1 A concise retrospection
In 1912, Young proved the following integral inequality in which an inverse function is involved.
Theorem 1.1
([29])
Anzeige
Let
\(h(x)\)
be a continuous and strictly increasing function on
\([0,c]\)
with
\(c>0\). If
\(h(0)=0\), \(a\in [0,c]\), and
\(b\in [0,h(c)]\), then
where
\(h^{-1}\)
is the inverse function of
h. The equality in (1.1) is valid if and only if
\(b=h(a)\).
$$ \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x\ge ab, $$
(1.1)
For avoiding confusion, we call inequality (1.1) Young’s integral inequality, because several other inequalities, such as
and the weighted arithmetic-geometric mean inequality, are also called Young’s inequality.
$$ \sum_{k=1}^{n}\frac{\cos (k\theta )}{k}>-1, \quad n\ge 2, \theta \in [0,\pi ] $$
In [9, Secton 2.7] and [10, Chapter XIV], many extensions, refinements, generalizations, and applications of Young’s integral inequality (1.1) in Theorem 1.1 were collected and reviewed.
In 2008, Hoorfar and Qi obtained the following double inequality which refines Young’s integral inequality (1.1).
Anzeige
Theorem 1.2
([4])
Let
\(h(x)\)
be a differentiable and strictly increasing function on
\([0,c]\)
for
\(c>0\), and let
\(h^{-1}\)
be the inverse function of
h. If
\(h(0)=0\), \(a\in [0,c]\), \(b\in [0,h(c)]\), and
\(h'(x)\)
is strictly monotonic on
\([0,c]\), then
where
The equalities in (1.2) are valid if and only if
\(b=h(a)\).
$$ \frac{m}{2} \bigl[a-h^{-1}(b) \bigr]^{2} \le \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab\le \frac{M}{2} \bigl[a-h^{-1}(b) \bigr]^{2}, $$
(1.2)
$$ m=\min \bigl\{ h'(a), h' \bigl(h^{-1}(b) \bigr) \bigr\} \quad \textit{and}\quad M=\max \bigl\{ h'(a), h' \bigl(h^{-1}(b) \bigr) \bigr\} . $$
In 2009 and 2010, motivated by Theorem 1.2 and its proof in [4], among other things, Jakšetić and Pečarić extended and generalized Young’s integral inequality (1.1) and Hoorfar–Qi’s double inequality (1.2) as the following theorems.
Theorem 1.3
Let
\(h(x)\)
be a differentiable and strictly increasing function on
\([0,c]\)
for
\(c>0\), \(h(0)=0\), \(a\in [0,c]\), \(b\in [0,h(c)]\), and
\(h^{-1}\)
be the inverse function of
h. Denote
$$ \alpha =\min \bigl\{ a,h^{-1}(b) \bigr\} \quad \textit{and} \quad \beta = \max \bigl\{ a,h^{-1}(b) \bigr\} . $$
(1.3)
1.
If
\(h'(x)\)
is increasing on
\([\alpha ,\beta ]\)
and
\(b< h(a)\), or if
\(h'(x)\)
is decreasing on
\([\alpha ,\beta ]\)
and
\(b>h(a)\), then
$$\begin{aligned} \bigl[a-h^{-1}(b) \bigr] \biggl[h \biggl( \frac{a+h^{-1}(b)}{2} \biggr)-b \biggr] \le& \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab \\ \le& \frac{1}{2} \bigl[a-h^{-1}(b) \bigr] \bigl[h(a)-b\bigr]. \end{aligned}$$
(1.4)
2.
If
\(h'(x)\)
is increasing on
\([\alpha ,\beta ]\)
and
\(b>h(a)\), or if
\(h'(x)\)
is decreasing on
\([\alpha ,\beta ]\)
and
\(b< h(a)\), then inequality (1.4) is reversed.
3.
Theorem 1.4
([6, Theorem 2.6])
Let
\(h(x)\)
be a differentiable and strictly increasing function on
\([0,c]\)
for
\(c>0\), and let
\(h^{-1}\)
be the inverse function of
h. If
\(h(0)=0\), \(a\in [0,c]\), \(b\in [0,h(c)]\), and
\(h'(x)\)
is convex on
\([\alpha ,\beta ]\), then
If
\(h'(x)\)
is concave, then the double inequality (1.5) is reversed.
