Approximating hypergeometric function F(1,1+a,2+a,z) for z->1

Question

in my studies a normalization constant for a pmf includes the hypergeometric function ${}_2F_1(1,1+a,2+a,z)$

The parameters are in the range $0.99<z<1$ and $0<a<5$.

I have tried some approximations for presentations including incomplete Beta and Lerch-Phi functions but z being near 1 does not give reasonable results.

Inserting the parameters yield for example

${}_2F_1(1,1+a,2+a,z)=(1+a) \Phi(z,1,1+a)=(1+a)\sum_{n=0}^\infty \frac{z^n}{1+a+n}$

$=(1+a) \int_0^\infty \frac{\exp(-(1+a)t)}{1-z \exp(-t)} dt$

$=(1+a) \int_0^1 \frac{t^a}{1-tz} dt$

Plotting the integrand from the last line, it looks like some power function could be a good fit, but maybe there are other alternatives, for example a transformation with $1-z$.

Best regards

Answer 1

Your last integral formula for the hypergeometric function is off; according to Wikipedia it should be

$$ {}_2F_1(1,1+a,2+a,z) = (1+a) \int_0^1 \frac{t^a}{1-tz}\,dt. $$

Now if we make the change of variables $tz = s$ we get

$$ {}_2F_1(1,1+a,2+a,z) = \frac{1+a}{z^{1+a}} \int_0^z \frac{s^a}{1-s}\,ds. \tag{1} $$

Heuristically, the integrand is largest when $s \approx z \approx 1$, and there we have

$$ \frac{s^a}{1-s} \approx \frac{1}{1-s}, $$

so we might expect that

$$ \int_0^z \frac{s^a}{1-s}\,ds \approx \int_0^z \frac{1}{1-s}\,ds = -\log(1-z). $$

To quantify this we'll essentially pull out the largest part of the integrand. To wit, we write

$$ \frac{s^a}{1-s} = \frac{s^a - 1 + 1}{1-s} = \frac{1}{1-s} + \frac{s^a-1}{1-s}, $$

and split the integral up like

$$ \begin{align} \int_0^z \frac{s^a}{1-s}\,ds &= \int_0^z \frac{1}{1-s}\,ds + \int_0^z \frac{s^a-1}{1-s}\,ds \\ &= -\log(1-z) + \int_0^z \frac{s^a-1}{1-s}\,ds. \tag{2} \end{align} $$

The remaining integral actually converges as $z \to 1^-$. In fact we have

$$ \lim_{z \to 1^-} \int_0^z \frac{s^a-1}{1-s}\,ds = \int_0^1 \frac{s^a-1}{1-s}\,ds = -H_a, $$

where $H_a$ is a harmonic number. So, we can actually rewrite $(2)$ as

$$ \begin{align} \int_0^z \frac{s^a}{1-s}\,ds &= -\log(1-z) + \int_0^1 \frac{s^a-1}{1-s}\,ds - \int_z^1 \frac{s^a-1}{1-s}\,ds \\ &= -\log(1-z) - H_a - \int_z^1 \frac{s^a-1}{1-s}\,ds, \tag{3} \end{align} $$

and now the last integral here tends to zero as $z \to 1^-$. Combining this with $(1)$ we see that, in this sense,

$$ {}_2F_1(1,1+a,2+a,z) \approx -\frac{1+a}{z^{1+a}} \Bigl(\log(1-z) + H_a\Bigr) $$ as $z \to 1^-$.

If more accuracy is required we could further study the remaining integral in $(3)$. The integrand is analytic near $s = 1$, so we can expand it in power series and integrate term-by-term. Indeed, by the binomial theorem we have

$$ s^a = \sum_{n=0}^{\infty} \binom{a}{n} (s-1)^n, $$

so

$$ \frac{s^a-1}{1-s} = -\sum_{n=1}^\infty \binom{a}{n} (s-1)^{n-1}, $$

thus

$$ \begin{align} \int_z^1 \frac{s^a-1}{1-s} \,ds &= - \sum_{n=1}^{\infty} \binom{a}{n} \int_z^1 (s-1)^{n-1}\,ds \\ &= \sum_{n=1}^{\infty} \binom{a}{n} \frac{(z-1)^n}{n}. \end{align} $$

So, combining this with $(3)$ and $(1)$,

$$ {}_2F_1(1,1+a,2+a,z) = -\frac{1+a}{z^{1+a}} \left(\log(1-z) + H_a + \sum_{n=1}^{\infty} \binom{a}{n} \frac{(z-1)^n}{n}\right). $$

The series inside may be truncated after as many terms as are needed to obtain the desired accuracy.