# Comptes Rendus
Mathématique

Bujar Xh. Fejzullahu

On the maximum value of a confluent hypergeometric function

Volume 359, issue 10 (2021), p. 1217-1224

Published online: 4 January 2022

https://doi.org/10.5802/crmath.256

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(see [3, 13], and references therein). For instance, Luke [13], among others, proved the following inequalities

$$ 
\begin{aligned}
e^{\frac{a}{b}x} < {}_1F_1(a; b; x) &< 1 - \frac{a}{b} + \frac{a}{b}e^x, & x > 0, b > a > 0, \\
{}_1F_1(a; b; x) &< \frac{(b-1)e^x}{(b-a-1)(1+x)}, & x > 0, b-1 > a > 0.
\end{aligned}
$$

We remark that when $x > 0$, $b > a$ and $b > \frac{1}{2}$, Love [11, Corollary 2] showed that

$$ |{}_1F_1(a; b; x)| \le \frac{\Gamma(b-a)e^x}{|\Gamma(1-a)|\Gamma(b)}. \quad (3) $$

Unfortunately, the conditions given for (3) do not seem to be correct. For instance, for $a \in \mathbb{N}$ and $b > a$, (3) gives ${}_1F_1(a; b; x) = 0$ for any $x > 0$.

When the parameters verify the so called oscillatory conditions $a < 0$ and $b-a > 1$, the estimate of ${}_1F_1(a; b; x)$ is much more complicated and, to the author's knowledge, has not been studied in the literature, except for the case when $a$ is a negative integer and $b > 0$ (see, for instance, [9, 10, 17, 21, 23], and references therein).

It is well known that, for $a < 0$ and $b > 0$, ${}_1F_1(a; b; x)$ has a finite number of real zeros (see [18, Section 13.9])

$$ x_{a,b}^1 < x_{a,b}^2 < \dots $$

Based on the Kummer transformation

$$ {}_1F_1(a; b; x) = e^{x} {}_1F_1(b - a; b; -x), $$

it follows that the real zeros of ${}_1F_1(a; b; x)$ are positive when $a < 0$ and $b > 0$.

In this paper, we use the Sonin-Pólya theorem as well as the Watson-Glaeske product formula for confluent hypergeometric functions to study the maximum value of ${}_1F_1(a; b; x)$ with oscillatory conditions of parameters.

Here is our main results.

**Theorem 1.** For $a < 0$ and $b > 1$

$$ \max_{x \ge 0} e^{-x} |{}_1F_1(a; b; x)| = 1. \quad (4) $$

Moreover, this maximum value is attained only when $x = 0$.

**Corollary 2.** When $a < 0$ and $b > 1$, let $\xi_k$, $k = 1, \dots$, be the successive maxima of $y(x) = e^{-x} {}_1F_1(a; b; x)$ arranged in increasing order, and let $j_{b,k}$ be the $k$-th positive zero of the Bessel function $J_b(x)$. Then,

$$ y^2(\xi_i) - y^2(\xi_j) < \frac{b-a}{b^2} \left( \frac{2b-1}{2} \Delta \xi_{ij}^2 + \frac{2}{3} \Delta \xi_{ij}^3 \right), \quad i < j, $$

where $\Delta \xi_{ij}^k = \xi_j^k - \xi_i^k$ and

$$ \xi_k = \frac{j_{b,k}^2}{2b - 4a + 2} \left( 1 + \frac{2(b^2 - 1) + j_{b,k}^2}{3(2b - 4a + 2)^2} \right) + O\left(\frac{1}{a^5}\right), \quad \text{as } a \to -\infty. $$

We remark that from the asymptotics (see [14, Section 6.8.2])

$$ e^{-x} {}_1F_1(a; b; x) = \frac{\Gamma(b)}{\Gamma(a)} x^{a-b} [1 + O(x^{-1})], \quad x \to +\infty, $$

where $a$ is not a negative integer or zero, it follows that $e^{-x}$ in Theorem 1 cannot be replaced by any $e^{-cx}$, $0 \le c < 1$. In the case when $a$ is a negative integer or zero, we have (see [23])

$$ \max_{x \ge 0} e^{-\frac{x}{2}} |{}_1F_1(a; b; x)| = 1, \quad (5) $$

where $b > 1$.
On the other hand, in the oscillation region of $_1F_1(a; b; x)$, we have the following result analogous to (5).

