1 Introduction
It is well known that the error function has numerous applications in probability, statistics, and partial differential equations theory. Recently, the bounds for the error function have attracted the attention of many researchers. In particular, many remarkable inequalities for the error function can be found in the literature [1‐13].
$$ \operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int _{0}^{x}e^{-t^{2}}\,dt=\frac {1}{\sqrt{\pi }} \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n! (n+\frac{1}{2} )}x^{2n+1} $$
Pólya [14] proved that the inequality holds for all \(x>0\).
$$ \operatorname{erf}(x)< \sqrt{1-e^{-4x^{2}/\pi}} $$
Anzeige
In [15], Chu proved that the double inequality holds for all \(x>0\) if and only if \(p\in(0, 1]\) and \(q\in[4/\pi, \infty)\).
$$ \sqrt{1-e^{-px^{2}}}< \operatorname{erf}(x)< \sqrt{1-e^{-qx^{2}}} $$
(1.1)
Alzer [16] presented the double inequality for \(x>0\) and \(p>0\) with \(p\neq1\), where \(\Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}\,dt\) is the classical gamma function, and \(\alpha(p)\) and \(\beta(p)\) are, respectively, given by and
$$ \bigl(1-e^{-\beta(p) x^{p}} \bigr)^{1/p}< \frac{1}{\Gamma (1+\frac{1}{p} )} \int _{0}^{x}e^{-t^{p}}\,dt< \bigl(1-e^{-\alpha(p) x^{p}} \bigr)^{1/p} $$
$$ \alpha(p)=\frac{1}{\Gamma^{p} (1+\frac{1}{p} )} \quad (p>1),\qquad \alpha(p)=1\quad (0< p< 1), $$
$$ \beta(p)=\frac{1}{\Gamma^{p} (1+\frac{1}{p} )} \quad (0< p< 1),\qquad \beta(p)=1\quad (p>1). $$
Let \(n\geq2\), and \(\alpha_{n}\), \(\beta_{n}\), \(\alpha^{\ast}_{n}\), \(\beta^{\ast}_{n}\) be, respectively, defined by
$$\begin{aligned}& \alpha_{2}=0.90686\ldots,\qquad \alpha_{n}=1 \quad (n\geq3), \qquad \beta_{n}=n-1, \\& \alpha^{\ast}_{n}=n+1 \quad (n=2k),\qquad \alpha^{\ast}_{n}=n-1\quad (n=2k-1), \qquad \beta^{\ast}_{n}=1. \end{aligned}$$
In [17, 18], Alzer proved that the double inequalities hold for all \(x_{i}\geq0\) and \(y\geq x>0\) if and only if \(\lambda_{n}\leq\alpha_{n}\), \(\mu_{n}\geq\beta_{n}\), \(\lambda \leq\operatorname{erf}(1)=0.8427\ldots\) , \(\mu\geq2/\sqrt{\pi }=1.1283\ldots\) , \(\lambda^{\ast}\leq0\) and \(\mu^{\ast}\geq1\), and inequality (1.2) holds for all \(x_{i}\leq0\) if and only if \(\lambda_{n}\geq \alpha^{\ast}_{n}\) and \(\mu_{n}\leq\beta^{\ast}_{n}\).
$$\begin{aligned}& \lambda_{n}\operatorname{erf} \Biggl(\sum _{i=1}^{n}x_{i} \Biggr)\leq \sum _{i=1}^{n}\operatorname{erf}(x_{i})-\prod _{i=1}^{n}\operatorname {erf}(x_{i}) \leq \mu_{n}\operatorname{erf} \Biggl(\sum_{i=1}^{n}x_{i} \Biggr), \\& \lambda\operatorname{erf}\bigl(y+\operatorname{erf}(x)\bigr)< \operatorname {erf} \bigl(x+\operatorname{erf}(y)\bigr)< \mu\operatorname{erf}\bigl(y+\operatorname {erf}(x)\bigr), \\& \lambda^{\ast}\operatorname{erf}\bigl(y\operatorname {erf}(x)\bigr)< \operatorname{erf}\bigl(x\operatorname{erf}(y)\bigr)\leq\mu^{\ast } \operatorname{erf} \bigl(y\operatorname{erf}(x)\bigr), \end{aligned}$$
(1.2)
Anzeige
Recently, Neuman [19] proved that the double inequality holds for all \(x>0\).
$$ \frac{2x}{\sqrt{\pi}}e^{-\frac{x^{2}}{3}}\leq\operatorname {erf}(x)\leq \frac {2x}{\sqrt {\pi}}\frac{e^{-x^{2}}+2}{3} $$
(1.3)
Let \(x\in(0, \infty)\), \(p\in(7/5, \infty)\), \(\lambda(p)\), \(\mu(p)\), and \(B_{p}(x)\) be, respectively, defined by
$$\begin{aligned}& \lambda(p)=\frac{16(5p-7)}{(15p^{2}-40p+28)(45p^{2}-60p-4)},\qquad \mu(p)=\frac{4(5p-7)}{5(3p-4)}, \end{aligned}$$
(1.4)
$$\begin{aligned}& B_{p}(x)=\sqrt{1-\lambda(p)e^{-px^{2}}-\bigl[1-\lambda(p) \bigr]e^{-\mu(p)x^{2}}}. \end{aligned}$$
(1.5)
The main purpose of this paper is to present the best possible parameters p and q on the interval \((7/5, \infty)\) such that the double inequality holds for all \(x>0\).
$$ B_{p}(x)< \operatorname{erf}(x)< B_{q}(x) $$
2 Lemmas
In order to prove our main results, we need to introduce an auxiliary function at first.
Let \(-\infty\leq a< b\leq\infty\), f and g be differentiable on \((a,b)\), and \(g'\neq0\) on \((a,b)\). Then the function \(H_{f, g}\) [20, 21] is defined by It is not difficult to verify that the auxiliary function \(H_{f,g}\) has the following properties: if \(g\neq0\) on \((a,b)\), and if both f and g are twice differentiable on \((a,b)\).
$$ H_{f,g}\equiv\frac{f'}{g'}g-f. $$
(2.1)
$$ \biggl(\frac{f}{g} \biggr)'= \frac{g'}{g^{2}}H_{f,g} $$
(2.2)
$$ H_{f,g}'= \biggl(\frac{f'}{g'} \biggr)'g $$
(2.3)
Lemma 2.1
[20], Theorem 8
Let
\(-\infty\leq a< b\leq\infty\), f
and
g
be differentiable on
\((a,b)\)
with
\(f(a^{+})=g(a^{+})=0\), \(g^{\prime}(x)\neq0\)
and
\(g^{\prime}(x)H_{f, g}(b^{-})< (>)\, 0\)
for all
\(x\in(a,b)\). If there exists
\(\lambda_{0}\in(a, b)\)
such that
\(f^{\prime}/g^{\prime}\)
is strictly increasing (decreasing) on
\((a, \lambda_{0})\)
and strictly decreasing (increasing) on
\((\lambda_{0}, b)\), then there exists
\(\mu_{0}\in(a, b)\)
such that
\(f/g\)
is strictly increasing (decreasing) on
\((a, \mu_{0})\)
and strictly decreasing (increasing) on
\((\mu_{0}, b)\).
Lemma 2.2
Let
\(p\in(7/5, \infty)\), \(p^{\ast}_{0}=(50+2\sqrt{30})/35=1.74155\ldots\) , \(\lambda(p)\)
and
\(\mu(p)\)
be defined by (1.4), and
\(u_{n}\)
be defined by
Then the following statements are true:
$$ u_{n}=(5p-6) (5p-8)n-\bigl(15p^{2}-40p+28 \bigr). $$
(2.4)
(1)
\(p>\mu(p)\), \(0<\lambda(p)\leq(8+\sqrt{14})/16=0.73385\ldots\)
and
\(0<\mu(p)<4/3\)
for
\(p\in(7/5, \infty)\), \(0<\mu(p)\leq1\)
for
\(p\in(7/5, 8/5]\)
and
\(1<\mu(p)<4/3\)
for
\(p\in(8/5, \infty)\);
(2)
\(u_{n}<0\)
for all
\(n\geq2\)
if
\(p\in(7/5, 8/5]\);
(3)
\(u_{2}\geq0\)
and
\(u_{n}> 0\)
for all
\(n\geq3\)
if
\(p\in [p^{\ast}, \infty)\);
(4)
there exists
\(n_{0}\geq2\)
such that
\(u_{n_{0}+1}\geq0\), \(u_{n}<0\)
for
\(2\leq n\leq n_{0}\)
and
\(u_{n}> 0\)
for
\(n\geq n_{0}+2\)
and
\(u_{n}>0\)
for
\(n>n_{0}\)
if
\(p\in(8/5, p^{\ast}_{0})\).