$$\begin{aligned} \frac{ [a-h^{-1}(b) ]^{2}}{2}h' \biggl(\frac{a+2h^{-1}(b)}{3} \biggr) \le& \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab \\ \le& \frac{ [a-h^{-1}(b) ]^{2}}{3} \biggl[\frac{h'(a)}{2}+h' \bigl(h^{-1}(b) \bigr) \biggr]. \end{aligned}$$
(1.5)
Theorem 1.5
([6, Theorem 2.1])
Let
\(h(x)\)
be a differentiable and strictly increasing function on
\([0,c]\)
for
\(c>0\), and let
\(h^{-1}\)
be the inverse function of
h. If
\(h(0)=0\), \(a\in [0,c]\), \(b\in [0,h(c)]\), and
\(h'(x)\)
is almost everywhere continuous with respect to Lebesgue measure on
\([\alpha , \beta ]\), then the double inequality
is valid for all
u, v
and
p, q
satisfying
where
and
$$ C_{u} \bigl\Vert h' \bigr\Vert _{v}\le \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab \le C_{p} \bigl\Vert h' \bigr\Vert _{q} $$
(1.6)
1.
\(\frac{1}{u}+\frac{1}{v}=1\)
for
\(u,v\in (-\infty ,0)\cup (0,1)\), or
\((u,v)=(1,-\infty )\), or
\((u,v)=(-\infty ,1)\);
2.
\(\frac{1}{p}+\frac{1}{q}=1\)
for
\(1< p\), \(q<\infty \), or
\((p,q)=(+\infty ,1)\), or
\((p,q)=(1,+\infty )\);
$$ C_{r}= \textstyle\begin{cases} [\frac{ \vert a-h^{-1}(b) \vert ^{r+1}}{r+1} ]^{1/r}, & r \ne 0,\pm \infty ; \\ \vert a-h^{-1}(b) \vert , & r=+\infty ; \\ 0, & r=-\infty \end{cases} $$
$$ \bigl\Vert h' \bigr\Vert _{r}= \textstyle\begin{cases} [\int _{\alpha }^{\beta }[h'(t)]^{r}\,\mathrm {d}t ]^{1/r}, & r \ne 0,\pm \infty ; \\ \sup \{h'(t),t\in [\alpha ,\beta ]\}, & r=+\infty ; \\ \inf \{h'(t),t\in [\alpha ,\beta ]\}, & r=-\infty . \end{cases} $$
We note that Hoorfar–Qi’s double inequality (1.2) has been applied and employed in the paper [7] and in the Undergraduate Texts in Mathematics [8].
In this paper, by virtue of Taylor’s theorems with different remainders, we establish some integral inequalities of Hoorfar–Qi’s type in terms of higher order derivatives and their norms, demonstrate that these newly-established integral inequalities generalize Young’s integral inequality (1.1), Hoorfar–Qi’s integral inequality (1.2), and Jakšetić–Pečarić’s integral inequalities (1.4), (1.5), and (1.6), and apply these integral inequalities to estimate several concrete definite integrals, including a definite integral of \(e^{-1/x}\) which plays an indispensable role in differential geometry and has a connection with the Lah numbers in combinatorics, the exponential integral \(\operatorname{Ei}(x)\), and the logarithmic integral \(\operatorname{li}(x)\).
2 Lemmas
For proving our main result, we need Taylor’s theorems below.
Lemma 2.1
([2, p. 113, Theorem 5.19])
Let
\(f(x)\)
be a function having finite
nth derivative
\(f^{(n)}(x)\)
everywhere in an open interval
\((\mu ,\nu )\)
and assume that
\(f^{(n-1)}(x)\)
is continuous on the closed interval
\([\mu ,\nu ]\). Then, for a fixed point
\(x_{0}\in [\mu ,\nu ]\)
and every
\(x\in [\mu ,\nu ]\)
with
\(x\ne x_{0}\), there exists a point
\(x_{1}\)
interior to the interval joining
x
and
\(x_{0}\)
such that
$$ f(x)=f(x_{0})+\sum_{k=1}^{n-1} \frac{f^{(k)}(x_{0})}{k!}(x-x_{0})^{k}+\frac{f ^{(n)}(x_{1})}{n!}(x-x_{0})^{n}. $$
(2.1)
Lemma 2.2
If
\(f(x)\in C^{n+1}[\mu ,\nu ]\)
and
\(x_{0}\in [\mu ,\nu ]\), then
$$ f(x)=\sum_{k=0}^{n} \frac{f^{(k)}(x_{0})}{k!}(x-x_{0})^{k}+ \frac{1}{n!} \int _{x_{0}}^{x}(x-t)^{n}f^{(n+1)}(t) \,\mathrm {d}t. $$
(2.2)
For proving our main results, we also need Hölder’s integral inequality and its reversed version, Čebyšev’s integral inequality, discrete and integral versions of Jensen’s inequality, and Hermite–Hadamard’s integral inequality.