**Theorem 3.** Let $a < 0$ and $b > 1$. For $0 \le x < \frac{(2b-1)(b-2a)}{b}$
$$e^{-\frac{x}{2}}|_1 F_1(a; b; x)| \le 1. \qquad (6)$$

Furthermore, for $0 \le x < \frac{(2b-1)(b-2a)}{2b}$
$$e^{-\frac{x}{2}}|_1 F_1(a; b; x)| \le \sqrt{\mathcal{M}(x, b)}, \qquad (7)$$

where

$$\mathcal{M}(x, b) = \frac{b-1}{\pi} \int_{0}^{\pi} \int_{0}^{\pi} |\cos(x \sin \theta \cos \psi)| (\sin \psi)^{2b-3} (\sin \theta)^{2b-2} d\psi d\theta$$

and it has the property

$$0 < \mathcal{M}(x, b) < \mathcal{M}(0, x) = 1, \quad x > 0.$$

Consequently, by applying (4) and (5), we obtain the following inequalities for the Gauss hypergeometric function $_2F_1$.

**Corollary 4.** Let $a < 0$, $b > 1$ and $\Re\sigma > 0$.

(i) If a is not a negative integer and $0 \le x < \Re e(z)$

$$| {}_2F_1(\sigma, a; b; \frac{x}{z}) | \le \sqrt{\cosh(\pi\Im m(\sigma))} |z^\sigma| |\Re e(z) - x|^{-\Re e(\sigma)}. \qquad (8)$$

In particular, for $\sigma > 0$ and $0 \le x < 1$

$$| {}_2F_1(\sigma, a; b; x) | \le (1-x)^{-\sigma}.$$

(ii) If a is a negative integer and $0 \le x < 2\Re e(z)$

$$| {}_2F_1(\sigma, a; b; \frac{x}{z}) | \le \sqrt{\cosh(\pi\Im m(\sigma))} |z^\sigma| |\Re e(z) - \frac{x}{2}|^{-\Re e(\sigma)}. $$

In particular, for $\sigma > 0$ and $0 \le x < 2$

$$| {}_2F_1(-n, \sigma; b; x) | \le \left(1 - \frac{x}{2}\right)^{-\sigma}, \quad n \in \mathbb{N} \cup \{0\}.$$

Under condition $a > 0$, several lower and upper bound inequalities for $_2F_1(\sigma, a; b; x)$ have been derived in the literature using different approaches (e.g. [2–4, 6, 13, 22] and references therein). For instance, in [13, Theorem 13], Luke gave the following two-sided bounds

$$(1 - \frac{a}{b}x)^{-\sigma} < {}_2F_1(\sigma, a; b; x) < 1 - \frac{a}{b} + \frac{a}{b}(1 - x)^{-\sigma}, \quad 0 < x < 1, 0 < \sigma, 0 < a < b,$$

whereas Karp and Sitnik [6, Theorem 5] showed that

$$ {}_2F_1(\sigma, a; b; x) < \left(1 - \frac{a}{b-1}x\right)^{-\sigma}, \quad 0 < x < 1, 0 < \sigma \le 1, 1 < a+1 < b. $$

On the other hand, in [22] the authors derived some inequalities for the Gauss hypergeometric function $_2F_1(\sigma, a; b, x)$ when $-1 < a < 0$, $1 < b < 2$, $0 < \sigma < 1$, and $x \in (0, 1)$. We remark that when $a$ is a negative integer or zero, the estimate of the polynomial $_2F_1(a, \sigma; b; x)$ has been considered in several papers from different point of views (see for instance [7, 8] and references therein)

## 2. Proof of the main results

One of the main tool that we need for our purpose is the well-known Sonin-Pólya theorem (see [21, footnote to Theorem 7.31.1]) in the following form given by Szegö. Notice that this theorem was used by Szegö [21] in a similar context to study the successive relative maxima of classical orthogonal polynomials.
## The Sonin-Pólya theorem

Suppose that a function $y = y(x)$ satisfies on an interval $I \subset \mathbb{R}$ the differential equation

$$ (py')' + qy = 0, \quad (9) $$

where $p = p(x) > 0$, $q = q(x) > 0$ and both $p'$ and $q'$ are continuous on that interval. Define Sonin's function by

$$ S(x) := y^2(x) + \frac{p(x)}{q(x)} y'^2(x), \quad (10) $$

then we observe that

$$ S'(x) = -[p(x)q(x)]' \left[ \frac{y'(x)}{q(x)} \right]^2, \quad (11) $$

by which successive relative maxima of $y^2$ form an increasing or decreasing sequence according as $pq$ decreases or increases on the corresponding interval.

Now, we can prove our main results.

