Regular Singular Point at \( x = 1 \)
As said before, apart from the origin \( x = 0 \), there is another regular singular point \( x = 1 \) for the hypergeometric equation ((1) of the last tutorial). To find the appropriate solution near it, we can simply utilize a change of variable \( t = 1-x \), so that the point \( t = 0 \) will correspond to \( x = 1\). Note that subsequently \( dy/dx = dy/d(1-t) = -dy/dt \) and similarly \( d^2y/dx^2 = d^2y/dt^2\). Then it becomes
\begin{align}
(1-t)t \frac{d^2y}{dt^2} + [c-(a+b+1)(1-t)](-\frac{dy}{dt}) -aby &= 0 \\
t(1-t) \frac{d^2y}{dt^2} + [(a+b+1-c)-(a+b+1)t]\frac{dy}{dt} -aby &= 0 \tag{1}
\end{align}
This is just another hypergeometric equation in \(t\) with the same \( a, b \), but \( c \) replaced by \( a+b+1-c \), and so by (18) of the last tutorial, the general solution at \(t = 0\) and hence \(x = 1\) will be
\begin{align}
y ={}& AF(a,b,a+b+1-c; t) + Bt^{1-(a+b+1-c)} \\
& F(a-(a+b+1-c)+1, b-(a+b+1-c)+1, 2-(a+b+1-c); t) \\
={}& AF(a,b,a+b+1-c; 1-x) \\
& + B(1-x)^{c-a-b}F(c-b, c-a, c-a-b+1; 1-x) \tag{2}
\end{align}
provided that \( c -a-b \) is not an integer. (Otherwise, we follow the discussion at the end of the last post in a similar fashion.)
Notable Special Values for Hypergeometric Function
Next, we briefly write down some special values or common functions that the hypergeometric equation can take at specific combinations of the parameters: (This list is mainly based on the reference book Mathematical Methods for Physics and Engineering by Riley, Hobson, and Bence.)
\begin{align}
\begin{aligned}
F(a,b,b;x) &= (1-x)^{-a} \\
F(1,1,2;-x) &= \frac{\ln(1+x)}{x} \\
F(-\frac{1}{2},b,b;\sin^2 x) &= |\cos x| \\
F(\frac{1}{2},b,b;\sin^2 x) &= |\sec x| \\
F(\frac{1}{2},\frac{1}{2},\frac{3}{2};x^2) &= \frac{\arcsin x}{x} \\
F(\frac{1}{2},1,\frac{3}{2};-x^2) &= \frac{\arctan x}{x} \\
F(\frac{1}{2},1,\frac{3}{2};x^2) &= \frac{1}{2x} \ln (\frac{1+x}{1-x}) = \frac{\tanh^{-1} x}{x} \\
\lim_{a\to\infty} F(a,a,\frac{1}{2};-\frac{x^2}{4a^2}) &= \cos x \\
\lim_{a\to\infty} F(a,a,\frac{3}{2};-\frac{x^2}{4a^2}) &= \frac{\sin x}{x} \\
\end{aligned}\tag{3}
\end{align}
Here we will prove some of them. For the first one, we recall (9) in the last tutorial:
\begin{align}
F(a,b,b;x) &= \frac{\Gamma(b)}{\Gamma(a)\Gamma(b)} \sum_{n=0}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(b+n)} \frac{x^n}{n!} \\
&= \sum_{n=0}^{\infty} \frac{\Gamma(a+n)}{\Gamma(a)} \frac{x^n}{n!} \\
&= \sum_{n=0}^{\infty} (a)(a+1)\cdots(a+n-1)\frac{x^n}{n!} \\
