} x^{r + 1 - gamma} = a_0 x^{1 - gamma} sum_{r = 0}^infty frac{(alpha + 1 - gamma)_r (eta + 1 - gamma)_r} {(1)_r (2 - gamma)_r} x^r \ &= a_0 x^{1 - gamma} }_2 F_1}(alpha - gamma + 1, eta - gamma + 1; 2 - gamma; x)end{align} Hence, $y = A' y_1 + B' y_2.$ Let "A"′ a₀ = "a" and "B"′ "a"₀ = "B". Then: $y = A$ }_2 F_1}(alpha, eta; gamma; x) + B x^{1 - gamma} }_2 F_1}(alpha - gamma + 1, eta - gamma + 1; 2 - gamma; x),

= γ = 1 =

Then "y"₁ = "y"|_{"c" = 0}. Since γ = 1, we have: $y = a_0 sum_{r = 0}^infty frac{(c + alpha)_r (c + eta)_r}{(c + 1)_r^2} x^{r + c}.$ Hence,: $egin{align} y_1 &= a_0 sum_{r = 0}^infty frac{(alpha)_r (eta)_r}{(1)_r (1)_r} x^r = a_0$ }_2 F_1}(alpha, eta; 1; x) \ y_2 &= left.frac{partial y}{partial c} ight|_{c = 0}.end{align}To calculate this derivative, let : $M_r = frac{(c + alpha)_r (c + eta)_r}{(c + 1)_r^2}.$ Then: $ln(M_r) = lnleft(frac{(c + alpha)_r (c + eta)_r}{(c + 1)_r^2}
ight) = ln(c + alpha)_r + ln(c + eta)_r - 2ln(c + 1)_r$ But: $ln(c + alpha)_r = lnigl((c + alpha)(c + alpha + 1) cdots (c + alpha + r - 1)igr) = sum_{k = 0}^{r - 1} ln(c + alpha + k).$ Hence,: $egin{align} ln(M_r) &= sum_{k = 0}^{r - 1} ln(c + alpha + k) + sum_{k = 0}^{r - 1} ln(c + eta + k) - 2 sum_{k = 0}^{r - 1} ln(c + 1 + k) \ &= sum_{k = 0}^{r - 1} igl(ln(c + alpha + k) + ln(c + eta + k) -2 ln(c + 1 + k)igr)end{align}$ Differentiating both sides of the equation with respect to "c", we get:: $frac{1}{M_r} frac{partial M_r}{partial c} = sum_{k = 0}^{r - 1} left(frac{1}{c + alpha + k} + frac{1}{c + eta +k } - frac{2}{c + 1 + k}
ight).$ Hence, : $frac{partial M_r}{partial c} = frac{(c + alpha)_r (c + eta)_r}{(c + 1)_r^2} sum_{k = 0}^{r - 1} left(frac{1}{c + alpha + k} + frac{1}{c + eta + k} - frac{2}{c + 1 + k}
ight).$ Now,: $y = a_0 x^c sum_{r = 0}^infty frac{(c + alpha)_r (c + eta)_r}{(c + 1)_r^2} x^r = a_0 x^c sum_{r = 0}^infty M_r x^r.$ Hence,: $egin{align} frac{partial y}{partial c} &= a_0 x^c ln(x) sum_{r = 0}^infty frac{(c + alpha)_r (c + eta)_r}{(c + 1)_r^2} x^r \ &quad + a_0 x^c sum_{r = 0}^infty left(frac{(c + alpha)_r (c + eta)_r}{(c + 1)_r^2} left{sum_{k = 0}^{r - 1} left(frac{1}{c + alpha + k} + frac{1}{c + eta + k} - frac{2}{c + 1 + k}
ight)
ight}
ight) x^r \ &= a_0 x^c sum_{r = 0}^infty frac{(c + alpha)_r (c + eta)_r}{(c + 1)_r)^2} left(ln x + sum_{k = 0}^{r - 1} left(frac{1}{c + alpha + k} +frac{1}{c + eta + k} - frac{2}{c + 1 + k}
ight)
ight) x^{r}.end{align}$ For "c" = 0, we get: $y_2 = a_0 sum_{r = 0}^infty frac{(alpha)_r (eta)_r}{(1)_r^{2$ left(ln x + sum_{k = 0}^{r - 1} left(frac{1}{alpha + k} + frac{1}{eta + k} - frac{2}{1 + k} ight) ight)x^{r}.Hence, "y" = "C"′ "y"₁ + "D"′ "y"₂. Let "C"′ "a"₀ = "C" and "D"′ "a"₀ = "D". Then: $y = C$ }_2 F_1}(alpha, eta; 1; x) + D sum_{r = 0}^infty frac{(alpha)_r (eta)_r}{(1)_r^2} left(ln x + sum_{k = 0}^{r - 1} left(frac{1}{alpha + k} + frac{1}{eta + k} - frac{2}{1 + k} ight) ight) x^r