**Proof of Theorem 1.** From (1), the corresponding differential equation for $y(x) = e^{-x} F_1(a; b; x)$ is

$$ xy'' + (b + x)y' + (b - a)y = 0. $$

By writing it in the self-adjoint form

$$ (x^b e^x y)' + (b - a)x^{b-1} e^x y = 0, $$

we see that

$$ p(x) = x^b e^x, \quad q(x) = (b - a)x^{b-1} e^x $$

and

$$ [p(x)q(x)]' = (b - a)(2b - 1 + 2x)x^{2b-1}e^{2x}. $$

Thus, if $a < 0$ and $b > 1$, the successive relative maxima of $|e^{-x} F_1(a; b; x)|$ are decreasing on $[0, \infty)$

and

$$ [e^{-x} F_1(a; b; x)]^2 \le S(x) \le S(0) = y^2(0) = 1, \quad x \ge 0. $$

This proves (4).

□

**Proof of Corollary 2.** We observe that, using the differential equation

$$ \frac{d}{dx}[e^{-x} F_1(a; b; x)] = -\frac{b-a}{b} e^{-x} F_1(a; b+1; x), \quad (12) $$

$\xi_k = x_{a,b+1}^k$, for all $k = 1, \dots$.

Thus, from (10) and (11) one has

$$ y^2(\xi_j) - y^2(\xi_i) = -\frac{1}{b-a} \int_{\xi_i}^{\xi_j} x(2b-1+2x)[y'(x)]^2 dx, \quad i < j. $$

Now we can apply (4) and (12) to yield

$$ \begin{aligned} y^2(\xi_i) - y^2(\xi_j) &< \frac{b-a}{b^2} \int_{\xi_i}^{\xi_j} x(2b-1+2x) dx \\ & = \frac{b-a}{b^2} \left( \frac{2b-1}{2} \Delta \xi_{ij}^2 + \frac{2}{3} \Delta \xi_{ij}^3 \right), \end{aligned} $$

where $\Delta \xi_{ij}^k = \xi_j^k - \xi_i^k$.

Finally, taking into account that the k-th positive zero $x_{a,b}^k$ can be approximated by (see [18, Section 13.9])

$$ \frac{j_{b-1,k}^2}{2b-4a} \left( 1 + \frac{2b(b-2) + j_{b-1,k}^2}{3(2b-4a)^2} \right) + O\left( \frac{1}{a^5} \right), \quad \text{as } a \to -\infty, $$

we can achieve the proof of the corollary.

□
**Proof of Theorem 3.** For the proof of (6), we proceed as in the proof of (4). According to (1), it is straightforward to check that the function $y(x) = e^{-\frac{x}{2}} {}_1F_1(a; b; x)$ satisfies

$$xy''(x) + by'(x) + \frac{2b-4a-x}{4}y(x) = 0. \quad (13)$$

In its self-adjoint form equation (13) becomes

$$(x^b y'(x))' + \frac{2b-4a-x}{4} x^{b-1} y(x) = 0,$$

which corresponds to equation (9) with

$$p(x) = x^b, \quad q(x) = \frac{2b-4a-x}{4} x^{b-1},$$

and

$$[p(x)q(x)]' = \frac{x^{2b-2}}{2} [(2b-1)(b-2a) - bx].$$

Thus, for $a < 0$ and $b > 1$, the successive relative maxima of $|e^{-\frac{x}{2}} {}_1F_1(a; b; x)|$ are decreasing if $0 < x < \frac{(2b-1)(b-2a)}{b}$ and increasing if $\frac{(2b-1)(b-2a)}{b} < x < 2b-4a$. This completes the proof of (6).

We now continue with the proof of (7). Our starting point in the proof is the Glaeske [5] product formula for Laguerre functions, which in terms of the confluent hypergeometric functions can be written as

$${}_1F_1(a; b; x) \cdot {}_1F_1(a; b; y) = \frac{\Gamma(b)}{\sqrt{\pi}} \int_0^\pi e^{-\sqrt{xy}\cos\theta} (\sin\theta)^{2b-2} \mathcal{J}_{b-\frac{3}{2}}(\sqrt{xy}\sin\theta) \\ \times {}_1F_1(a; b; x+y+2\sqrt{xy}\cos\theta) \, d\theta, \quad (14)$$

where $x, y \ge 0$, $\Re(e) > \frac{1}{2}$, and $\mathcal{J}_v(z) := \left(\frac{z}{2}\right)^{-v} J_v(z)$. For $a = -n$, $n \in \mathbb{N}$, equation (14) was first obtained by Watson [23] and later on by several authors using quite different methods (see [5, 7, 16, 19, 20]), whereas in [15] Markett gave another, analytic proof of Glaeske's result. In Appendix A, we give another simple proof of (14).