Proof
For part (1), from (1.4) we clearly see that
for \(p>7/5\).
$$\begin{aligned}& p>\mu(p), \\& \lambda \biggl(\frac{7}{5} \biggr)=\lambda(\infty)=0,\qquad \lambda \biggl(\frac{14+2\sqrt{14}}{15} \biggr)=\frac{8+\sqrt {14}}{16}=0.73385\ldots, \end{aligned}$$
(2.5)
$$\begin{aligned}& \mu \biggl(\frac{7}{5} \biggr)=0, \qquad \mu(\infty)=\frac{4}{3}, \qquad \mu \biggl(\frac{8}{5} \biggr)=1, \end{aligned}$$
(2.6)
$$\begin{aligned}& \lambda^{\prime}(p)=-\frac {80(3p-4)^{2}}{9(15p^{2}-40p+28)^{2}(45p^{2}-60p-4)^{2}} \\& \hphantom{\lambda^{\prime}(p)={}}{}\times\biggl(p-\frac{14-2\sqrt{14}}{15} \biggr) \biggl(p-\frac{14+2\sqrt {14}}{15} \biggr), \end{aligned}$$
(2.7)
$$\begin{aligned}& \mu^{\prime}(p)=\frac{4}{5(3p-4)^{2}}>0 \end{aligned}$$
(2.8)
Equation (2.7) implies that \(\lambda(p)\) is strictly increasing on \((7/5, (14+2\sqrt{14})/15]\) and strictly decreasing on \([(14+2\sqrt{14})/15, \infty)\). Therefore, \(0<\lambda(p)\leq(8+\sqrt{14})/16\) for \(p\in(7/5, \infty)\) as follows from (2.5) and the piecewise monotonicity of \(\lambda(p)\) on the interval \((7/5, \infty)\), and the remaining desired results for \(\mu(p)\) follow easily from (2.6) and (2.8).
For parts (2) and (3), let \(x\geq2\), \(p_{1}(x)\) and \(p_{2}(x)\) be defined by
$$ p_{1}(x)=\frac{35x-20-\sqrt{5}\sqrt{5x^{2}+4x-4}}{5(5x-3)},\qquad p_{2}(x)= \frac{35x-20+\sqrt{5}\sqrt{5x^{2}+4x-4}}{5(5x-3)}. $$
Then simple computations lead to
for \(x\geq2\).
$$\begin{aligned}& p_{1}(2)=\frac{50-2\sqrt{30}}{35}=1.11558\ldots,\qquad p_{1}( \infty)=\frac{6}{5}, \end{aligned}$$
(2.9)
$$\begin{aligned}& p_{2}(2)=p^{\ast}_{0},\qquad p_{2}(3)= \frac{85+\sqrt{265}}{60}=1.68798\ldots,\qquad p_{2}(\infty)= \frac{8}{5}, \end{aligned}$$
(2.10)
$$\begin{aligned}& u_{n}=5(5n-3)\bigl[p-p_{1}(n)\bigr] \bigl[p-p_{2}(n) \bigr], \end{aligned}$$
(2.11)
$$\begin{aligned}& p^{\prime}_{1}(x)=\frac{25x-14-\sqrt{5}\sqrt{5x^{2}+4x-4}}{\sqrt {5}(5x-3)^{2}\sqrt{5x^{2}+4x-4}}>0, \end{aligned}$$
(2.12)
$$\begin{aligned}& p^{\prime}_{2}(x)=-\frac{25x-14+\sqrt{5}\sqrt{5x^{2}+4x-4}}{\sqrt {5}(5x-3)^{2}\sqrt{5x^{2}+4x-4}}< 0 \end{aligned}$$
(2.13)
It follows from (2.9)-(2.13) that
for \(n\geq2\) and for \(n\geq3\)
$$\begin{aligned}& u_{2}=35\bigl[p-p^{\ast}_{0}\bigr] \biggl[p- \frac{50-2\sqrt{30}}{35} \biggr], \end{aligned}$$
(2.14)
$$\begin{aligned}& \frac{50-2\sqrt{30}}{35}\leq p_{1}(n)< \frac{6}{5},\qquad \frac{8}{5}< p_{2}(n)\leq p^{\ast}_{0} \end{aligned}$$
(2.15)
$$ \frac{8}{5}< p_{2}(n)\leq\frac{85+\sqrt{265}}{60} $$
(2.16)
For part (4), if \(p\in(8/5, p^{\ast}_{0})\), then from (2.4) and (2.14) we clearly see that the sequence \(\{u_{n}\}_{n=2}^{\infty}\) is strictly increasing and Therefore, part (4) follows from (2.17) and the monotonicity of the sequence \(\{u_{n}\}_{n=2}^{\infty}\). □
$$ u_{2}< 0,\qquad u_{\infty}=\infty. $$
(2.17)
Lemma 2.3
Let
\(x\in(0, \infty)\), \(p\in(7/5, \infty)\), \(\lambda(p)\), \(\mu(p)\)
and
\(B_{p}(x)\)
be defined by (1.4) and (1.5), and
\(C_{p}(x)\)
and
\(\alpha(p)\)
be defined by
and
Then the following statements are true:
$$ C_{p}(x)=\frac{p\lambda(p)e^{-px^{2}}+\mu(p)(1-\lambda(p))e^{-\mu (p)x^{2}}}{p\lambda(p)+\mu(p)(1-\lambda(p))} $$
$$ \alpha(p)=\sqrt{\frac{4}{\pi[p\lambda(p)+\mu(p)(1-\lambda (p))]}}=\sqrt {\frac{45p^{2}-60p-4}{3\pi p(5p-7)}}. $$
(2.18)
(1)
the function
\(p\rightarrow B_{p}(x)\)
is strictly increasing on
\((7/5, \infty)\);
(2)
the function
\(p\rightarrow C_{p}(x)\)
is strictly decreasing on
\((7/5, \infty)\);
(3)
the function
\(p\rightarrow\alpha(p)B_{p}(x)\)
is strictly decreasing on
\((7/5, \infty)\).
Proof
For part (1), it suffices to show that \(\partial B^{2}_{p}(x)/\partial p>0\) for \(x\in(0, \infty)\) and \(p\in(7/5, \infty)\). Let \(t=(p-\mu(p))x^{2}\) and
$$ F_{1}(t)=-\bigl(p-\mu(p)\bigr)\lambda^{\prime}(p)+ \lambda(p)t+\bigl(p-\mu (p)\bigr)\lambda ^{\prime}(p)e^{t}+ \mu^{\prime}(p) \bigl(1-\lambda(p)\bigr)te^{t}. $$
(2.19)
Then it follows from (1.4), (1.5), (2.19), and Lemma 2.2(1) that
for \(p\in(7/5, \infty)\) and \(t>0\).
$$\begin{aligned}& \frac{\partial B^{2}_{p}(x)}{\partial p}=\frac{e^{-px^{2}}}{p-\mu(p)}F_{1}(t), \end{aligned}$$
(2.20)
$$\begin{aligned}& p-\mu(p)>0, \quad t>0, \end{aligned}$$
(2.21)
$$\begin{aligned}& F_{1}(0)=0, \end{aligned}$$
(2.22)
$$\begin{aligned}& F^{\prime}_{1}(t)=\lambda(p)+\bigl(p-\mu(p)\bigr) \lambda^{\prime }(p)e^{t}+\mu ^{\prime}(p) \bigl(1- \lambda(p)\bigr)e^{t}+\mu^{\prime}(p) \bigl(1-\lambda(p) \bigr)te^{t}, \\& F^{\prime}_{1}(0)=\frac{12(15p^{2}-40p+28)}{(45p^{2}-60p-4)^{2}}>0, \end{aligned}$$
(2.23)
$$\begin{aligned}& F^{\prime\prime}_{1}(t)=\frac {40(3p-2)(3p-4)}{(45p^{2}-60p-4)^{2}}e^{t}+ \frac {20p(3p-4)}{(15p^{2}-40p+28)(45p^{2}-60p-4)}te^{t}>0 \end{aligned}$$
(2.24)
From (2.22)-(2.24) we clearly see that for \(p\in(7/5, \infty)\) and \(t>0\).