γ an integer and γ ≠ 1

γ ≤ 0

From the recurrence relation: $a_r = frac{(r + c + alpha - 1)(r + c + eta - 1)}{(r + c)(r + c + gamma - 1)} a_{r - 1},$ we see that when "c" = 0 (the smaller root), "a"_{1 − γ} → ∞. Hence, we must make the substitution "a"₀ = "b"₀ ("c" - "c"_"i"), where "c"_"i" is the root for which our solution is infinite. Hence, we take "a"₀ = "b"₀ "c" and our assumed solution takes the new form : $y_b = b_0 x^c sum_{r = 0}^infty frac{c(c + alpha)_r (c + eta)_r}{(c + 1)_r (c + gamma)_r} x^r$ Then "y"₁ = "y"_"b"|_{"c" = 0}. As we can see, all terms before: $frac{c(c + alpha)_{1 - gamma}(c + eta)_{1 - gamma$ {(c + 1)_{1 - gamma}(c + gamma)_{1 - gamma x^{1 - gamma} vanish because of the "c" in the numerator. Starting from this term however, the "c" in the numerator vanishes. To see this, note that: $(c + gamma)_{1 - gamma} = (c + gamma)(c + gamma + 1) cdots c.$ Hence, our solution takes the form: $egin{align} y_1 &= b_0 left( frac{(alpha)_{1 - gamma} (eta)_{1 - gamma$ {(1)_{1 - gamma} (gamma)_{-gamma x^{1 - gamma} + frac{(alpha)_{2 - gamma} (eta)_{2 - gamma{(1)_{2 - gamma} (gamma)_{-gamma}(1)} x^{2 - gamma} + frac{(alpha)_{3 - gamma} (eta)_{3 - gamma{(1)_{3 - gamma} (gamma)_{-gamma}(1)(2)} x^{3 - gamma} + cdots ight) \ &= frac{b_0}{(gamma)_{-gamma sum_{r = 1 - gamma}^infty frac{(alpha)_r (eta)_r}{(1)_r (1)_{r + gamma - 1 x^r.end{align}Now,: $y_2 = left.frac{partial y_b}{partial c}
ight|_{c = 1 - gamma}.$ To calculate this derivative, let: $M_r = frac{c(c + alpha)_r (c + eta)_r}{(c + 1)_r(c + gamma)_r}.$ Then following the method in the previous case, we get : $frac{partial M_r}{partial c} = frac{c(c + alpha)_r (c + eta )_r}{(c + 1)_r (c + gamma)_r} left{ frac{1}{c} + sum_{k = 0}^{r - 1} left(frac{1}{c + alpha + k} + frac{1}{c + eta + k} - frac{1}{c + 1 + k} -frac{1}{c + gamma + k}
ight)
ight}.$ Now,: $y_b = b_0 sum{r = 0}^infty frac{c(c + alpha)_r (c + eta)_r}{(c + 1)_r (c + gamma)_r} x^{r + c} = b_0 x^c sum_{r = 0}^infty M_r x^r.$ Hence,: $egin{align} frac{partial y}{partial c} &= b_0 x^c ln(x) sum_{r = 0}^infty frac{c(c + alpha)_r (c + eta)_r}{(c + 1)_r (c + gamma)_r} x^r \ &quad + b_0 x^c sum_{r = 0}^infty frac{c(c + alpha)_r (c + eta)_r}{(c + 1)_r (c + gamma)_r} left{ frac{1}{c} + sum_{k = 0}^{r - 1} left(frac{1}{c + alpha + k} + frac{1}{c + eta + k} - frac{1}{c + 1 + k} - frac{1}{c + gamma + k}
ight)
ight} x^rend{align}$ Hence,: $egin{align} frac{partial y}{partial c} = b_0 x^c sum_{r = 0}^infty &frac{c(c + alpha)_r (c + eta)_r}{(c + 1)_r (c + gamma)_r} Biggl(ln x + frac{1}{c} + \ &quad + sum_{k = 0}^{r - 1} left(frac{1}{c + alpha + k} + frac{1}{c + eta + k} - frac{1}{c + 1 + k} - frac{1}{c + gamma + k}
ight) Biggr) x^r.end{align}$ At "c" = 1- γ, we get "y"₂. Hence, "y" = "E"′ "y"₁ + "F"′ "y"₂. Let "E"′ "b"₀ = "E" and "F"′ "b"₀ = "F". Then: $egin{align} y &= frac{E}{(gamma)_{-gamma sum_{r = 1 - gamma}^infty frac{(alpha)_r (eta)_r} {(1)_r (1)_{r + gamma - 1 x^r \ &quadegin{align} {}+ F x^{1 - gamma} sum_{r = 0}^infty &frac{(1 - gamma) (alpha + 1 - gamma)_r (eta + 1 - gamma)_r} {(2 - gamma)_r (1)_r} Biggl(ln x + frac{1}{1 - gamma} + \ &+ sum_{k = 0}^{r - 1} left(frac{1}{alpha + k + 1 - gamma} + frac{1}{eta + k + 1 - gamma} - frac{1}{2 + k - gamma} - frac{1}{1 + k}
ight) Biggr) x^r. end{align}end{align}$