Using Poisson's integral (see [23, 3.3])

$$\mathcal{J}_v(z) = \frac{1}{\sqrt{\pi\Gamma\left(v+\frac{1}{2}\right)}} \int_0^\pi e^{iz\cos\psi} (\sin\psi)^{2v} d\psi, \quad \Re(e) > -\frac{1}{2},$$

the product formula (14) becomes

$$ \begin{aligned} {}_1F_1(a; b; x) \cdot {}_1F_1(a; b; y) &= \frac{b-1}{\pi} \int_0^\pi \int_0^\pi e^{-\sqrt{xy}\cos\theta + i\sqrt{xy}\sin\theta\cos\psi} (\sin\psi)^{2b-3}(\sin\theta)^{2b-2} \\ & \qquad \times {}_1F_1(a, b, x+y+2\sqrt{xy}\cos\theta) \, d\psi \, d\theta \\ &= \frac{b-1}{\pi} \int_0^\pi \int_0^\pi e^{-\sqrt{xy}\cos\theta} \cos(\sqrt{xy}\sin\theta\cos\psi) (\sin\psi)^{2b-3} \\ & \qquad \times (\sin\theta)^{2b-2} {}_1F_1(a, b, x+y+2\sqrt{xy}\cos\theta) \, d\psi \, d\theta. \end{aligned} \quad (15) $$

We put $x = y$ in (15) and multiply the obtained relation by $e^{-x}$. As a result, we obtain

$$\left[e^{-\frac{x}{2}} {}_1F_1(a; b; x)\right]^2 = \frac{b-1}{\pi} \int_0^\pi \int_0^\pi \cos(x \sin\theta \cos\psi) (\sin\psi)^{2b-3} (\sin\theta)^{2b-2} \\ \times e^{-x(1+\cos\theta)} {}_1F_1(a, b, 2x(1+\cos\theta)) \, d\psi \, d\theta.$$

Then, taking into account (6), for $0 \le x < \frac{(2b-1)(b-2a)}{2b}$

$$\left[e^{-\frac{x}{2}} {}_1F_1(a; b; x)\right]^2 \le \frac{b-1}{\pi} \int_0^\pi \int_0^\pi |\cos(x \sin\theta \cos\psi)| (\sin\psi)^{2b-3} (\sin\theta)^{2b-2} \, d\psi \, d\theta = \mathcal{M}(x, b).$$

Finally, based on equation (20), we have

$$0 < \mathcal{M}(x, b) < \mathcal{M}(0, b), \quad x > 0.$$

The proof of Theorem 3 is completed. □
**Proof of Corollary 4. (i).** Applying inequality (4) to the confluent hypergeometric function appearing in the Laplace transform of the Gauss hypergeometric function (see [12, p. 59])

$$ {}_2F_1\left(\sigma, a; b; \frac{x}{z}\right) = \frac{z^{\sigma}}{\Gamma(\sigma)} \int_0^{\infty} e^{-zt} t^{\sigma-1} {}_1F_1(a; b; xt) dt, $$

where $0 \le x < \Re(e(z))$ and $\Re(e(\sigma)) > 0$, we have

$$ \begin{aligned} \left| {}_2F_1\left(\sigma, a; b; \frac{x}{z}\right) \right| & \le \frac{|z^{\sigma}|}{|\Gamma(\sigma)|} \int_0^{\infty} e^{-t[\Re(e(z)-x)]} t^{\Re(e(\sigma))-1} dt \\ & = \frac{\Gamma(\Re(e(\sigma)))}{|\Gamma(\sigma)|} |z^{\sigma}| |\Re(e(z)-x)|^{-\Re(e(\sigma))}. \end{aligned} $$

Finally, using inequality (see [18, Section 5.6])

$$ |\Gamma(p+iq)| \ge \frac{|\Gamma(p)|}{\sqrt{\cosh(q\pi)}} $$

we get (8).

(ii). By making use of (5), the proof of case (ii) can be completed by following the proof of case (i). $\square$