$$ F_{1}(t)>0 $$
(2.25)
For part (2), it is enough to prove that \(\partial C_{p}(x)/\partial p<0\) for \(x\in(0, \infty)\) and \(p\in(7/5, \infty)\). Let \(t=(p-\mu(p))x^{2}\) and
$$ F_{2}(t)=-\frac{15p^{2}-40p+28}{p-\mu(p)}t-10(3p-4)+ \biggl[-\frac {15p^{2}-40p+28}{p-\mu(p)}t+10(3p-4) \biggr]e^{t}. $$
Then elaborated computations lead to
for \(p\in(7/5,\infty)\) and \(t>0\).
$$\begin{aligned}& C_{p}(x)=\frac{1}{3} \biggl[\frac{4e^{-px^{2}}}{15p^{2}-40p+28}+ \frac {5(3p-4)^{2}e^{-\mu(p)x^{2}}}{15p^{2}-40p+28} \biggr], \\& \frac{\partial C_{p}(x)}{\partial p}= -\frac{1}{3} \biggl[\frac{4x^{2}}{15p^{2}-40p+28}+ \frac {40(3p-4)}{(15p^{2}-40p+28)^{2}} \biggr]e^{-px^{2}} \\& \hphantom{\frac{\partial C_{p}(x)}{\partial p}={}}{}+\frac{1}{3} \biggl[-\frac{4x^{2}}{15p^{2}-40p+28}+ \frac {40(3p-4)}{(15p^{2}-40p+28)^{2}} \biggr]e^{-\mu(p)x^{2}} \\& \hphantom{\frac{\partial C_{p}(x)}{\partial p}}= \frac{4e^{-px^{2}}}{3(15p^{2}-40p+28)^{2}}F_{2}(t), \end{aligned}$$
(2.26)
$$\begin{aligned}& F_{2}(t)= -5(3p-4)t-10(3p-4)+\bigl[-5(3p-4)t+10(3p-4) \bigr]e^{t} \\& \hphantom{F_{2}(t)}= -5(3p-4)\bigl[(t+2)+(t-2)e^{t}\bigr] \\& \hphantom{F_{2}(t)}= -5(3p-4)\sum_{n=2}^{\infty} \frac{(n-2)}{n!}t^{n}< 0 \end{aligned}$$
(2.27)
For part (3), let \(G_{p}(x)\) be defined by Then elaborated computations lead to
$$ G_{p}(x)=\frac{\pi}{4}\alpha^{2}(p)B^{2}_{p}(x)= \frac{1-\lambda (p)e^{-px^{2}}-(1-\lambda(p))e^{-\mu(p)x^{2}}}{p\lambda(p)+\mu (p)(1-\lambda(p))}. $$
(2.28)
$$ \frac{\partial G_{p}(x)}{\partial x}=\frac{2x [p\lambda(p)e^{-px^{2}}+\mu(p)(1-\lambda (p))e^{-\mu (p)x^{2}} ]}{p\lambda(p)+\mu(p)(1-\lambda(p))} =2xC_{p}(x). $$
(2.29)
It follows from Lemma 2.3(2) and (2.29) that for \(x>0\) and \(p\in(7/5, \infty)\).
$$ \frac{\partial}{\partial x} \biggl(\frac{\partial G_{p}(x)}{\partial p} \biggr) = \frac{\partial}{\partial p} \biggl(\frac{\partial G_{p}(x)}{\partial x} \biggr)=2x\frac{\partial C_{p}(x)}{\partial p}< 0 $$
(2.30)
Inequality (2.30) implies that the function \(x\rightarrow\partial G_{p}(x)/\partial p\) is strictly decreasing on \((0, \infty)\) and for \(x>0\) and \(p\in(7/5, \infty)\).
$$\begin{aligned} \frac{\partial G_{p}(x)}{\partial p} < &\frac{\partial G_{p}(x)}{\partial p}\bigg|_{x=0} \\ =& \biggl[-\frac{1-\lambda(p)e^{-px^{2}}-(1-\lambda(p))e^{-\mu (p)x^{2}}}{(p\lambda(p)+\mu(p)(1-\lambda(p)))^{2}} \frac{d(p\lambda(p)+\mu(p)(1-\lambda(p)))}{dp} \biggr]_{x=0} \\ &{}+ \biggl[\frac{-\lambda^{\prime}(p)e^{-px^{2}}+\lambda (p)x^{2}e^{-px^{2}}+\lambda^{\prime}(p)e^{-\mu(p)x^{2}} +\mu^{\prime}(p)x^{2}(1-\lambda(p))e^{-\mu(p)x^{2}}}{p\lambda (p)+\mu (p)(1-\lambda(p))} \biggr]_{x=0} \\ =& 0 \end{aligned}$$
(2.31)
Lemma 2.4
Let
\(p\in(7/5, \infty)\), \(x\in(0, \infty)\), \(\lambda(p)\), \(\mu(p)\), \(B_{p}(x)\)
and
\(H_{f, g}(x)\)
be, respectively, defined by (1.4), (1.5) and (2.1), and
\(f_{1}(x)\), \(g_{1}(x)\), \(f_{2}(x)\)
and
\(g_{2}(x)\)
be, respectively, defined by
Then
$$\begin{aligned}& f_{1}(x)=B^{2}_{p}(x)=1-\lambda(p)e^{-px^{2}}- \bigl(1-\lambda(p)\bigr)e^{-\mu(p)x^{2}},\qquad g_{1}(x)= \operatorname{erf}^{2}(x), \end{aligned}$$
(2.32)
$$\begin{aligned}& f_{2}(x)= \bigl[p\lambda(p)e^{(1-p)x^{2}}+\mu(p) \bigl(1-\lambda (p) \bigr)e^{(1-\mu (p))x^{2}} \bigr]x,\qquad g_{2}(x)=\frac{2}{\sqrt{\pi}} \operatorname{erf}(x). \end{aligned}$$
(2.33)
$$\begin{aligned}& H_{f_{2}, g_{2}}(\infty)=\lim_{x\rightarrow \infty} \biggl( \frac{f^{\prime}_{2}(x)}{g^{\prime }_{2}(x)}g_{2}(x)-f_{2}(x) \biggr)= \textstyle\begin{cases} \infty, & p\in (\frac{7}{5}, \frac{8}{5} ], \\ -\infty, & p\in (\frac{8}{5}, \infty ), \end{cases}\displaystyle \end{aligned}$$
(2.34)
$$\begin{aligned}& H_{f_{1}, g_{1}}(\infty)=\lim_{x\rightarrow \infty} \biggl( \frac{f^{\prime}_{1}(x)}{g^{\prime }_{1}(x)}g_{1}(x)-f_{1}(x) \biggr)= \textstyle\begin{cases} \infty, & p\in (\frac{7}{5}, \frac{8}{5} ], \\ -1, & p\in (\frac{8}{5}, \infty ). \end{cases}\displaystyle \end{aligned}$$
(2.35)
Proof
Let \(t=(p-\mu(p))x^{2}\) and Then (2.1) and (2.33) lead to
$$\begin{aligned} h_{1}(t) =&p\lambda(p) \bigl(p-\mu(p)\bigr)-2p \lambda(p) (p-1)t \\ &{}+\mu(p) \bigl(p-\mu(p)\bigr) \bigl(1-\lambda(p)\bigr)e^{t}-2 \mu(p) \bigl(\mu(p)-1\bigr) \bigl(1-\lambda (p)\bigr)te^{t}. \end{aligned}$$
(2.36)
$$\begin{aligned}& f_{2}(x)=\frac{\sqrt{t}}{\sqrt{p-\mu(p)}} \bigl(p\lambda (p)e^{\frac {1-p}{p-\mu(p)}t}+\mu(p) \bigl(1-\lambda(p)\bigr)e^{\frac{1-\mu(p)}{p-\mu (p)}t} \bigr), \end{aligned}$$
(2.37)
$$\begin{aligned}& \frac{f^{\prime}_{2}(x)}{g^{\prime}_{2}(x)}=\frac{\pi}{4(p-\mu (p))}e^{\frac{2-p}{p-\mu(p)}t}h_{1}(t), \\& H_{f_{2},g_{2}}(x)=\frac{f^{\prime}_{2}(x)}{g^{\prime }_{2}(x)}g_{2}(x)-f_{2}(x) \\& \hphantom{H_{f_{2},g_{2}}(x)}=\frac{\sqrt{\pi}}{2(p-\mu(p))}\operatorname{erf}(x)e^{\frac {2-p}{p-\mu (p)}t}h_{1}(t) \\& \hphantom{H_{f_{2},g_{2}}(x)={}}{}-\frac{\sqrt{t}}{\sqrt{p-\mu(p)}} \bigl(p\lambda(p)e^{\frac {1-p}{p-\mu (p)}t}+\mu(p) \bigl(1-\lambda(p)\bigr)e^{\frac{1-\mu(p)}{p-\mu(p)}t} \bigr). \end{aligned}$$
(2.38)
If \(p\in(8/5, \infty)\), then Lemma 2.2(1), (2.36), and (2.37) lead to
$$\begin{aligned}& p>\mu(p),\qquad 0< \lambda(p)< 1,\qquad 1< \mu(p)< \frac{4}{3}, \end{aligned}$$
(2.39)
$$\begin{aligned}& f_{2}(\infty)=0, \end{aligned}$$
(2.40)
$$\begin{aligned}& \lim_{t\rightarrow \infty}\frac{h_{1}(t)}{te^{t}}=-2\mu(p) \bigl(\mu(p)-1\bigr) \bigl(1-\lambda(p)\bigr)< 0, \\& \lim_{t\rightarrow \infty}e^{\frac{2-p}{p-\mu(p)}t}h_{1}(t)=\lim _{t\rightarrow \infty}te^{\frac{2-\mu(p)}{p-\mu(p)}t}\lim_{t\rightarrow \infty} \frac{h_{1}(t)}{te^{t}}=-\infty. \end{aligned}$$
(2.41)
Therefore, \(H_{f_{2}, g_{2}}(\infty)=-\infty\) for \(p\in(8/5, \infty)\) follows from (2.38), (2.39), and (2.41).