Lemma 2.3
Let
\(\frac{1}{p}+\frac{1}{q}=1\)
with
\(p>0\)
and
\(p\ne 1\), let
f
and
g
be real functions on
\([\mu ,\nu ]\), and let
\(|f|^{p}\)
and
\(|g|^{q}\)
be integrable on
\([\mu ,\nu ]\).
1.
If
\(p>1\), then
The equality in (2.3) holds if and only if
\(A|f(x)|^{p}=B|g(x)|^{q}\)
almost everywhere for two constants
A
and
B.
$$ \int _{\mu }^{\nu } \bigl\vert f(x)g(x) \bigr\vert \,\mathrm {d}x\le \biggl[ \int _{\mu }^{\nu } \bigl\vert f(x) \bigr\vert ^{p} \,\mathrm {d}x \biggr]^{1/p} \biggl[ \int _{\mu }^{\nu } \bigl\vert g(x) \bigr\vert ^{q}\,\mathrm {d}x \biggr]^{1/q}. $$
(2.3)
Lemma 2.4
Let
\(f,g:[\mu ,\nu ]\to \mathbb{R}\)
be integrable functions satisfying that they are both increasing or both decreasing. Then
If one of the functions
f
or
g
is nonincreasing and the other nondecreasing, then the inequality in (2.4) is reversed.
$$ \int _{\mu }^{\nu }f(x)\,\mathrm {d}x \int _{\mu }^{\nu }g(x)\,\mathrm {d}x \le (\nu - \mu ) \int _{\mu }^{\nu }f(x)g(x)\,\mathrm {d}x. $$
(2.4)
Lemma 2.5
If
f
is a convex function on an interval
\(I\subseteq \mathbb{R}\)
and if
\(n\ge 2\)
and
\(x_{k}\in I\)
for
\(1\le k\le n\), then
where
\(p_{k}>0\)
for
\(1\le k\le n\). If
f
is concave, inequality (2.5) is reversed.
$$ f \Biggl(\frac{1}{\sum_{k=1}^{n}p_{k}}\sum_{k=1}^{n}p_{k}x_{k} \Biggr) \le \frac{1}{\sum_{k=1}^{n}p_{k}}\sum_{k=1}^{n}p_{k}f(x_{k}), $$
(2.5)
Let
ϕ
be a convex function on
\([\mu ,\nu ]\), \(f\in L_{1}(\mu , \nu )\), and
σ
be a nonnegative measure. Then
If
ϕ
is a concave function, then inequality (2.6) is reversed.
$$ \phi \biggl(\frac{\int _{\mu }^{\nu }f(x)\,\mathrm {d}\sigma }{\int _{\mu }^{ \nu }\,\mathrm {d}\sigma } \biggr) \le \frac{\int _{\mu }^{\nu }\phi (f(x)) \,\mathrm {d}\sigma }{\int _{\mu }^{\nu }\,\mathrm {d}\sigma }. $$
(2.6)
These five lemmas are general knowledge in mathematics and they will continue to play important roles in this paper.
Lemma 2.6
Let
\(f(x)\)
and
\(g(x)\)
be nonnegative and convex functions on
\([\mu ,\nu ]\). Then
where
$$\begin{aligned}& 2f \biggl(\frac{\mu +\nu }{2} \biggr)g \biggl(\frac{\mu +\nu }{2} \biggr)- \frac{1}{6}M(\mu ,\nu )-\frac{1}{3}N(\mu ,\nu ) \\& \quad \le \frac{1}{\nu -\mu } \int _{\mu }^{\nu }f(x)g(x)\,\mathrm {d}x \le \frac{1}{3}M( \mu ,\nu )+\frac{1}{6}N(\mu ,\nu ), \end{aligned}$$
(2.7)
$$ M(\mu ,\nu )=f(\mu )g(\mu )+f(\nu )g(\nu )\quad \textit{and}\quad N( \mu ,\nu )=f(\mu )g(\nu )+f(\nu )g(\mu ). $$
3 Main results and proofs
Now we are in a position to state and prove our main results.