**Appendix A. Proof of the Watson-Glaeske formula**

Substituting the integral representation for ${}_1F_1$ (see [14, Section 6.5])

$$ {}_1F_1(a; b; z) = \frac{2^{1-b}\Gamma(b)e^{\frac{z}{2}}}{\Gamma(a)\Gamma(b-a)} \int_{-1}^{1} e^{\frac{zt}{2}} (1-t)^{b-2} \left(\frac{1+t}{1-t}\right)^{a-1} dt, \quad (16) $$

where $\Re(e(b)) > \Re(e(a)) > 0$, into Bailey's product formula for confluent hypergeometric functions (see [1])

$$ {}_1F_1(a; b; x) \cdot {}_1F_1(a; b; y) = \sum_{k=0}^{\infty} \frac{(-1)^k a_k (b-a)_k}{k! (b)_k (b)_{2k}} (xy)^k {}_1F_1(a+k; b+2k; x+y) $$

we get, after interchanging the order of summation and integration

$$ {}_1F_1(a; b; x) \cdot {}_1F_1(a; b; y) = \frac{2^{1-b}[\Gamma(b)]^2 e^{\frac{x+y}{2}}}{\Gamma(a)\Gamma(b-a)} \int_{-1}^{1} e^{\frac{x+y}{2}t} (1-t)^{b-2} \left(\frac{1+t}{1-t}\right)^{a-1} \\ \times \left[ \sum_{k=0}^{\infty} \frac{(-1)^k \left(\frac{\sqrt{xy(1-t^2)}}{2}\right)^{2k}}{k!\Gamma(b+k)} \right] dt. $$

Using the series expansion

$$ \mathcal{J}_v(z) = \sum_{k=0}^{\infty} \frac{(-1)^k \left(\frac{z}{2}\right)^{2k}}{k! \Gamma(v+k+1)}, $$

we obtain

$$ {}_1F_1(a; b; x) \cdot {}_1F_1(a; b; y) = \frac{2^{1-b}[\Gamma(b)]^2 e^{\frac{x+y}{2}}}{\Gamma(a)\Gamma(b-a)} \int_{-1}^{1} e^{\frac{x+y}{2}t} (1-t)^{b-2} \left(\frac{1+t}{1-t}\right)^{a-1} \\ \times \mathcal{J}_{b-1}(\sqrt{xy(1-t^2)}) dt. \quad (17) $$

On the other hand, from Gegenbauer's double integral representation for $J_v$ (see [23, Section 3.33])

$$ \mathcal{J}_v(\omega)) = \frac{1}{\pi \Gamma(v)} \int_0^\pi \int_0^\pi e^{iZ \cos\theta - iz(\cos\Phi \cos\theta + \sin\Phi \sin\theta \cos\psi)} (\sin\psi)^{2v-1} (\sin\theta)^{2v} d\psi d\theta, $$
where $\omega^2 = Z^2 + z^2 - zZ \cos\Phi$ and $\nu > 0$, for $Z = -it\sqrt{xy}$, $z = -\sqrt{xy}$, $\Phi = \frac{\pi}{2}$ and $\nu = b-1$ we have

$$ \mathcal{J}_{b-1}(\sqrt{xy}(1-t^2)) = \frac{1}{\pi\Gamma(b-1)} \int_0^\pi \int_0^\pi e^{t\sqrt{xy}\cos\theta + i\sqrt{xy}\sin\theta\cos\psi} \\ \times (\sin\psi)^{2b-3}(\sin\theta)^{2b-2} d\psi d\theta. \quad (18) $$

Now substituting (18) into (17) and taking into account (16) yields

$$ \begin{aligned} {}_1F_1(a; b; x) {}_1F_1(a; b; y) &= \frac{2^{1-b}(b-1)\Gamma(b)e^{\frac{x+y}{2}}}{\pi\Gamma(a)\Gamma(b-a)} \int_0^\pi \int_0^\pi e^{i\sqrt{xy}\sin\theta\cos\psi} (\sin\psi)^{2b-3}(\sin\theta)^{2b-2} \\ & \times \left[ \int_{-1}^1 e^{\left(\frac{x+y}{2} + \sqrt{xy}\cos\theta\right)t} (1-t)^{b-2} \left(\frac{1+t}{1-t}\right)^{a-1} dt \right] d\psi d\theta \\ &= \frac{b-1}{\pi} \int_0^\pi \int_0^\pi e^{-\sqrt{xy}\cos\theta + i\sqrt{xy}\sin\theta\cos\psi} (\sin\psi)^{2b-3}(\sin\theta)^{2b-2} \\ & \times {}_1F_1(a, b, x+y+2\sqrt{xy}\cos\theta) d\psi d\theta. \end{aligned} \quad (19) $$

By using analytic continuation, equation (19) can be extended to $a \in \mathbb{C}$ and $\Re(e(b) > 1$. This proves equation (15) and completes the proof of Glaeske's result.

In particular, putting $x = y = 0$ in (19) yields

$$ 1 = \frac{b-1}{\pi} \int_0^\pi \int_0^\pi (\sin \psi)^{2b-3} (\sin \theta)^{2b-2} d\psi d\theta. \quad (20) $$

## Acknowledgements

The author thanks the referee for a careful reading of the manuscript and for his suggestions.

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