If \(p\in(7/5, 8/5]\), then it follows from Lemma 2.2(1) and (2.36) together with (2.37) that
$$\begin{aligned}& p>\mu(p),\qquad 0< \lambda(p)< 1,\qquad 0< \mu(p)\leq1, \end{aligned}$$
(2.42)
$$\begin{aligned}& \lim_{t\rightarrow \infty}\frac{f_{2}(x)}{te^{\frac{1-\mu(p)}{p-\mu(p)}t}}=\lim_{t\rightarrow \infty} \frac{p\lambda(p)e^{-t}+\mu(p)(1-\lambda(p))}{\sqrt{(p-\mu (p))t}}=0, \end{aligned}$$
(2.43)
$$\begin{aligned}& \lim_{t\rightarrow \infty}\frac{h_{1}(t)}{te^{\frac{1-\mu(p)}{p-\mu(p)}t}} \\& \quad =\lim_{t\rightarrow \infty} \biggl[\frac{p\lambda(p)(p-\mu(p))}{t}e^{\frac{\mu (p)-1}{p-\mu (p)}t}-2p \lambda(p) (p-1)e^{\frac{\mu(p)-1}{p-\mu(p)}t} \biggr] \\& \qquad {}+\lim_{t\rightarrow \infty} \biggl[\frac{\mu(p)(1-\lambda(p))(p-\mu(p))}{t}e^{\frac {p-1}{p-\mu (p)}t}+2 \mu(p) \bigl(1-\lambda(p)\bigr) \bigl(1-\mu(p)\bigr)e^{\frac{p-1}{p-\mu (p)}t} \biggr] \\& \quad =\infty. \end{aligned}$$
(2.44)
Therefore, for \(p\in(7/5, 8/5]\) as follows from (2.38) and (2.42)-(2.44).
$$ H_{f_{2}, g_{2}}(\infty)=\lim_{t\rightarrow \infty}te^{\frac{1-\mu(p)}{p-\mu(p)}t} \biggl[ \frac{\sqrt{\pi}e^{\frac{2-p}{p-\mu(p)}t}}{2(p-\mu (p))}\operatorname{erf} (x)\frac{h_{1}(t)}{te^{\frac{1-\mu(p)}{p-\mu(p)}t}} -\frac{f_{2}(x)}{te^{\frac{1-\mu(p)}{p-\mu(p)}t}} \biggr]=\infty $$
Similarly, from (2.1) and (2.32) we have
$$\begin{aligned} H_{f_{1}, g_{1}}(x) =&\frac{\sqrt{\pi}}{2}x\operatorname{erf}(x) \bigl[p\lambda (p)e^{(1-p)x^{2}}+\mu(p) \bigl(1-\lambda(p)\bigr)e^{(1-\mu(p))x^{2}} \bigr] \\ &{}- \bigl[1-\lambda(p)e^{-px^{2}}-\bigl(1-\lambda(p)\bigr)e^{-\mu (p)x^{2}} \bigr]. \end{aligned}$$
(2.45)
Lemma 2.5
Let
\(p\in(7/5, \infty)\), \(p^{\ast}_{0}=(50+2\sqrt{30})/35=1.74155\ldots\) , \(x\in(0, \infty)\), \(\lambda(p)\), \(\mu(p)\), \(B_{p}(x)\), \(H_{f, g}(x)\), \(f_{1}(x)\), \(g_{1}(x)\), \(f_{2}(x)\)
and
\(g_{2}(x)\)
be, respectively, defined by (1.4), (1.5), (2.1), (2.32) and (2.33). Then the following statements are true:
(1)
if
\(f_{1}(x)/g_{1}(x)\)
is strictly increasing on
\((0, \infty)\), then
\(p\in(7/5, 8/5]\);
(2)
if
\(f_{1}(x)/g_{1}(x)\)
is strictly decreasing on
\((0, \infty)\), then
\(p\in[p^{\ast}_{0}, \infty)\).
Proof
(1) It follows from (2.1) and (2.32) that If \(f_{1}(x)/g_{1}(x)\) is strictly increasing on \((0, \infty)\), then (2.48) leads to and we assert that \(p\in(7/5, 8/5]\). Otherwise, \(p>8/5\) and (2.35) lead to the conclusion \(H_{f_{1}, g_{1}}(\infty)=-1\), which contradicts with (2.49).