&= \sum_{n=0}^{\infty} (-a)(-(a+1))\cdots(-(a+n-1))\frac{(-x)^n}{n!} \\
&= (1-x)^{-a} \tag{4}
\end{align}
The third one (as well as the fourth) is a direct application of (4):
\begin{align}
F(-\frac{1}{2},b,b;\sin^2 x) = (1-\sin^2 x)^{1/2} = (\cos^2 x)^{1/2} = |\cos x| \tag{5}
\end{align}
Finally, the second last one: (assuming we can put the limit inside the summation and using (20) in the last post)
\begin{align}
&\quad \lim_{a\to\infty} F(a,a,\frac{1}{2};-\frac{x^2}{4a^2}) \\
&= \lim_{a\to\infty} \frac{\Gamma(\frac{1}{2})}{\Gamma(a)\Gamma(a)} \sum_{n=0}^{\infty} \frac{\Gamma(a+n)\Gamma(a+n)}{\Gamma(\frac{1}{2}+n)} \frac{(-x^2/4a^2)^n}{n!} \\
&= \sum_{n=0}^{\infty} \frac{1}{(1/2)(3/2)\cdots(1/2+n-1)} \\
&\quad \left(\lim_{a\to\infty} \frac{\Gamma(a+n)\Gamma(a+n)}{\Gamma(a)\Gamma(a)} \frac{(-1)^n(x/2a)^{2n}}{n!} \right) \\
&= \sum_{n=0}^{\infty} \frac{1}{(1/2)(3/2)\cdots(1/2+n-1)(2^n)} \\
&\quad \left(\lim_{a\to\infty} \frac{[(a)(a+1)\cdots(a+n-1)]^2}{(a^n)^2} \frac{(-1)^nx^{2n}}{(2^n)n!} \right) \\
&= \sum_{n=0}^{\infty} \frac{1}{(1)(3)\cdots(2n-1)} (1) \frac{(-1)^nx^{2n}}{(2)(4)\cdots(2n)} \\
&= \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!} = \cos x \tag{6}
\end{align}
Integral Representation
The hypergeometric function can also be expressed as an integral. Recall (9) in the last tutorial, we can derive that
\begin{align}
&\quad F(a,b,c;x) \\
&= \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \sum_{n=0}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)} \frac{x^n}{n!} \\
&= \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)\Gamma(c-b)} \sum_{n=0}^{\infty} \Gamma(a+n)B(b+n,c-b)\frac{x^n}{n!} \\
&= \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)\Gamma(c-b)} \sum_{n=0}^{\infty} \Gamma(a+n)\frac{x^n}{n!} \int_0^1 t^{b+n-1}(1-t)^{c-b-1} dt \\
&= \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_0^1 t^{b-1}(1-t)^{c-b-1} \sum_{n=0}^{\infty} \frac{\Gamma(a+n)}{\Gamma(a)}\frac{(tx)^n}{n!} dt \\
&= \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_0^1 t^{b-1}(1-t)^{c-b-1} (1-tx)^{-a} dt \tag{7}
\end{align}
In the third line, we have used the relation between the Gamma/Beta function \( B(m,n) = \Gamma(m)\Gamma(n)/\Gamma(m+n)\). And in the fourth line, we invoke the integral form of the Beta Function \( B(m,n) = \int_0^1 t^{m-1}(1-t)^{n-1} dt \). Finally, we exchange the summation and integral, and rewrite the summation using the Taylor series of \( (1-tx)^{-a} \). Same as the Beta function, it requires \( c > b > 0 \) to converge due to the factor \(t^{b-1}\) and \((1-t)^{c-b-1}\) in the integral.