γ > 1

From the recurrence relation: $a_r = frac{(r + c + alpha - 1)(r + c + eta - 1)}{(r + c)(r + c + gamma - 1)} a_{r - 1},$ we see that when "c" = 1 - γ (the smaller root), "a"_{γ − 1} → ∞. Hence, we must make the substitution "a"₀ = "b"₀("c" − "c"_"i"), where "c"_"i" is the root for which our solution is infinite. Hence, we take "a"₀ = "b"₀("c" + γ - 1) and our assumed solution takes the new form:: $y_b = b_0 x^c sum_{r = 0}^infty frac{(c + gamma -1 )(c + alpha )_r (c + eta)_r} {(c + 1)_r (c + gamma)_r} x^r.$ Then "y"₁ = "y"_"b"|_{"c" = 1 - γ}. All terms before: $frac{(c + gamma - 1)(c + alpha)_{gamma - 1} (c + eta)_{gamma - 1 {(c + 1)_{gamma - 1}(c + gamma)_{gamma - 1 x^{gamma - 1}$ vanish because of the "c" + γ - 1 in the numerator. Starting from this term, however, the "c" + γ - 1 in the numerator vanishes. To see this, note that : $(c + 1)_{gamma - 1} = (c + 1)(c + 2)cdots(c + gamma - 1).$ Hence, our solution takes the form : $egin{align} y_1 &= b_0 x^{1 - gamma} left(frac{(alpha + 1 - gamma)_{gamma - 1} (eta + 1 - gamma)_{gamma - 1 {(2 - gamma)_{gamma - 2} (1)_{gamma - 1 x^{gamma - 1} + frac{(alpha + 1 - gamma)_{gamma} (eta + 1 - gamma)_{gamma {(2 - gamma)_{gamma - 2} (1) (1)_{gamma x^{gamma} + cdots
ight) \ &= frac{b_0}{(2 - gamma)_{gamma - 2 x^{1 - gamma} sum_{r = gamma - 1}^infty frac{(alpha + 1 - gamma)_r (eta + 1 - gamma)_r} {(1)_r (1)_{r + 1 - gamma x^r.end{align}$ Now,: $y_2 = left.frac{partial y_b}{partial c}
ight|_{c = 0}.$ To calculate this derivative, let: $M_r = frac{(c + gamma - 1)(c + alpha)_r (c + eta)_r}{(c + 1)_r (c + gamma)_r}.$ Then following the method in the second case above, : $egin{align} frac{partial M_r}{partial c} &= frac{(c + gamma - 1) (c + alpha)_r (c + eta)_r}{(c + 1)_r (c + gamma)_r} Biggl(frac{1}{c + gamma - 1} + \ &qquad + sum_{k = 0}^{r - 1} left(frac{1}{c + alpha + k} + frac{1}{c + eta + k} - frac{1}{c + 1 + k} - frac{1}{c + gamma + k}
ight) Biggr)end{align}$ Now,: $y_b = b_0 sum_{r = 0}^infty frac{(c + gamma - 1) (c + alpha)_r (c + eta)_r} {(c + 1)_r (c + gamma)_r} x^{r + c} = b_0 x^c sum_{r = 0}^infty M_r x^r.$ Hence,: $egin{align} frac{partial y}{partial c} &= b_0 x^c ln(x) sum_{r = 0}^infty frac{(c + gamma - 1) (c + alpha)_r (c + eta)_r} {(c + 1)_r (c + gamma)_r} x^r +\ &qquadegin{align} {}+ b_0 x^c sum_{r = 0}^infty &frac{(c + gamma - 1) (c + alpha)_r (c + eta)_r} {(c + 1)_r (c + gamma)_r} Biggl(frac{1}{c + gamma - 1} + \ &+ sum_{k = 0}^{r - 1} left(frac{1}{c + alpha + k} + frac{1}{c + eta + k} - frac{1}{c + 1 + k} - frac{1}{c + gamma + k}
ight) Biggr)x^r end{align} \ &egin{align} {}= b_0 x^c sum_{r = 0}^infty &frac{(c + gamma - 1) (c + alpha)_r (c + eta)_r} {(c + 1)_r (c + gamma)_r} Biggl(ln x + frac{1}{c + gamma - 1} + \ &+ sum_{k = 0}^{r - 1} left(frac{1}{c + alpha + k} + frac{1}{c +eta + k} - frac{1}{c + 1 + k} - frac{1}{c + gamma + k}
ight) Biggr) x^r. end{align}end{align}$ At "c" = 0 we get "y"₂. Hence, "y" = "G"′"y"₁ + "H"′"y"₂. Let "G"′"b"₀ = "E" and "H"′"b"₀ = "F". Then: $egin{align} y = frac{G}{(2 - gamma)_{gamma - 2 &x^{1 - gamma} sum_{r = gamma - 1}^infty frac{(alpha + 1 - gamma)_r (eta + 1 - gamma)_r} {(1)_r (1)_{r + 1 - gamma x^r \ &egin{align} {} + H sum_{r = 0}^infty &frac{(1 - gamma) (alpha + 1 - gamma)_r (eta + 1 - gamma)_r} {(2 - gamma)_r (1)_r} Biggl(ln x + frac{1}{gamma - 1} + \ &+ sum_{k = 0}^{r - 1} left(frac{1}{alpha + k} + frac{1}{eta + k} - frac{1}{1 + k} - frac{1}{gamma + k}
ight) Biggr) x^r. end{align}end{align}$