Theorem 3.1
Let
\(h(0)=0\)
and
\(h(x)\)
be strictly increasing on
\([0,c]\)
for
\(c>0\), let
\(h^{(n)}(x)\)
for
\(n\ge 0\)
be continuous on
\([0,c]\), let
\(h^{(n+1)}(x)\)
be finite and strictly monotonic on
\((0,c)\), and let
\(h^{-1}\)
be the inverse function of
h. For
\(a\in [0,c]\)
and
\(b\in [0,h(c)]\),
1.
if
\(b< h(a)\), then
where
and
$$ \begin{gathered}[b] \sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} +m_{n}(a,b) \frac{ [a-h^{-1}(b) ]^{n+2}}{(n+2)!} \\ \quad \le \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab \\ \quad \le \sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} +M_{n}(a,b) \frac{ [a-h^{-1}(b) ]^{n+2}}{(n+2)!}, \end{gathered} $$
(3.1)
$$ m_{n}(a,b)=\min \bigl\{ h^{(n+1)} \bigl(h^{-1}(b) \bigr),h^{(n+1)}(a) \bigr\} $$
$$ M_{n}(a,b)=\max \bigl\{ h^{(n+1)} \bigl(h^{-1}(b) \bigr),h^{(n+1)}(a) \bigr\} ; $$
2.
if
\(b>h(a)\), then
(b)
when
\(n=2\ell +1\)
for
\(\ell \ge 0\), we have
$$ \begin{gathered}[b] \sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} -M_{n}(a,b) \frac{ [a-h^{-1}(b) ]^{n+2}}{(n+2)!} \\ \quad \le \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab \\ \quad \le \sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} -m_{n}(a,b) \frac{ [a-h^{-1}(b) ]^{n+2}}{(n+2)!}; \end{gathered} $$
(3.2)
Proof
Employing several basic properties of definite integrals, such as substitution of variables and integration by parts, reveals
By virtue of formula (2.1), we have
where ξ is a point interior to the interval joining y and \(h^{-1}(b)\). As a result,
$$ \begin{gathered}[b] \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x \\ \quad = \int _{0}^{a}h(x) \,\mathrm {d}x+ \int _{0}^{h^{-1}(b)}y\,\mathrm {d}h(y) \\ \quad = \int _{0}^{a}h(x)\,\mathrm {d}x+yh(y)|_{0}^{h^{-1}(b)}- \int _{0}^{h^{-1}(b)}h(y) \,\mathrm {d}y =bh^{-1}(b)+ \int _{h^{-1}(b)}^{a}h(y)\,\mathrm {d}y \\ \quad =ab+ \int _{h^{-1}(b)}^{a}\bigl[h(y)-b\bigr]\,\mathrm {d}y =ab+ \int _{h^{-1}(b)}^{a} \bigl[h(y)-h \bigl(h^{-1}(b) \bigr) \bigr]\,\mathrm {d}y. \end{gathered} $$
(3.3)
$$\begin{aligned} h(y)-h \bigl(h^{-1}(b) \bigr) &=\sum_{k=1}^{n} \frac{h^{(k)} (h^{-1}(b) )}{k!} \bigl[y-h^{-1}(b) \bigr]^{k} \\ &\quad {} +\frac{h^{(n+1)}(\xi )}{(n+1)!} \bigl[y-h^{-1}(b) \bigr]^{n+1}, \end{aligned}$$
$$\begin{aligned}& \int _{h^{-1}(b)}^{a} \bigl[h(y)-h \bigl(h^{-1}(b) \bigr) \bigr]\,\mathrm {d}y \\& \quad = \sum_{k=1}^{n} \frac{h^{(k)} (h^{-1}(b) )}{k!} \int _{h^{-1}(b)} ^{a} \bigl[y-h^{-1}(b) \bigr]^{k}\,\mathrm {d}y \\& \qquad {}+\frac{1}{(n+1)!} \int _{h^{-1}(b)}^{a}h^{(n+1)}(\xi ) \bigl[y-h^{-1}(b) \bigr]^{n+1}\,\mathrm {d}y \\& \quad =\sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} + \int _{h^{-1}(b)}^{a}h^{(n+1)}(\xi ) \frac{ [y-h^{-1}(b) ]^{n+1}}{(n+1)!}\,\mathrm {d}y. \end{aligned}$$
When \(h^{-1}(b)< a\), if \(h^{(n+1)}(x)\) is strictly increasing on \((0,c)\), then
if \(h^{(n+1)}(x)\) is strictly decreasing on \((0,c)\), then
Consequently, the double inequality \(m_{n}(a,b)< h^{(n+1)}(\xi )< M_{n}(a,b)\) is valid. Thus, it follows that
Hence, the double inequality (3.1) is proved.