Relationships between Hypergeometric Functions
Here we state two of the most frequently used relationships between hypergeometric functions, as provided by the book Mathematical Methods for Physics and Engineering by Riley, Hobson, and Bence. The first one starts from (7) and undergoes a change of variable \( t = (1-u)/(1-ux) \) (so that when \(t = 0\), \( u=1\), and \(t = 1\), \( u = 0\), and \( dt = -(1-x)/(1-ux)^2 du\)):
\begin{align}
&\quad F(a,b,c;x) \\
&= \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_0^1 t^{b-1}(1-t)^{c-b-1} (1-tx)^{-a} dt \\
&= \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \\
&\quad \int_0^1 (\frac{1-u}{1-ux})^{b-1}(1-\frac{1-u}{1-ux})^{c-b-1} (1-\frac{x(1-u)}{1-ux})^{-a} d(\frac{1-u}{1-ux}) \\
&= \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \\
&\quad \int_1^0 \frac{(1-u)^{b-1}}{(1-ux)^{b-1}}\frac{(u(1-x))^{c-b-1}}{(1-ux)^{c-b-1}} \frac{(1-x)^{-a}}{(1-ux)^{-a}} \frac{-(1-x)}{(1-ux)^2} du \\
&= \frac{\Gamma(c)}{\Gamma(c-b)\Gamma(b)} \int_0^1 (1-x)^{c-b-a}u^{c-b-1} (1-u)^{b-1} (1-ux)^{-(c-a)} du \\
&= (1-x)^{c-b-a}F(c-a,c-b,c;x) \tag{8}
\end{align}
In the last, we apply (7) again, supplied with new parameters and dummy variable. The second one is by differentiating (9) in the last tutorial (term-by-term, enabled by analyticity):
\begin{align}
&\quad F'(a,b,c;x) \\
&= \frac{d}{dx} \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \sum_{n=0}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)} \frac{x^n}{n!} \\
&= \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \sum_{n=1}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)} \frac{nx^{n-1}}{n!} \\
&= \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \sum_{n=0}^{\infty} \frac{\Gamma(a+n+1)\Gamma(b+n+1)}{\Gamma(c+n+1)} \frac{(n+1)x^n}{(n+1)!} \\
&= \frac{ab}{c} \frac{\Gamma(c+1)}{\Gamma(a+1)\Gamma(b+1)} \sum_{n=0}^{\infty} \frac{\Gamma((a+1)+n)\Gamma((b+1)+n)}{\Gamma((c+1)+n)} \frac{x^n}{n!} \\
&= \frac{ab}{c}F(a+1,b+1,c+1;x) \tag{9}
\end{align}
where we have adjusted the index \(n\) by \(1\), used the fact that \(\Gamma(n+1) = n\Gamma(n) \), and applied (9) again with the parameters all increased by \(1\).
Exercise
Verify the second formula in (3) by considering the ODE
\begin{align}
-x(1+x)y^{\prime\prime} -(2+3x)y’ -y = 0 \tag{10}
\end{align}
Answer
Note that (10) is a hypergeometric equation in disguise by making the substitution \( x \to -z\):
\begin{align}
z(1+(-z))y^{\prime\prime} -(2+3(-z))(-y’) -y &= 0 \\
z(1-z)y^{\prime\prime} + (2-3z)y’ -y &= 0
\end{align}
with \(a=b=1\), \(c=2\), and hence by (18) in the last post, the two linearly independent solutions will consist of
\begin{align}
y_1 &= F(1,1,2;z) \\
y_2 &= z^{-1}F(1-2+1,1-2+1,2-2;z) \\
&= z^{-1}F(0,0,0;z) = \frac{1}{z} = -\frac{1}{x}
\end{align}
By a direct verification, we can see that \(\ln(1+x)/x = -\ln(1-z)/z\) is a solution to the equation above:
\begin{align}
&\quad \begin{aligned}&z(1-z)(-\frac{2\ln(1-z)}{z^3}-\frac{1}{(1-z)z^2}-\frac{1}{z^2(1-z)}+\frac{1}{z(1-z)^2}) \\
&+ (2-3z)(\frac{\ln(1-z)}{z^2}+\frac{1}{z(1-z)}) -(-\frac{\ln(1-z)}{z})\end{aligned} \\
&= -\frac{(2-2z)\ln(1-z)}{z^2} -\frac{2}{z} + \frac{1}{1-z} + (2-3z)\frac{\ln(1-z)}{z^2} \\
&\quad + \frac{2}{z(1-z)} -\frac{3}{1-z} + \frac{z\ln(1-z)}{z^2} \\
&= -\frac{2}{z} -\frac{2}{1-z} + (\frac{2}{z} + \frac{2}{1-z}) = 0
\end{align}
Therefore, \(\ln(1+x)/x = -\ln(1-z)/z\), as a series solution, must coincide with the first solution \(F(1,1,2;-x) = F(1,1,2;z)\) to (10) and we are done.
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