$$ \lim_{x\rightarrow\infty}e^{x^{2}} \biggl( \frac {f_{1}(x)}{g_{1}(x)} \biggr)^{\prime} =\lim_{x\rightarrow\infty}e^{x^{2}} \frac{g^{\prime }_{1}(x)}{g^{2}_{1}(x)}H_{f_{1}, g_{1}}(x) =\frac{4}{\sqrt{\pi}}\lim _{x\rightarrow\infty}H_{f_{1}, g_{1}}(x). $$
(2.48)
$$ \lim_{x\rightarrow\infty}H_{f_{1}, g_{1}}(x)\geq0 $$
(2.49)
(2) Let \(t=(p-\mu(p))x^{2}\), \(u_{n}\) and \(h_{1}(t)\) be, respectively, defined by (2.4) and (2.36), and \(h_{2}(t)\) and \(v_{n}\) be, respectively, defined by
$$\begin{aligned}& h_{2}(t)= 2\mu(p) \bigl(1-\lambda(p)\bigr) \bigl(\mu(p)-1\bigr) \bigl( \mu(p)-2\bigr)te^{t}-\mu (p) \bigl(1-\lambda(p)\bigr) \bigl(3\mu(p)-4 \bigr) \\& \hphantom{h_{2}(t)={}}{}\times \bigl(p-\mu(p)\bigr)e^{t}+2p\lambda(p) (p-1) (p-2)t \\& \hphantom{h_{2}(t)={}}{}-p\lambda(p) (3p-4) \bigl(p-\mu(p)\bigr), \end{aligned}$$
(2.50)
$$\begin{aligned}& v_{n}=-\frac{4\mu(p)(1-\lambda(p))}{25(3p-4)^{2}}u_{n}. \end{aligned}$$
(2.51)
Then from (2.1)-(2.3), (2.32), (2.33), (2.36), and (2.50) we have
$$\begin{aligned}& \biggl(\frac{f^{\prime}_{2}(x)}{g^{\prime}_{2}(x)} \biggr)^{\prime }= \frac {\pi}{4(p-\mu(p))} \frac{d}{dt} \bigl[e^{\frac{2-p}{p-\mu (p)}t}h_{1}(t) \bigr] \frac{dt}{dx} \\& \hphantom{\biggl(\frac{f^{\prime}_{2}(x)}{g^{\prime}_{2}(x)} \biggr)^{\prime }}= \frac{\pi x}{2(p-\mu(p))}e^{\frac{2-p}{p-\mu(p)}t}h_{2}(t), \end{aligned}$$
(2.52)
$$\begin{aligned}& h_{2}(t)= 2\mu(p) \bigl(1-\lambda(p)\bigr) \bigl(\mu(p)-1\bigr) \bigl( \mu(p)-2\bigr)\sum_{n=1}^{\infty } \frac{t^{n}}{(n-1)!} \\& \hphantom{h_{2}(t)={}}{}-\mu(p) \bigl(1-\lambda(p)\bigr) \bigl(3\mu(p)-4\bigr) \bigl(p- \mu(p)\bigr)\sum_{n=0}^{\infty }\frac {t^{n}}{n!} \\& \hphantom{h_{2}(t)={}}{}+2p\lambda(p) (p-1) (p-2)t-p\lambda(p) (3p-4) \bigl(p-\mu(p) \bigr) \\& \hphantom{h_{2}(t)}= -\bigl(p-\mu(p)\bigr)\bigl[\bigl(p-\mu(p)\bigr) \bigl(3p+3 \mu(p)-4\bigr)\lambda(p)+\mu(p) \bigl(3\mu (p)-4\bigr)\bigr] \\& \hphantom{h_{2}(t)={}}{}+\bigl(p-\mu(p)\bigr) \bigl(2p^{2}+5 \mu^{2}(p)+2p\mu(p)-6p-10\mu(p)+4\bigr)\lambda (p) \\& \hphantom{h_{2}(t)={}}{}+\mu(p) \bigl(4p-3p\mu(p)+5\mu^{2}(p)-10\mu(p)+4 \bigr)+\sum_{n=2}^{\infty }\frac {v_{n}t^{n}}{n!} \\& \hphantom{h_{2}(t)}= \sum_{n=2}^{\infty} \frac{v_{n}t^{n}}{n!}, \\& \biggl(\frac{f^{\prime}_{1}(x)}{g^{\prime}_{1}(x)} \biggr)^{\prime} =\frac{g^{\prime}_{1}(x)}{g^{2}_{1}(x)} \biggl[ \frac{f^{\prime }_{1}(x)}{g^{\prime}_{1}(x)}g_{1}(x)-f_{1}(x) \biggr] = \frac{g^{\prime}_{1}(x)}{g^{2}_{1}(x)}H_{f_{1}, g_{1}}(x), \\& H^{\prime}_{f_{1}, g_{1}}(x)= \biggl(\frac{f^{\prime}_{1}(x)}{g^{\prime}_{1}(x)} \biggr)^{\prime}g_{1}(x) = \biggl(\frac{f_{2}(x)}{g_{2}(x)} \biggr)^{\prime}g_{1}(x) =\frac{g^{\prime}_{2}(x)}{g^{2}_{2}(x)}H_{f_{2}, g_{2}}(x)g_{1}(x), \\& H^{\prime}_{f_{2}, g_{2}}(x)= \biggl(\frac{f^{\prime}_{2}(x)}{g^{\prime}_{2}(x)} \biggr)^{\prime}g_{2}(x) =\frac{\pi x}{2(p-\mu(p))}e^{\frac{2-p}{p-\mu(p)}t}h_{2}(t)g_{2}(x), \\& \frac{g^{\prime}_{1}(x)}{g^{2}_{1}(x)}=\frac{4}{\sqrt{\pi}}\frac {e^{-x^{2}}}{\operatorname{erf}^{3}(x)}\sim \frac{\pi}{2x^{3}}, \qquad \frac{g^{\prime}_{2}(x)}{g^{2}_{2}(x)}g_{1}(x)=e^{-x^{2}}\sim1\quad \bigl(x \rightarrow0^{+}\bigr), \\& g_{2}(x)=\frac{2}{\sqrt{\pi}}\operatorname{erf}(x)\sim\frac {4x}{\pi}, \qquad h_{2}(t)\sim\frac{v_{2}}{2}t^{2}= \frac{(p-\mu(p))^{2}}{2}v_{2}x^{4} \quad \bigl(x \rightarrow0^{+}\bigr), \\& \biggl(\frac{f_{1}(x)}{g_{1}(x)} \biggr)^{\prime}\sim\frac{\pi }{2x^{3}}H_{f_{1}, g_{1}}(x) \quad \bigl(x\rightarrow0^{+}\bigr), \\& H^{\prime}_{f_{1}, g_{1}}(x)\sim H_{f_{2}, g_{2}}(x), \\& H^{\prime}_{f_{2}, g_{2}}(x)\sim\bigl(p-\mu(p)\bigr)v_{2}x^{6} \quad \bigl(x\rightarrow0^{+}\bigr). \end{aligned}$$
(2.53)
Note that \(H_{f_{1}, g_{1}}(0^{+})=H_{f_{2}, g_{2}}(0^{+})=0\). Making use of the L’Hôspital rule we get
$$\begin{aligned}& \lim_{x\rightarrow 0^{+}}x^{-5} \biggl(\frac{f_{1}(x)}{g_{1}(x)} \biggr)^{\prime } \\& \quad = \frac{\pi}{2}\lim_{x\rightarrow0^{+}}\frac{H_{f_{1}, g_{1}(x)}(x)}{x^{8}}= \frac{\pi}{16}\lim_{x\rightarrow 0^{+}}\frac{H_{f_{2}, g_{2}(x)}(x)}{x^{7}} \\& \quad = \frac{\pi}{112}\lim_{x\rightarrow0^{+}}\frac{H^{\prime}_{f_{2}, g_{2}(x)}(x)}{x^{6}}= \frac{\pi(p-\mu(p))}{112}v_{2} \\& \quad = -\frac{\pi\mu(p)(p-\mu(p))(1-\lambda(p))}{700} \bigl(35p^{2}-100p+68 \bigr). \end{aligned}$$
(2.54)
3 Main results
Theorem 3.1
Let
\(p\in(7/5, \infty)\), \(x>0\), \(p^{\ast}_{0}=(50+2\sqrt{30})/35\), \(\lambda(p)\), \(\mu(p)\), \(B_{p}(x)\)
and
\(\alpha(p)\)
be, respectively, defined by (1.4), (1.5), and (2.18), \(x_{0}\)
be the unique solution of the equation
on the interval
\((0, \infty)\)
and
\(\beta(p)=\operatorname{erf}(x_{0})/B_{p}(x_{0})\). Then the following statements are true:
$$ \frac{d}{dx} \biggl(\frac{B^{2}_{p}(x)}{\operatorname {erf}^{2}(x)} \biggr)=0 $$
(1)
the function
\(x\rightarrow Q_{p}(x)=\operatorname {erf}(x)/B_{p}(x)\)
is strictly decreasing on
\((0, \infty)\)
if and only if
\(p\in(7/5, 8/5]\), and the double inequality
holds for all
\(x>0\)
with the best possible parameters 1 and
\(\alpha(p)\)
if
\(p\in(7/5, 8/5]\);
$$ 1< \frac{\operatorname{erf}(x)}{B_{p}(x)}< \alpha(p) $$
(3.1)
(2)
the function
\(x\rightarrow Q_{p}(x)=\operatorname {erf}(x)/B_{p}(x)\)
is strictly increasing on
\((0, \infty)\)
if and only if
\(p\in [p^{\ast}_{0}, \infty)\), and the double inequality
holds for all
\(x>0\)
with the best possible parameters 1 and
\(\alpha(p)\)
if
\(p\in[p^{\ast}_{0}, \infty)\);
$$ \alpha(p)< \frac{\operatorname{erf}(x)}{B_{p}(x)}< 1 $$
(3.2)
(3)
if
\(p\in(8/5, p^{\ast}_{0})\), then
\(Q_{p}(x)\)
is strictly decreasing on
\((0, x_{0}]\)
and strictly increasing on
\([x_{0}, \infty)\), and the double inequality
for all
\(x>0\).