$$ h^{(n+1)} \bigl(h^{-1}(b) \bigr)< h^{(n+1)}(\xi )< h^{(n+1)}(y)\le h^{(n+1)}(a); $$
$$ h^{(n+1)}(a)\le h^{(n+1)}(y)< h^{(n+1)}(\xi )< h^{(n+1)} \bigl(h^{-1}(b) \bigr). $$
$$\begin{aligned} m_{n}(a,b)\frac{ [a-h^{-1}(b) ]^{n+2}}{(n+2)!} \le& \int _{h^{-1}(b)}^{a}h^{(n+1)}(\xi ) \frac{ [y-h^{-1}(b) ]^{n+1}}{(n+1)!}\,\mathrm {d}y \\ \le& M_{n}(a,b)\frac{ [a-h^{-1}(b) ]^{n+2}}{(n+2)!}. \end{aligned}$$
When \(h^{-1}(b)>a\), if \(h^{(n+1)}(x)\) is increasing on \((0,c)\), then
if \(h^{(n+1)}(x)\) is decreasing on \((0,c)\), then
Consequently, the double inequality \(m_{n}(a,b)< h^{(n+1)}(\xi )< M_{n}(a,b)\) is still valid. Hence, since
we acquire
and
for \(\ell \ge 0\). The double inequality (3.2) is thus proved. The proof of Theorem 3.1 is complete. □
$$ h^{(n+1)} \bigl(h^{-1}(b) \bigr)\ge h^{(n+1)}(\xi )\ge h^{(n+1)}(y) \ge h^{(n+1)}(a); $$
$$ h^{(n+1)}(a)\ge h^{(n+1)}(y)\ge h^{(n+1)}(\xi )\ge h^{(n+1)} \bigl(h ^{-1}(b) \bigr). $$
$$\begin{aligned}& \int _{h^{-1}(b)}^{a}h^{(n+1)}(\xi ) \frac{ [y-h^{-1}(b) ]^{n+1}}{(n+1)!}\,\mathrm {d}y \\& \quad =\frac{(-1)^{n}}{(n+1)!} \int _{a}^{h^{-1}(b)}h^{(n+1)}(\xi ) \bigl[h ^{-1}(b)-y \bigr]^{n+1}\,\mathrm {d}y, \end{aligned}$$
$$\begin{aligned} \frac{(-1)^{n}m_{n}(a,b)}{(n+2)!} \bigl[h^{-1}(b)-a \bigr]^{n+2} \le& \int _{h^{-1}(b)}^{a}h^{(n+1)}(\xi ) \frac{ [y-h^{-1}(b) ]^{n+1}}{(n+1)!}\,\mathrm {d}y \\ \le& \frac{(-1)^{n}M_{n}(a,b)}{(n+2)!} \bigl[h^{-1}(b)-a \bigr]^{n+2},\quad n=2\ell, \end{aligned}$$
$$\begin{aligned} \frac{(-1)^{n}m_{n}(a,b)}{(n+2)!} \bigl[h^{-1}(b)-a \bigr]^{n+2} \ge& \int _{h^{-1}(b)}^{a}h^{(n+1)}(\xi ) \frac{ [y-h^{-1}(b) ]^{n+1}}{(n+1)!}\,\mathrm {d}y \\ \ge& \frac{(-1)^{n}M_{n}(a,b)}{(n+2)!} \bigl[h^{-1}(b)-a \bigr]^{n+2}, \quad n=2\ell +1 \end{aligned}$$
Theorem 3.2
Let
\(n\ge 0\)
and
\(h(x)\in C^{n+1}[0,c]\)
such that
\(h(0)=0\), \(h^{(n+1)}(x)\ge 0\)
on
\([\alpha ,\beta ]\), and
\(h(x)\)
is strictly increasing on
\([0,c]\)
for
\(c>0\), let
\(h^{-1}\)
be the inverse function of
h, and let
\(a\in [0,c]\)
and
\(b\in [0,h(c)]\). Then
where
α, β
are defined as in (1.3),
and
u, v, p, q
satisfy
1.
when
\(b>h(a)\)
and
\(n=2\ell \)
for
\(\ell \ge 0\)
or when
\(b< h(a)\), we have
$$\begin{aligned} \frac{C_{u,n}}{(n+1)!} \bigl\Vert h^{(n+1)} \bigr\Vert _{v} \le& \int _{0}^{a}h(x) \,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab \\ &{}-\sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} \le \frac{C_{p,n}}{(n+1)!} \bigl\Vert h^{(n+1)} \bigr\Vert _{q}; \end{aligned}$$
2.