= Solution around "x" = 1 =

Let us now study the singular point "x" = 1. To see if it is regular, : $egin{align} &lim_{x o a} frac{(x - a) P_1(x)}{P_2(x)} = lim_{x o 1} frac{(x - 1) (gamma - (1 + alpha + eta)x)}{x(1 - x)} \ &quad = lim_{x o 1} frac{-(gamma - (1 + alpha + eta)x)}{x} = 1 + alpha + eta - gamma \ &lim_{x o a} frac{(x - a)^2 P_0(x)}{P_2(x)} = lim_{x o 1} frac{(x - 1)^2 (-alphaeta)}{x(1 - x)} = lim_{x o 1} frac{(x - 1) alpha eta}{x} = 0end{align}$ Hence, both limits exist and "x" = 1 is a regular singular point. Now, instead of assuming a solution on the form: $y = sum_{r = 0}^infty a_r (x - 1)^{r + c},$ we will try to express the solutions of this case in terms of the solutions for the point "x" = 0. We proceed as follows: we had the hypergeometric equation: $x(1 - x)y" + (gamma - (1 + alpha + eta)x)y' - alphaeta y = 0.,$ Let "z" = 1 - "x". Then: $egin{align} &frac{dy}{dx} = frac{dy}{dz} imes frac{dz}{dx} = -frac{dy}{dz} = -y' \ &frac{d^2 y}{dx^2} = frac{d}{dx}left( frac{dy}{dx}
ight) = frac{d}{dx}left( -frac{dy}{dz}
ight) = frac{d}{dz}left( -frac{dy}{dz}
ight) imes frac{dz}{dx} =frac{d^{2}y}{dz^{2 = y"end{align}$ Hence, the equation takes the form: $z(1 - z) y" + (alpha + eta - gamma + 1 - (1 + alpha + eta)z) y' - alphaeta y = 0.,$ Since "z" = 1 - "x", the solution of the hypergeometric equation at "x" = 1 is the same as the solution for this equation at "z" = 0. But the solution at z = 0 is identical to the solution we obtained for the point "x" = 0, if we replace each γ by α + β - γ + 1. Hence, to get the solutions, we just make this substitution in the previous results. Note also that for "x" = 0, "c"₁ = 0 and "c"₂ = 1 - γ. Hence, in our case, "c"₁ = 0 while "c"₂ = γ - α - β. Let us now write the solutions. It should be noted in the following we replaced each "z" by 1 - "x".

Analysis of the solution in terms of the difference γ − α − β of the two roots

γ − α − β not an integer

$egin{align} y &= A cdot$ }_2 F_1}(alpha, eta; alpha + eta - gamma + 1; 1 - x) \ &quad + B (1 - x)^{gamma - alpha - eta} }_2 F_1}(gamma - alpha, gamma - eta; gamma - alpha - eta + 1; 1 - x)end{align}

=γ − α − β = 0=

$egin{align} y &= C cdot$ }_2 F_1}(alpha, eta; 1; 1 - x) \ &quad + D sum_{r = 0}^infty frac{(alpha)_r (eta)_r}{(1)_r^2} left(ln(1 - x) + sum_{k = 0}^{r - 1} left(frac{1}{alpha + k} + frac{1}{eta + k} - frac{2}{1 + k} ight) ight) (1 - x)^r end{align}

γ − α − β is an integer and γ − α − β ≠ 0

γ − α − β > 0

$egin{align} y &= frac{E}{(alpha + eta - gamma + 1)_{gamma - alpha - eta - 1 sum_{r = 1 - gamma}^infty frac{(alpha)_r (eta)_r}{(1)_r (1)_{r + alpha + eta - gamma (1 - x)^r + {}\ &quadegin{align} {} + F(1 - x)^{gamma - alpha - eta} sum_{r = 0}^infty & frac{(gamma - alpha - eta)(gamma - eta)_r (gamma - alpha)_r} {(1 + gamma - alpha - eta)_r (1)_r} Biggl(ln(1 - x) + frac{1}{gamma - alpha - eta} + {} \ &+ sum_{k = 0}^{r - 1} left(frac{1}{k + gamma - eta} + frac{1}{k + gamma - alpha} - frac{1}{1 + k + gamma - alpha - eta} - frac{1}{1 + k}
ight) Biggr) (1 - x)^r end{align}end{align}$

γ − α − β < 0

$egin{align} y &= frac{G}{(1 + gamma - alpha - eta)_{alpha + eta - gamma - 1 (1 - x)^{gamma - alpha - eta} sum_{r = alpha + eta - gamma}^infty frac{(gamma - eta )_r (gamma - alpha)_r} {(1)_r (1)_{r + gamma - alpha - eta (1 - x)^r + {}\ &quadegin{align} {} + H sum_{r = 0}^infty & frac{(gamma - alpha - eta)(gamma - eta)_r (gamma - alpha)_r} {(1 + gamma - alpha - eta)_r (1)_r} Biggl(ln(1 - x) + frac{1}{alpha + eta - gamma} + {} \ & + sum_{k = 0}^{r - 1} left(frac{1}{alpha + k} + frac{1}{eta + k} - frac{1}{1 + k} - frac{1}{alpha + eta - gamma + 1 + k}
ight) Biggr) (1 - x)^r end{align}end{align}$

Solution around infinity

Finally, we study the singularity as "x" → ∞. Since we can't study this directly, we let "x" = "s"⁻¹. Then the solution of the equation as "x" → ∞ is identical to the solution of the modified equation when "s" = 0. We had