$$ \beta(p)\leq\frac{\operatorname{erf}(x)}{B_{p}(x)}< \max\bigl\{ 1, \alpha (p)\bigr\} $$
(3.3)
Proof
Let \(t=(p-\mu(p))x^{2}\), \(f_{1}(x)\), \(g_{1}(x)\), \(f_{2}(x)\), \(g_{2}(x)\), \(u_{n}\), \(v_{n}\), and \(h_{2}(t)\) be defined by (2.32) and (2.33), (2.4), (2.51), and (2.53). Then
$$\begin{aligned}& Q^{-2}_{p}(x)=\frac{f_{1}(x)}{g_{1}(x)}, \\& f_{1}\bigl(0^{+}\bigr)=g_{1} \bigl(0^{+}\bigr)=0,\qquad g^{\prime}_{1}(x)>0, \end{aligned}$$
(3.4)
$$\begin{aligned}& f_{2}\bigl(0^{+}\bigr)=g_{2}\bigl(0^{+} \bigr)=0,\qquad g^{\prime}_{2}(x)>0, \end{aligned}$$
(3.5)
$$\begin{aligned}& \frac{f^{\prime}_{1}(x)}{g^{\prime}_{1}(x)}=\frac{f_{2}(x)}{g_{2}(x)}. \end{aligned}$$
(3.6)
(1) If \(Q_{p}(x)=\operatorname{erf}(x)/B_{p}(x)\) is strictly decreasing on \((0, \infty)\), then \(f_{1}(x)/g_{1}(x)\) is strictly increasing on \((0, \infty)\) and \(p\in(7/5, 8/5]\) by Lemma 2.5(1).
If \(p\in(7/5, 8/5]\), then it follows from Lemma 2.2(1) and (2) together with (2.51)-(2.53) that the function \(f^{\prime}_{2}(x)/g^{\prime}_{2}(x)\) is strictly increasing on \((0, \infty)\). Then from the monotone form of L’Hôpital’s rule [22], Theorem 1.25, and (3.5) together with (3.6) we know that the function \(f^{\prime}_{1}(x)/g^{\prime}_{1}(x)\) is strictly increasing on \((0, \infty)\). Therefore, \(Q_{p}(x)\) is strictly decreasing or \(f_{1}(x)/g_{1}(x)\) is strictly increasing on \((0, \infty)\) as follows from the monotone form of L’Hôpital’s rule [22], Theorem 1.25, and (3.4) together with the monotonicity of the function \(f^{\prime}_{1}(x)/g^{\prime}_{1}(x)\) on the interval \((0, \infty)\).
Note that Therefore, the double inequality (3.1) holds for all \(x>0\) and \(p\in (7/5, 8/5]\) with the best possible parameters 1 and \(\alpha(p)\) as follows from (3.7) and the monotonicity of \(f_{1}(x)/g_{1}(x)\) on the interval \((0, \infty)\).
$$ \lim_{x\rightarrow 0^{+}}\frac{f_{1}(x)}{g_{1}(x)}= \frac{\pi}{4}\bigl[p\lambda(p)+\mu (p) \bigl(1-\lambda(p)\bigr)\bigr]= \frac{1}{\alpha^{2}(p)},\qquad \lim_{x\rightarrow\infty}\frac{f_{1}(x)}{g_{1}(x)}=1. $$
(3.7)
(2) If \(Q_{p}(x)=\operatorname{erf}(x)/B_{p}(x)\) is strictly increasing on \((0, \infty)\), then \(f_{1}(x)/g_{1}(x)\) is strictly decreasing on \((0, \infty)\) and \(p\in[p^{\ast}_{0}, \infty)\) by Lemma 2.5(2).
If \(p\in[p^{\ast}_{0}, \infty)\), then it follows from Lemma 2.2(1) and (3) together with (2.51)-(2.53) that the function \(f^{\prime}_{2}(x)/g^{\prime}_{2}(x)\) is strictly decreasing on \((0, \infty)\). Therefore, \(Q_{p}(x)\) is strictly increasing or \(f_{1}(x)/g_{1}(x)\) is strictly decreasing on \((0, \infty)\) as follows from the monotone form of L’Hôpital’s rule and (3.4)-(3.6) together with the monotonicity of the function \(f^{\prime}_{2}(x)/g^{\prime}_{2}(x)\) on the interval \((0, \infty)\), and the double inequality (3.2) holds for all \(x>0\) and \(p\in [p^{\ast}_{0}, \infty)\) with the best possible parameters 1 and \(\alpha(p)\) as follows from (3.7) and the monotonicity of \(f_{1}(x)/g_{1}(x)\) on the interval \((0, \infty)\).
(3) If \(p\in(8/5, p^{\ast}_{0})\), then it follows from [23], Lemma 6.4, or [24], Lemma 7, Lemma 2.2(1) and (4), (2.34), (2.35), and (2.51)-(2.53) that there exists \(x_{1}\in(0, \infty)\) such that \(f^{\prime}_{2}(x)/g^{\prime}_{2}(x)\) is strictly increasing on \((0, x_{1})\) and strictly decreasing on \((x_{1}, \infty)\), and
$$\begin{aligned}& H_{f_{1}, g_{1}}(\infty)=-1, \end{aligned}$$
(3.8)
$$\begin{aligned}& H_{f_{2}, g_{2}}(\infty)=-\infty. \end{aligned}$$
(3.9)
From Lemma 2.1, (3.5), (3.6), (3.9) and the piecewise monotonicity of \(f^{\prime}_{2}(x)/g^{\prime}_{2}(x)\) on the interval \((0, \infty)\) we known that there exists \(x_{2}\in(0, \infty)\) such that \(f^{\prime}_{1}(x)/g^{\prime}_{1}(x)\) is strictly increasing on \((0, x_{2})\) and strictly decreasing on \((x_{2}, \infty)\). Then (3.4), (3.8) and Lemma 2.1 lead to the conclusion that there exists \(x_{0}\in(0, \infty)\) such that the function \(f_{1}(x)/g_{1}(x)=B^{2}_{p}(x)/\operatorname{erf}^{2}(x)\) is strictly increasing on \((0, x_{0})\) and strictly decreasing on \((x_{0}, \infty)\). We clearly see that \(x_{0}\) is the unique solution of the equation on the interval \((0, \infty)\). Therefore, \(Q_{p}(x)=\operatorname{erf}(x)/B_{p}(x)=(f_{1}(x)/g_{1}(x))^{-1/2}\) is strictly decreasing on \((0, x_{0}]\) and strictly increasing on \([x_{0}, \infty)\), and inequality (3.3) holds for all \(x>0\) as follows easily from (3.7). □
$$ \frac{d}{dx} \biggl(\frac{B^{2}_{p}(x)}{\operatorname {erf}^{2}(x)} \biggr)=0 $$
Let \(p\in(7/5, \infty)\) and \(\alpha(p)=\sqrt{ (45p^{2}-60p-4 )/[3p\pi(5p-7)]}=1\), then \(p=p_{0}=(21\pi-60+\sqrt{3}\sqrt{147\pi^{2}-920\pi+1440})/[30(\pi -3)]=1.71318\ldots\in (7/5, p^{\ast}_{0})\). Numerical computations show that \(x_{0}=1.68913\ldots\) is the unique solution of the equation on the interval \((0, \infty)\), \(\beta(p_{0})=\operatorname{erf}(x_{0})/B_{p_{0}}(x_{0})=0.9998\ldots \) . Therefore, Theorem 3.1(3) leads to Corollary 3.1 immediately.
$$ \frac{d}{dx} \biggl(\frac{B^{2}_{p_{0}}(x)}{\operatorname {erf}^{2}(x)} \biggr)=0 $$
(3.10)
Corollary 3.1
Let
\(p_{0}=(21\pi-60+\sqrt{3}\sqrt{147\pi^{2}-920\pi+1440})/[30(\pi-3)]\), \(B_{p}(x)\)
be defined by (1.5) and
\(x_{0}=1.68913\ldots\)
be the unique solution of equation (3.10) on the interval
\((0, \infty)\). Then the double inequality
holds for all
\(x>0\).
$$ 0.9998< \frac{\operatorname{erf}(x_{0})}{B_{p_{0}}(x_{0})}\leq\frac {\operatorname{erf} (x)}{B_{p_{0}}(x)}< 1 $$
(3.11)
Theorem 3.2
Let
\(p_{0}=(21\pi-60+\sqrt{3}\sqrt{147\pi^{2}-920\pi+1440})/[30(\pi-3)]\), \(p\in(7/5, \infty)\), \(x>0\), \(\lambda(p)\), \(\mu(p)\)
and
\(B_{p}(x)\)
be, respectively, defined by (1.4) and (1.5). Then the inequality
holds for all
\(x>0\)
if and only if
\(p\in(7/5, 8/5]\), and inequality (3.12) is reversed if and only if
\(p\in[p_{0}, \infty)\).