when
\(b>h(a)\)
and
\(n=2\ell +1\)
for
\(\ell \ge 0\), we have
$$\begin{aligned} -\frac{C_{p,n}}{(n+1)!} \bigl\Vert h^{(n+1)} \bigr\Vert _{q} \le& \int _{0}^{a}h(x) \,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab \\ &{}-\sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} \le -\frac{C_{u,n}}{(n+1)!} \bigl\Vert h^{(n+1)} \bigr\Vert _{v}; \end{aligned}$$
$$\begin{aligned}& C_{r,n} = \textstyle\begin{cases} {[\frac{ \vert a-h^{-1}(b) \vert ^{r(n+1)+1}}{r(n+1)+1} ]}^{1/r}, & r\ne 0,\pm \infty ; \\ \vert a-h^{-1}(b) \vert ^{n+1}, & r=+\infty ; \\ 0, & r=-\infty , \end{cases}\displaystyle \\& \bigl\Vert h^{(n+1)} \bigr\Vert _{r} = \textstyle\begin{cases} {[\int _{\alpha }^{\beta } [h^{(n+1)}(t) ]^{r}\,\mathrm {d}t ]}^{1/r}, & r\ne 0,\pm \infty ; \\ \sup \{h^{(n+1)}(t),t\in [ \alpha ,\beta ] \}, & r=+\infty ; \\ \inf \{h^{(n+1)}(t),t \in [\alpha ,\beta ] \}, & r=-\infty , \end{cases}\displaystyle \end{aligned}$$
1.
\(u<1\)
and
\(u\ne 0\)
with
\(\frac{1}{u}+\frac{1}{v}=1\), or
\((u,v)=(- \infty ,1)\), or
\((u,v)=(1,-\infty )\);
2.
\(1< p\), \(q<\infty \)
with
\(\frac{1}{p}+\frac{1}{q}=1\), or
\((p,q)=(+\infty ,1)\), or
\((p,q)=(1,+\infty )\).
Proof
Applying formula (2.2) to the last term in (3.3) yields
$$\begin{aligned}& \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab \\& \quad =\sum _{k=1}^{n}h ^{(k)} \bigl(h^{-1}(b) \bigr) \frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!}+ \int _{h^{-1}(b)}^{a}\frac{1}{n!} \int _{h^{-1}(b)}^{x}(x-t)^{n}h^{(n+1)}(t) \,\mathrm {d}t\,\mathrm {d}x \\& \quad =\sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} + \int _{h^{-1}(b)}^{a}\frac{1}{n!} \int _{t}^{a}(x-t)^{n}h ^{(n+1)}(t) \,\mathrm {d}x\,\mathrm {d}t \\& \quad =\sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} + \int _{h^{-1}(b)}^{a} \frac{(a-t)^{n+1}h^{(n+1)}(t)}{(n+1)!}\,\mathrm {d}t. \end{aligned}$$
(3.4)
It is not difficult to verify that
Since
and, by Hölder’s integral inequality (2.3),
for \(p>1\) and \(\frac{1}{p}+\frac{1}{q}=1\), it follows that where \(\frac{1}{p}+\frac{1}{q}=1\) for \(1< p\), \(q<\infty \), or \((p,q)=(+ \infty ,1)\), or \((p,q)=(1,+\infty )\).