: $egin{align} & x(1-x)y"+left{ gamma -(1+alpha +eta )x
ight}y'-alpha eta y=0 \ & frac{dy}{dx}=frac{dy}{ds} imes frac{ds}{dx}=-s^{^{2 imes frac{dy}{ds}=-s^{^{2y' \ & frac{d^{2}y}{dx^{2=frac{d}{dx}left( frac{dy}{dx}
ight)=frac{d}{dx}left( -s^{^{2 imes frac{dy}{ds}
ight)=frac{d}{ds}left( -s^{^{2 imes frac{dy}{ds}
ight) imes frac{ds}{dx} \ & left( (-2s) imes frac{dy}{ds}+(-s^{2})frac{d^{2}y}{ds^{2
ight) imes (-s^{2})=2s^{3}y'+s^{4}y" end{align}$ Hence, the equation takes the new form

: $frac{1}{s} left(1 - frac{1}{s}
ight) left(2 s^3 y' + s^4 y"
ight) + left(gamma - (1 + alpha + eta)frac{1}{s}
ight) (-s^2 y') - alpha eta y = 0$

which reduces to

: $(s^{3}-s^{2})y"+ igl((2-gamma )s^{2}+(alpha +eta -1)sigr)y'-alpha eta y = 0.$

Let

: $P_{0}(s)=-alpha eta, qquad P_{1}(s)=((2-gamma )s^{2}+(alpha +eta -1)s), qquad P_{2}(s)=(s^{3}-s^{2}).$

As we said, we shall only study the solution when "s" = 0. As we can see, this is a singular point since "P"₂(0) = 0. To see if it's regular,

: $egin{align} & underset{s o a}{mathop{lim ,frac{left( s-a
ight)P_{1}(s)}{P_{2}(s)}=underset{s o 0}{mathop{lim ,frac{left( s-0
ight)((2-gamma )s^{2}+(alpha +eta -1)s)}{(s^{3}-s^{2})}=underset{s o 0}{mathop{lim ,frac{((2-gamma )s^{2}+(alpha +eta -1)s)}{s^{2}-s} \ & =underset{s o 0}{mathop{lim ,frac{((2-gamma )s+(alpha +eta -1))}{s-1}=1-alpha -eta ext{ } ext{. } \ & underset{s o a}{mathop{lim ,frac{left( s-a
ight)^{2}P_{0}(s)}{P_{2}(s)}=underset{s o 0}{mathop{lim ,frac{left( s-0
ight)^{2}left( -alpha eta
ight)}{(s^{3}-s^{2})}=underset{x o 0}{mathop{lim ,frac{left( -alpha eta
ight)}{s-1}=alpha eta ext{ } ext{.} end{align}$

Hence, both limits exist and "s" = 0 is a regular singular point. Therefore, we assume the solution takes the form

: $y=sumlimits_{r=0}^{infty }{a_{r}s^{r+c$ with "a"₀ ≠ 0.

Hence,

: $y'=sumlimits_{r=0}^{infty }{a_{r}(r+c)s^{r+c-1$ and $y"=sumlimits_{r=0}^{infty }{a_{r}(r+c)(r+c-1)s^{r+c-2.$

Substituting in the modified hypergeometric equation we get

: $egin{align} & (s^{3}-s^{2})y"+((2-gamma )s^{2}+(alpha +eta -1)s)y'-alpha eta y=0 \ & s^{3}sumlimits_{r=0}^{infty }{a_{r}(r+c)(r+c-1)s^{r+c-2-s^{2}sumlimits_{r=0}^{infty }{a_{r}(r+c)(r+c-1)x^{r+c-2 \ & +(2-gamma )s^{2}sumlimits_{r=0}^{infty }{a_{r}(r+c)s^{r+c-1+(alpha +eta -1)ssumlimits_{r=0}^{infty }{a_{r}(r+c)s^{r+c-1-alpha eta sumlimits_{r=0}^{infty }{a_{r}s^{r+c=0end{align}$

i.e.,

: $egin{align}sumlimits_{r=0}^{infty }&{a_{r}(r+c)(r+c-1)s^{r+c+1-sumlimits_{r=0}^{infty }{a_{r}(r+c)(r+c-1)x^{r+c \ & +(2-gamma )sumlimits_{r=0}^{infty }{a_{r}(r+c)s^{r+c+1+(alpha +eta -1)sumlimits_{r=0}^{infty }{a_{r}(r+c)s^{r+c-alpha eta sumlimits_{r=0}^{infty }{a_{r}s^{r+c=0 end{align}$

In order to simplify this equation, we need all powers to be the same, equal to "r" + "c", the smallest power. Hence, we switch the indices as follows: $egin{align} & sumlimits_{r=1}^{infty }{a_{r-1}(r+c-1)(r+c-2)s^{r+c-sumlimits_{r=0}^{infty }{a_{r}(r+c)(r+c-1)x^{r+c \ & +(2-gamma )sumlimits_{r=1}^{infty }{a_{r-1}(r+c-1)s^{r+c+(alpha +eta -1)sumlimits_{r=0}^{infty }{a_{r}(r+c)s^{r+c-alpha eta sumlimits_{r=0}^{infty }{a_{r}s^{r+c=0 end{align}$

Thus, isolating the first term of the sums starting from 0 we get: $egin{align} & a_{0}left{ -(c)(c-1)+(alpha +eta -1)(c)-alpha eta
ight}s^{c}+sumlimits_{r=1}^{infty }{a_{r-1}(r+c-1)(r+c-2)s^{r+c \ & -sumlimits_{r=1}^{infty }{a_{r}(r+c)(r+c-1)x^{r+c+(2-gamma )sumlimits_{r=1}^{infty }{a_{r-1}(r+c-1)s^{r+c \ & +(alpha +eta -1)sumlimits_{r=1}^{infty }{a_{r}(r+c)s^{r+c-alpha eta sumlimits_{r=1}^{infty }{a_{r}s^{r+c=0 \ end{align}$

Now, from the linear independence of all powers of "s" (i.e., of the functions 1, "s", "s"², ..., the coefficients of "s"^k vanish for all "k". Hence, from the first term we have

: $a_{0}left{ -(c)(c-1)+(alpha +eta -1)(c)-alpha eta
ight}=0$

which is the indicial equation. Since "a"₀ ≠ 0, we have

: $(c)(-c+1+alpha +eta -1)-alpha eta )=0.,$

Hence, "c"₁ = α and "c"₂ = β.