$$ \operatorname{erf}(x)>B_{p}(x) $$
(3.12)
Proof
Making use of the L’Hôspital rule and Lemma 2.2(1) we have
$$\begin{aligned}& \lim_{x\rightarrow \infty}\frac{\operatorname{erf}^{2}(x)-B^{2}_{p}(x)}{e^{-\mu (p)x^{2}}} \\& \quad = \lim_{x\rightarrow \infty} \biggl[-\frac{2\operatorname{erf}(x)}{\sqrt{\pi}\mu (p)} \frac{e^{(\mu (p)-1)x^{2}}}{x}+\frac{p\lambda(p)}{ \mu(p)}e^{-(p-\mu(p))x^{2}}+1-\lambda(p) \biggr], \end{aligned}$$
(3.13)
$$\begin{aligned}& p>\mu(p),\qquad 0< \lambda(p)\leq\frac{8+\sqrt{14}}{16}=0.73385\ldots, \end{aligned}$$
(3.14)
$$\begin{aligned}& 0< \mu(p)\leq1\quad (7/5< p\leq8/5), \end{aligned}$$
(3.15)
$$\begin{aligned}& 1< \mu(p)< \frac{4}{3} \quad (8/5< p< \infty). \end{aligned}$$
(3.16)
It follows from (3.13)-(3.16) that
$$ \lim_{x\rightarrow \infty}\frac{\operatorname{erf}^{2}(x)-B^{2}_{p}(x)}{e^{-\mu(p)x^{2}}}= \textstyle\begin{cases} 1-\lambda(p)>0, & p\in (\frac{7}{5}, \frac{8}{5} ], \\ -\infty, & p\in (\frac{8}{5}, \infty ). \end{cases} $$
(3.17)
If inequality (3.12) holds for all \(x>0\), then and \(p\in(7/5, 8/5]\) as follows easily from (3.17) and (3.18).
$$ \lim_{x\rightarrow \infty}\frac{\operatorname{erf}^{2}(x)-B^{2}_{p}(x)}{e^{-\mu (p)x^{2}}}\geq0, $$
(3.18)
If \(p\in(7/5, 8/5]\), then inequality (3.12) holds for all \(x>0\) as follows directly from Theorem 3.1(1).
If \(\operatorname{erf}(x)< B_{p}(x)\) for all \(x>0\), then \(p\geq p_{0}\) as follows easily from
$$ \lim_{x\rightarrow 0^{+}}\frac{\operatorname{erf}(x)}{B_{p}(x)}=\alpha(p)=\sqrt{ \frac {45p^{2}-60p-4}{3\pi p(5p-7)}}\leq1. $$
If \(p\in[p_{0}, \infty)\), then we divide the proof into two cases.
Case 1. \(p\in[p^{\ast}_{0}, \infty)\). Then \(\operatorname {erf}(x)< B_{p}(x)\) for all \(x>0\) as follows from Theorem 3.1(2).
Remark 3.1
Let \(p^{\ast}_{0}=(50+2\sqrt{30})/35\), and \(f_{2}(x)\) and \(g_{2}(x)\) be defined by (2.33). Then from (3.6) and the proof of Theorem 3.1 we know that the function \(f_{2}(x)/g_{2}(x)\) is strictly increasing on \((0, \infty)\) if \(p\in(7/5, 8/5]\) and strictly decreasing on \((0, \infty)\) if \(p\in[p^{\ast}_{0}, \infty)\). Therefore, we have for all \(x\in(0, \infty)\) and \(p\in(7/5, 8/5]\), and for all \(x\in(0, \infty)\) and \(p\in[p^{\ast}_{0}, \infty)\).
$$ \frac{\pi}{4}\bigl[p\lambda(p)+\mu(p) \bigl(1-\lambda(p)\bigr)\bigr]=\lim _{x\rightarrow 0^{+}}\frac{f_{2}(x)}{g_{2}(x)}< \frac{f_{2}(x)}{g_{2}(x)}< \lim _{x\rightarrow \infty}\frac{f_{2}(x)}{g_{2}(x)}=\infty $$
$$ 0=\lim_{x\rightarrow \infty}\frac{f_{2}(x)}{g_{2}(x)}< \frac{f_{2}(x)}{g_{2}(x)}< \lim _{x\rightarrow 0^{+}}\frac{f_{2}(x)}{g_{2}(x)}=\frac{\pi}{4}\bigl[p\lambda(p)+ \mu (p) \bigl(1-\lambda(p)\bigr)\bigr] $$
Theorem 3.3
Let
\(p^{\ast}_{0}=(50+2\sqrt{30})/35\)
and
\(C_{p}(x)\)
be defined by Lemma
2.3. Then the inequality
holds for all
\(x>0\)
if
\(p\in(7/5, 8/5]\), and inequality (3.20) is reversed for all
\(x>0\)
if
\(p\in[p^{\ast}_{0}, \infty)\).
$$ \operatorname{erf}(x)< \frac{2xe^{x^{2}}}{\sqrt{\pi}}C_{p}(x) $$
(3.20)
Remark 3.2
Let \(p_{0}=(21\pi-60+\sqrt{3}\sqrt{147\pi^{2}-920\pi+1440})/[30(\pi-3)]\), \(p, q\in(7/5, \infty)\), \(x>0\), \(\lambda(p)\), \(\mu(p)\) and \(B_{p}(x)\) be, respectively, defined by (1.4) and (1.5). Then it follows from Lemma 2.3(1) and Theorem 3.2 that the double inequality holds for all \(x>0\) with the best possible parameters \(p=8/5\) and \(q=p_{0}\).
$$\begin{aligned}& \sqrt{1-\lambda(p)e^{-px^{2}}-\bigl[1-\lambda(p)\bigr]e^{-\mu (p)x^{2}}} \\& \quad =B_{p}(x)< \operatorname{erf}(x)< B_{q}(x)=\sqrt{1- \lambda(q)e^{-qx^{2}}-\bigl[1-\lambda(q)\bigr]e^{-\mu(q)x^{2}}} \end{aligned}$$
Corollary 3.2
Let
\(p_{0}=(21\pi-60+\sqrt{3}\sqrt{147\pi^{2}-920\pi+1440})/[30(\pi-3)]\), and
\(\lambda(p)\)
and
\(\mu(p)\)
be defined by (1.4). Then the inequalities
hold for all
\(x>0\).
$$\begin{aligned} \sqrt{1-e^{-x^{2}}}&< \sqrt{1-\frac{25}{57}e^{-8x^{2}/5}- \frac {32}{57}e^{-x^{2}}}< \operatorname{erf}(x) \\ &< \sqrt{1-\lambda(p_{0})e^{-p_{0}x^{2}}-\bigl[1- \lambda(p_{0})\bigr]e^{-\mu (p_{0})x^{2}}}< \sqrt{1-e^{4x^{2}/\pi}} \end{aligned}$$
Proof
From (1.4) one has Note that \(p_{0}\) satisfies the identity
$$ \lambda \biggl(\frac{8}{5} \biggr)=\frac{25}{57}, \qquad \mu \biggl(\frac{8}{5} \biggr)=1. $$
(3.21)
$$ \frac{45p^{2}_{0}-60p_{0}-4}{p_{0}(5p_{0}-7)}=3\pi. $$
(3.22)
It follows from Remark 3.2 and (3.21) that for all \(x>0\). Therefore, it suffices to prove that
for all \(x>0\).