$$ \int _{h^{-1}(b)}^{a}(a-t)^{n+1}h^{(n+1)}(t) \,\mathrm {d}t = \textstyle\begin{cases} \int _{\alpha }^{\beta } \vert a-t \vert ^{n+1}h^{(n+1)}(t)\,\mathrm {d}t, & b< h(a); \\ (-1)^{n}\int _{\alpha }^{\beta } \vert a-t \vert ^{n+1}h^{(n+1)}(t)\,\mathrm {d}t, & b>h(a). \end{cases} $$
$$\begin{aligned}& \begin{aligned} \int _{\alpha }^{\beta } \vert a-t \vert ^{n+1}h^{(n+1)}(t)\,\mathrm {d}t &\le \bigl\vert a-f ^{-1}(b) \bigr\vert ^{n+1} \int _{\alpha }^{\beta }h^{(n+1)}(t)\,\mathrm {d}t \\ &= \bigl\vert a-f^{-1}(b) \bigr\vert ^{n+1} \bigl\Vert h^{(n+1)} \bigr\Vert _{1}, \end{aligned} \\& \begin{aligned} \int _{\alpha }^{\beta } \vert a-t \vert ^{n+1}h^{(n+1)}(t)\,\mathrm {d}t &\le \sup \bigl\{ h^{(n+1)}(t),t\in [\alpha ,\beta ] \bigr\} \int _{\alpha }^{\beta } \vert a-t \vert ^{n+1} \,\mathrm {d}t \\ &=\frac{(\beta -\alpha )^{n+2}}{n+2} \bigl\Vert h^{(n+1)} \bigr\Vert _{+ \infty } = \frac{ \vert a-h^{-1}(b) \vert ^{n+2}}{n+2} \bigl\Vert h^{(n+1)} \bigr\Vert _{+\infty }, \end{aligned} \end{aligned}$$
$$\begin{aligned}& \int _{\alpha }^{\beta } \vert a-t \vert ^{n+1}h^{(n+1)}(t)\,\mathrm {d}t \\& \quad \le \biggl( \int _{\alpha }^{\beta } \vert a-t \vert ^{p(n+1)}\,\mathrm {d}t \biggr)^{1/p} \biggl( \int _{\alpha }^{\beta } \bigl[h^{(n+1)}(t) \bigr]^{q}\,\mathrm {d}t \biggr)^{1/q} \\& \quad = \biggl[\frac{(\beta -\alpha )^{p(n+1)+1}}{p(n+1)+1} \biggr]^{1/p} \bigl\Vert h^{(n+1)} \bigr\Vert _{q} = \biggl[ \frac{ \vert a-h^{-1}(b) \vert ^{p(n+1)+1}}{p(n+1)+1} \biggr]^{1/p} \bigl\Vert h^{(n+1)} \bigr\Vert _{q} \end{aligned}$$
1.
when \(b>h(a)\) and \(n=2\ell \) for \(\ell \ge 0\) or when \(b< h(a)\), we have
$$\begin{aligned}& \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab \\& \quad \le \sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!}+\frac{C_{p,n}}{(n+1)!} \bigl\Vert h^{(n+1)} \bigr\Vert _{q}; \end{aligned}$$
2.
when \(b>h(a)\) and \(n=2\ell +1\) for \(\ell \ge 0\), we have
$$\begin{aligned}& \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab \\& \quad \ge \sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!}-\frac{C_{p,n}}{(n+1)!} \bigl\Vert h^{(n+1)} \bigr\Vert _{q}; \end{aligned}$$
On the other hand, since
and, by the reversed version of Hölder’s integral inequality (2.3),
for \(p<1\), \(p\ne 0\), and \(\frac{1}{p}+\frac{1}{q}=1\), it follows that where \(p<1\) and \(p\ne 0\) with \(\frac{1}{p}+\frac{1}{q}=1\), or \((p,q)=(-\infty ,1)\), or \((p,q)=(1,-\infty )\). The proof of Theorem 3.2 is complete. □
$$\begin{aligned}& \int _{\alpha }^{\beta } \vert a-t \vert ^{n+1}h^{(n+1)}(t)\,\mathrm {d}t \ge 0 \int _{\alpha }^{\beta }h^{(n+1)}(t)\,\mathrm {d}t =0 \bigl\Vert h^{(n+1)} \bigr\Vert _{1}, \\& \begin{aligned} \int _{\alpha }^{\beta } \vert a-t \vert ^{n+1}h^{(n+1)}(t)\,\mathrm {d}t &\ge \inf \bigl\{ h^{(n+1)}(t),t\in [\alpha ,\beta ] \bigr\} \int _{\alpha }^{\beta } \vert a-t \vert ^{n+1} \,\mathrm {d}t \\ &=\frac{(\beta -\alpha )^{n+2}}{n+2} \bigl\Vert h^{(n+1)} \bigr\Vert _{- \infty } = \frac{ \vert a-h^{-1}(b) \vert ^{n+2}}{n+2} \bigl\Vert h^{(n+1)} \bigr\Vert _{-\infty }, \end{aligned} \end{aligned}$$
$$\begin{aligned}& \int _{\alpha }^{\beta } \vert a-t \vert ^{n+1}h^{(n+1)}(t)\,\mathrm {d}t \\ & \quad \ge \biggl( \int _{\alpha }^{\beta } \vert a-t \vert ^{p(n+1)}\,\mathrm {d}t \biggr)^{1/p} \biggl( \int _{\alpha }^{\beta } \bigl[h^{(n+1)}(t) \bigr]^{q}\,\mathrm {d}t \biggr)^{1/q} \\ & \quad = \biggl[\frac{(\beta -\alpha )^{p(n+1)+1}}{p(n+1)+1} \biggr]^{1/p} \bigl\Vert h^{(n+1)} \bigr\Vert _{q} = \biggl[ \frac{ \vert a-h^{-1}(b) \vert ^{p(n+1)+1}}{p(n+1)+1} \biggr]^{1/p} \bigl\Vert h^{(n+1)} \bigr\Vert _{q} \end{aligned}$$
1.