Also, from the rest of the terms we have: $egin{align} & left{ (r+c-1)(r+c-2)+(2-gamma )(r+c-1)
ight}a_{r-1} \ & +left{ -(r+c)(r+c-1)+(alpha +eta -1)(r+c)-alpha eta
ight}a_{r}=0 end{align}$

Hence,: $egin{align} & a_{r}=-frac{left{ (r+c-1)(r+c-2)+(2-gamma )(r+c-1)
ight{left{ -(r+c)(r+c-1)+(alpha +eta -1)(r+c)-alpha eta
ighta_{r-1} \ & ext{ }=frac{left{ (r+c-1)(r+c-gamma )
ight{left{ (r+c)(r+c-alpha -eta )+alpha eta
ighta_{r-1}end{align}$ But: $egin{align} (r+c)(r+c-alpha -eta )+alpha eta &=(r+c-alpha )(r+c)-eta (r+c)+alpha eta \ & =(r+c-alpha )(r+c)-eta (r+c-alpha ).end{align}$

Hence, we get the recurrence relation: $a_{r}=frac{(r+c-1)(r+c-gamma )}{(r+c-alpha )(r+c-eta )}a_{r-1},,forall r ge 1$

Let's now simplify this relation by giving "a"_r in terms of "a"₀ instead of "a"_{"r" − 1}. From the recurrence relation,: $egin{align} & a_{1}=frac{(c)(c+1-gamma )}{(c+1-alpha )(c+1-eta )}a_{0} \ & a_{2}=frac{(c+1)(c+2-gamma )}{(c+2-alpha )(c+2-eta )}a_{1}=frac{(c+1)(c)(c+2-gamma )(c+1-gamma )}{(c+2-alpha )(c+1-alpha )(c+2-eta )(c+1-eta )}a_{0} \ & = ext{ }frac{(c)_{2}(c+1-gamma )_{2{(c+1-alpha )_{2}(c+1-eta )_{2a_{0} end{align}$

As we can see,: $a_{r}=frac{(c)_{r}(c+1-gamma )_{r{(c+1-alpha )_{r}(c+1-eta )_{ra_{0} ext{ }forall ext{r}ge ext{0}$

Hence, our assumed solution takes the form: $y=a_{0}sumlimits_{r=0}^{infty }{left( frac{(c)_{r}(c+1-gamma )_{r{(c+1-alpha )_{r}(c+1-eta )_{rs^{r+c}
ight)}$

We are now ready to study the solutions corresponding to the different cases for "c"₁ − "c"₂ = α − β.

Analysis of the solution in terms of the difference α - β of the two roots

α − β not an integer

Then "y"₁ = "y"|_{"c" = α} and "y"₂ = "y"|_{"c" = β}. Since : $y=a_{0}sumlimits_{r=0}^{infty }{left( frac{(c)_{r}(c+1-gamma )_{r{(c+1-alpha )_{r}(c+1-eta )_{rs^{r+c}
ight)}$ ,

we have : $egin{align}y_{1}&=a_{0}sumlimits_{r=0}^{infty }{left( frac{(alpha )_{r}(alpha +1-gamma )_{r{(1)_{r}(alpha +1-eta )_{rs^{r+alpha }
ight)}=a_{0}s^{alpha }_{2}F_{1}(alpha , ext{ }alpha +1-gamma ; ext{ }alpha +1-eta ; ext{ s)} \ y_{2}&=a_{0}sumlimits_{r=0}^{infty }{left( frac{(eta )_{r}(eta +1-gamma )_{r{(eta +1-alpha )_{r}(1)_{rs^{r+eta }
ight)=a_{0}s^{eta }_{2}F_{1}(eta , ext{ }eta +1-gamma ; ext{ }eta +1-alpha ; ext{ s) end{align}$

Hence, "y" = "A"′"y"₁ + "B"′"y"₂. Let "A"′"a"₀ = "A" and "B"′"a"₀ = "B". Then, noting that "s" = "x"^-1,: $egin{align} & y=Ax^{-alpha }_{2}F_{1}(alpha , ext{ }alpha +1-gamma ; ext{ }alpha +1-eta ; ext{ x}^{-1})+Bx^{-eta }_{2}F_{1}(eta , ext{ }eta +1-gamma ; ext{ }eta +1-alpha ; ext{ x}^{-1}) \ end{align}$

=α − β = 0=

Then "y"₁ = "y"|_{"c" = α}. Since α = β, we have : $y=a_{0}sumlimits_{r=0}^{infty }{left( frac{(c)_{r}(c+1-gamma )_{r{left( (c+1-alpha )_{r}
ight)^{2s^{r+c}
ight)}$