$$ \sqrt{1-\frac{25}{57}e^{-8x^{2}/5}-\frac {32}{57}e^{-x^{2}}}< \operatorname{erf} (x)< \sqrt{1-\lambda(p_{0})e^{-p_{0}x^{2}}-\bigl[1- \lambda(p_{0})\bigr]e^{-\mu (p_{0})x^{2}}} $$
$$\begin{aligned}& \frac{25}{57}e^{-8x^{2}/5}+\frac{32}{57}e^{-x^{2}}< e^{-x^{2}}, \end{aligned}$$
(3.23)
$$\begin{aligned}& \lambda(p_{0})e^{-p_{0}x^{2}}+\bigl[1-\lambda(p_{0}) \bigr]e^{-\mu (p_{0})x^{2}}>e^{-4x^{2}/\pi} \end{aligned}$$
(3.24)
Inequality (3.23) follows easily from
$$ \frac{25}{57}e^{-8x^{2}/5}+\frac{32}{57}e^{-x^{2}}-e^{-x^{2}}= \frac {25}{57}e^{-8x^{2}/5} \bigl(1-e^{3x^{2}/5} \bigr). $$
Making use of (1.4) and (3.22) together with the arithmetic-geometric mean inequality one has for all \(x>0\). □
$$\begin{aligned} \lambda(p_{0})e^{-p_{0}x^{2}}+\bigl[1-\lambda(p_{0}) \bigr]e^{-\mu (p_{0})x^{2}}&>e^{- [p_{0}\lambda(p_{0})+\mu(p_{0})(1-\lambda (p_{0})) ]x^{2}} \\ &=e^{-\frac {12p_{0}(5p_{0}-7)}{45p^{2}_{0}-60p_{0}-4}x^{2}}=e^{-4x^{2}/\pi} \end{aligned}$$
Remark 3.3
Let \(p^{\ast}_{0}=(50+2\sqrt{30})/35\) and \(\alpha(p)\) be defined by (2.18). Then
$$\begin{aligned}& \alpha \biggl(\frac{8}{5} \biggr)=\sqrt{\frac{19}{6\pi }}=1.00398 \ldots,\qquad \alpha \biggl(\frac{3}{2} \biggr)=\sqrt{ \frac{29}{9\pi }}=1.01275\ldots, \end{aligned}$$
(3.25)
$$\begin{aligned}& \alpha(\infty)=\sqrt{\frac{3}{\pi}}=0.97720\ldots,\qquad \alpha(2)=\sqrt{ \frac{28}{9\pi}}=0.99513\ldots, \\& \alpha\bigl(p^{\ast}_{0}\bigr)=\sqrt{\frac{160}{51\pi}}=0.99930 \ldots. \end{aligned}$$
(3.26)
Corollary 3.3
Let
\(p=3/2, 8/5\)
in Theorem
3.1(1) and
\(p=p^{\ast}_{0}, 2, \infty\)
in Theorem
3.1(2). Then Lemma
2.3(1) and (3) together with (3.25) and (3.26) leads to
$$\begin{aligned}& \sqrt{1-\frac{128}{203}e^{-3x^{2}/2}-\frac {75}{203}e^{-4x^{2}/5}} \\& \quad < \sqrt {1-\frac{25}{57}e^{-8x^{2}/5}-\frac{32}{57}e^{-x^{2}}}< \operatorname{erf} (x) \\& \quad < \sqrt{\frac{19}{6\pi}}\sqrt{1-\frac{25}{57}e^{-8x^{2}/5}- \frac {32}{57}e^{-x^{2}}} \\& \quad < \sqrt{\frac{29}{9\pi}}\sqrt{1- \frac{128}{203}e^{-3x^{2}/2}-\frac {75}{203}e^{-4x^{2}/5}}, \end{aligned}$$
(3.27)
$$\begin{aligned}& \sqrt{\frac{3}{\pi}}\sqrt{1-e^{-4x^{2}/3}} \\& \quad < \sqrt{\frac{28}{9\pi }}\sqrt {1-\frac{3}{28}e^{-2x^{2}}- \frac{25}{28}e^{-6x^{2}/5}} \\& \quad < \sqrt{\frac{160}{51\pi}}\sqrt{1-\frac{480-43\sqrt {30}}{960}e^{-(50+2\sqrt{30})x^{2}/35}- \frac{480+43\sqrt {30}}{960}e^{-(50-2\sqrt{30})x^{2}/35}} \\& \quad < \operatorname{erf}(x)< \sqrt{1-\frac{480-43\sqrt {30}}{960}e^{-(50+2\sqrt {30})x^{2}/35}- \frac{480+43\sqrt{30}}{960}e^{-(50-2\sqrt {30})x^{2}/35}} \\& \quad < \sqrt{1-\frac{3}{28}e^{-2x^{2}}-\frac{25}{28}e^{-6x^{2}/5}} \\& \quad < \sqrt {1-e^{-4x^{2}/3}}. \end{aligned}$$
(3.28)
Remark 3.4
From inequality (3.27) we clearly see that the double inequalities hold for all \(x>0\).
$$\begin{aligned}& 0< \frac{\operatorname{erf}(x)-B_{3/2}(x)}{\operatorname {erf}(x)}< \sqrt{\frac{29}{9\pi }}-1=0.01275\ldots, \\& 0< \frac{\operatorname{erf}(x)-B_{3/2}(x)}{\operatorname {erf}(x)}< \sqrt{\frac{19}{6\pi }}-1=0.00398\ldots , \end{aligned}$$
Remark 3.5
Let \(p_{0}=(21\pi-60+\sqrt{3}\sqrt{147\pi^{2}-920\pi+1440})/[30(\pi-3)]\), \(p^{\ast}_{0}=(50+2\sqrt{30})/35\), \(B_{p}(x)\) be defined by (1.5) and \(x_{0}=1.68913\ldots\) be the unique solution of equation (3.10) on the interval \((0, \infty)\), \(\beta(p_{0})=\operatorname{erf}(x_{0})/B_{p_{0}}(x_{0})=0.9998\ldots \) . Then Corollary 3.1 and (3.28) lead to for all \(x>0\).
$$\begin{aligned}& -0.00013\ldots=1-\frac{1}{\beta(p_{0})}< \frac{\operatorname{erf} (x)-B_{p_{0}}(x)}{\operatorname{erf} (x)}\leq 0, \\& -0.00069\ldots=1-\sqrt{\frac{51\pi}{160}}< \frac{\operatorname{erf} (x)-B_{p^{\ast }_{0}}(x)}{\operatorname{erf}(x)}< 0, \\& -0.00488\ldots=1-\sqrt{\frac{9\pi}{28}}< \frac{\operatorname{erf} (x)-B_{2}(x)}{\operatorname{erf}(x)}< 0, \\& -0.02332\ldots=1-\sqrt{\frac{\pi}{3}}< \frac{\operatorname {erf}(x)-\sqrt {1-e^{-4x^{2}/3}}}{\operatorname{erf}(x)}< 0 \end{aligned}$$
Corollary 3.4
Let
\(p=(7/5)^{+}, 3/2, 8/5\)
and
\(p=2, \infty\)
in Theorem
3.3. Then it follows from Lemma
2.3(2) that the inequalities
hold for all
\(x>0\).
$$\begin{aligned} \frac{2x}{\sqrt{\pi}}e^{-x^{2}/3}&< \frac{2x}{\sqrt{\pi}}\frac {e^{-x^{2}}+5e^{-x^{2}/5}}{6}< \operatorname{erf}(x) < \frac{2x}{\sqrt{\pi}}\frac{5e^{-3x^{2}/5}+4}{9} \\ &< \frac{2x}{\sqrt{\pi}}\frac{16e^{-x^{2}/2}+5e^{x^{2}/5}}{21}< \frac {2x}{\sqrt{\pi}}\frac{20e^{-2x^{2}/5}+e^{x^{2}}}{21} \end{aligned}$$
Remark 3.6
From the identities and we know that the results given in Theorem 3.3 or Corollary 3.4 are better than that in (1.3).
$$\begin{aligned}& \frac{e^{-x^{2}}+5e^{-x^{2}/5}}{6}-e^{-x^{2}/3} \\& \quad =\frac {1}{6}e^{-x^{2}} \bigl(e^{2x^{2}/15}-1 \bigr)^{2} \bigl(5e^{8x^{2}/15}+4e^{2x^{2}/5}+3e^{4x^{2}/15}+2e^{2x^{2}/15}+1 \bigr) \end{aligned}$$
$$\begin{aligned}& \frac{5e^{-3x^{2}/5}+4}{9}-\frac{e^{-x^{2}}+2}{3} \\& \quad = -\frac{e^{-x^{2}}}{9} \bigl(e^{x^{2}/5}-1 \bigr)^{2} \bigl(2e^{3x^{2}/5}+4e^{2x^{2}/5}+6e^{x^{2}/5}+3 \bigr) \end{aligned}$$
Acknowledgements
The research was supported by the Natural Science Foundation of China under Grants 61374086, 11371125 and 11401191.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.