Hence,: $egin{align} & y_{1}=a_{0}sumlimits_{r=0}^{infty }{left( frac{(alpha )_{r}(alpha +1-gamma )_{r{(1)_{r}(1)_{rs^{r+alpha }
ight)}=a_{0}s^{alpha }_{2}F_{1}(alpha , ext{ }alpha +1-gamma ; ext{ 1; s)} \ & y_{2}=frac{partial y}{partial c} ext{ at }c=alpha end{align}$

To calculate this derivative, let: $egin{align} & M_{r}=frac{(c)_{r}(c+1-gamma )_{r{left( (c+1-alpha )_{r}
ight)^{2end{align}$

Then using the method in the case γ = 1 above, we get: $egin{align} & frac{partial M_{r{partial c}=frac{(c)_{r}(c+1-gamma )_{r{left( (c+1-alpha )_{r}
ight)^{2left{ sumlimits_{k=0}^{r-1}{left( frac{1}{c+k}+frac{1}{c+1-gamma +k}-frac{2}{c+1-alpha +k}
ight)}
ight} \ end{align}$

Now, : $egin{align}y=a_{0}s^{c}sumlimits_{r=0}^{infty }{left( frac{(c)_{r}(c+1-gamma )_{r{left( (c+1-alpha )_{r}
ight)^{2s^{r}
ight)}=a_{0}s^{c}sumlimits_{r=0}^{infty }{M_{r}s^{rend{align}$

Hence: $egin{align} & =a_{0}s^{c} ext{ ln(s)}sumlimits_{r=0}^{infty }{left( frac{(c)_{r}(c+1-gamma )_{r{left( (c+1-alpha )_{r}
ight)^{2s^{r}
ight)} \ & ext{ }+a_{0}s^{c}sumlimits_{r=0}^{infty }{left( frac{(c)_{r}(c+1-gamma )_{r{left( (c+1-alpha )_{r}
ight)^{2left{ sumlimits_{k=0}^{r-1}{left( frac{1}{c+k}+frac{1}{c+1-gamma +k}-frac{2}{c+1-alpha +k}
ight)}
ight}s^{r}
ight)} \ end{align}$

Hence, : $egin{align} & frac{partial y}{partial c}=a_{0}s^{c}sumlimits_{r=0}^{infty }{left( left( frac{(c)_{r}(c+1-gamma )_{r{left( (c+1-alpha )_{r}
ight)^{2
ight)left( ln s+sumlimits_{k=0}^{r-1}{left( frac{1}{c+k}+frac{1}{c+1-gamma +k}-frac{2}{c+1-alpha +k}
ight)}
ight)s^{r}
ight)} \ end{align}$

For "c" = α we get: $egin{align} & y_{2}=a_{0}s^{alpha }sumlimits_{r=0}^{infty }{left( left( frac{(alpha )_{r}(alpha +1-gamma )_{r{left( (1)_{r}
ight)^{2
ight)left( ln s+sumlimits_{k=0}^{r-1}{left( frac{1}{alpha +k}+frac{1}{alpha +1-gamma +k}-frac{2}{1+k}
ight)}
ight)s^{r}
ight)} \ &end{align}$ Hence, "y" = "C"′"y"₁ + "D"′"y"₂. Let "C"′"a"₀ = "C" and "D"′"a"₀ = "D". Noting that "s" = "x"^-1,: $y=Cx^{-alpha }_{2}F_{1}(alpha ,alpha +1-gamma ; 1; x^{-1})$

: $+Dx^{-alpha }sumlimits_{r=0}^{infty }{left( left( frac{(alpha )_{r}(alpha +1-gamma )_{r{left( (1)_{r}
ight)^{2
ight)left( ln x^{-1}+sumlimits_{k=0}^{r-1}{left( frac{1}{alpha +k}+frac{1}{alpha +1-gamma +k}-frac{2}{1+k}
ight)}
ight)x^{-r}
ight)}$

α − β an integer and α − β ≠ 0

α − β > 0

From the recurrence relation

: $a_{r}=frac{(r+c-1)(r+c-gamma )}{(r+c-alpha )(r+c-eta )}a_{r-1}$ we see that when "c" = β (the smaller root), "a"_{α - β} → ∞. Hence, we must make the substitution "a"₀ = "b"₀("c" − "c"_"i"), where "c"_"i" is the root for which our solution is infinite. Hence, we take "a"₀ = "b"₀("c" − β) and our assumed solution takes the new form

: $y_{b}=b_{0}sumlimits_{r=0}^{infty }{left( frac{(c-eta )(c)_{r}(c+1-gamma )_{r{(c+1-alpha )_{r}(c+1-eta )_{rs^{r+c}
ight)}$

Then "y"₁ = "y"_b|_{"c" = β}. As we can see, all terms before

: $frac{(c-eta )(c)_{alpha -eta }(c+1-gamma )_{alpha -eta {(c+1-alpha )_{alpha -eta }(c+1-eta )_{alpha -eta s^{alpha -eta }$

vanish because of the "c" − β in the numerator.

But starting from this term, the "c" − β in the numerator vanishes. To see this, note that : $(c+1-alpha )_{alpha -eta } =(c+1-alpha )(c+2-alpha )cdots(c-eta ).$

Hence, our solution takes the form : $egin{align} & y_{1}=b_{0}left( frac{(eta )_{alpha -eta }(eta +1-gamma )_{alpha -eta {(eta +1-alpha )_{alpha -eta -1}(1)_{alpha -eta s^{alpha -eta }+frac{(eta )_{alpha -eta +1}(eta +1-gamma )_{alpha -eta +1{(eta +1-alpha )_{alpha -eta -1}(1)(1)_{alpha -eta +1s^{alpha -eta +1}+...
ight) \ & =frac{b_{0{(eta +1-alpha )_{alpha -eta -1sumlimits_{r=alpha -eta }^{infty }{left( frac{(eta )_{r}(eta +1-gamma )_{r{(1)_{r}(1)_{r+eta -alpha s^{r}
ight)} \ & \ end{align}$

Now,: $y_{2}=left.frac{partial y_{b{partial c}
ight|_{c=alpha}.$ To calculate this derivative, let : $M_{r}=frac{(c-eta )(c)_{r}(c+1-gamma )_{r{(c+1-alpha )_{r}(c+1-eta )_{r.$

Then using the method in the case γ = 1 above we get: $egin{align} frac{partial M_r}{partial c} &= frac{(c - eta)(c)_r (c + 1 - gamma)_r}{(c + 1 - alpha)_r (c + 1 - eta)_r}Biggl(frac{1}{c - eta} + \ &qquad + sum_{k = 0}^{r - 1} left(frac{1}{c + k} + frac{1}{c + 1 - gamma + k} -frac{1}{c + 1 - alpha + k} - frac{1}{c + 1 - eta + k}
ight) Biggr)end{align}$

Now,: $y_{b}=b_{0}sumlimits_{r=0}^{infty }{left( frac{(c-eta )(c)_{r}(c+1-gamma )_{r{(c+1-alpha )_{r}(c+1-eta )_{rs^{r+c}
ight)}=b_{0}x^{c}sumlimits_{r=0}^{infty }{M_{r}s^{r$

Hence,: $egin{align} frac{partial y}{partial c} &= b_0 s^c ln(s) sum_{r = 0}^infty frac{(c - eta)(c)_r (c + 1 - gamma)_r} {(c + 1 - alpha)_r (c + 1 - eta)_r} s^r \ &quadegin{align} {}+ b_0 s^c sum_{r = 0}^infty & frac{(c - eta) (c)_r (c + 1 - gamma)_r} {(c + 1 - alpha)_r (c + 1 - eta)_r} Biggl(frac{1}{c - eta} + \ &+ sum_{k = 0}^{r - 1} left(frac{1}{c + k} + frac{1}{c + 1 - gamma + k} - frac{1}{c + 1 - alpha + k} - frac{1}{c + 1 - eta + k}
ight) Biggr) s^r end{align}end{align}$

Hence,: $egin{align} frac{partial y}{partial c} = b_0 s^c sum_{r = 0}^infty &frac{(c - eta)(c)_r (c + 1 - gamma)_r} {(c + 1 - alpha)_r (c + 1 - eta)_r} Biggl(ln s + frac{1}{c - eta } + \ &+ sum_{k = 0}^{r - 1} left(frac{1}{c + k} + frac{1}{c + 1 - gamma + k} - frac{1}{c + 1 - alpha + k} - frac{1}{c + 1 - eta + k}
ight) Biggr) s^{r}end{align}$

At "c" = α we get "y"₂. Hence, "y" = "E"′"y"₁ + "F"′"y"₂. Let "E"′"b"₀ = "E" and "F"′"b"₀ = "F". Noting that "s" = "x"^-1 we get: $egin{align} y &= frac{E}{(eta + 1 - alpha)_{alpha - eta - 1 sum_{r = alpha - eta}^infty frac{(eta)_r (eta + 1 - gamma)_r} {(1)_r (1)_{r + eta - alpha x^{-r} \ &quadegin{align} {} + F x^{-alpha} sum_{r = 0}^infty & frac{(alpha - eta) (alpha)_r (alpha + 1 - gamma)_r} {(1)_r (alpha + 1 - eta)_r} Biggl(ln x^{-1} + frac{1}{alpha -eta } \ &+ sum_{k = 0}^{r - 1} left(frac{1}{alpha + k} + frac{1}{alpha + 1 + k - gamma} -frac{1}{1 + k} -frac{1}{alpha + 1 + k - eta}
ight) Biggr) x^{-r} end{align}end{align}$

α − β < 0

From the symmetry of the situation here, we see that

: $egin{align} y &= frac{G}{(alpha + 1 - eta)_{eta - alpha - 1 sum_{r = eta - alpha}^infty frac{(alpha)_r (alpha + 1 - gamma)_r} {(1)_r (1)_{r + alpha - eta x^{-r} \ &quadegin{align} {} + H x^{-eta} sum_{r = 0}^infty &frac{(eta - alpha) (eta)_r (eta + 1 - gamma)_r} {(1)_r (eta + 1 - alpha)_r} Biggl(ln x^{-1} + frac{1}{eta - alpha } \ &+ sum_{k = 0}^{r - 1} left(frac{1}{eta + k} + frac{1}{eta + 1 + k - gamma} - frac{1}{1 + k} - frac{1}{eta + 1 + k - alpha}
ight) Biggr) x^{-r} end{align}end{align}$

Reference

* cite book|author=Ian Sneddon
title=Special functions of mathematical physics and chemistry
year=1966
publisher=OLIVER B
id=ISBN 978-0050013342

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