when \(b>h(a)\) and \(n=2\ell \) for \(\ell \ge 0\) or when \(b< h(a)\), we have
$$\begin{aligned}& \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab \\ & \quad \ge \sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!}+\frac{C_{p,n}}{(n+1)!} \bigl\Vert h^{(n+1)} \bigr\Vert _{q}; \end{aligned}$$
2.
when \(b>h(a)\) and \(n=2\ell +1\) for \(\ell \ge 0\), we have
$$\begin{aligned}& \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab \\ & \quad \le \sum_{k=1}^{n}h^{(k)} \bigl(h^{-1}(b) \bigr)\frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!}-\frac{C_{p,n}}{(n+1)!} \bigl\Vert h^{(n+1)} \bigr\Vert _{q}; \end{aligned}$$
Theorem 3.3
Let
\(n\ge 0\)
and
\(h(x)\in C^{n+1}[0,c]\)
such that
\(h(0)=0\)
and
\(h(x)\)
is strictly increasing on
\([0,c]\)
for
\(c>0\), let
\(h^{-1}\)
be the inverse function of
h, let
\(a\in [0,c]\)
and
\(b\in [0,h(c)]\), and let
\(\ell \ge 0\)
be an integer. Then
where
α, β
are defined as in (1.3).
1.
when
the inequality
is valid;
(a)
either
\(h(a)>b\)
and
\(h^{(n+1)}(x)\)
is increasing on
\([\alpha ,\beta ]\);
(b)
or
\(h(a)< b\), \(h^{(n+1)}(x)\)
is increasing on
\([\alpha ,\beta ]\), and
\(n=2\ell +1\);
(c)
or
\(h(a)< b\), \(h^{(n+1)}(x)\)
is decreasing on
\([\alpha ,\beta ]\), and
\(n=2\ell \);
$$\begin{aligned}& \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab -\sum _{k=1}^{n}h ^{(k)} \bigl(h^{-1}(b) \bigr) \frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} \\ & \quad \le \frac{ [a-h^{-1}(b) ]^{n+1}}{(n+2)!} \bigl[h^{(n)}(a)-h ^{(n)} \bigl(h^{-1}(b) \bigr) \bigr] \end{aligned}$$
(3.5)
2.
when
inequality (3.5) is reversed;
(a)
either
\(h(a)>b\)
and
\(h^{(n+1)}(x)\)
is decreasing on
\([\alpha ,\beta ]\);
(b)
or
\(h(a)< b\), \(h^{(n+1)}(x)\)
is increasing on
\([\alpha ,\beta ]\), and
\(n=2\ell \);
(c)
or
\(h(a)< b\), \(h^{(n+1)}(x)\)
is decreasing on
\([\alpha ,\beta ]\), and
\(n=2\ell +1\);
Proof
When \(a>h^{-1}(b)\) and \(h^{(n+1)}(x)\) is increasing, applying Lemma 2.4 gives
when \(a>h^{-1}(b)\) and \(h^{(n+1)}(x)\) is decreasing, the above inequality is reversed.
$$\begin{aligned}& \int _{h^{-1}(b)}^{a}(a-t)^{n+1}h^{(n+1)}(t) \,\mathrm {d}t \\& \quad \le \frac{1}{a-f^{-1}(b)} \int _{h^{-1}(b)}^{a}(a-t)^{n+1}\,\mathrm {d}t \int _{h^{-1}(b)}^{a}h^{(n+1)}(t)\,\mathrm {d}t \\& \quad =\frac{ [a-h^{-1}(b) ]^{n+1}}{n+2} \bigl[h^{(n)}(a)-h^{(n)} \bigl(h^{-1}(b) \bigr) \bigr]; \end{aligned}$$
When \(a< h^{-1}(b)\), \(h^{(n+1)}(x)\) is increasing, and \(n=2\ell \) for \(\ell \ge 0\), applying Lemma 2.4 reveals
when \(a< h^{-1}(b)\), \(h^{(n+1)}(x)\) is increasing, and \(n=2\ell +1\) for \(\ell \ge 0\), inequality (3.6) is reversed.
$$ \begin{gathered}[b] \int _{h^{-1}(b)}^{a}(a-t)^{n+1}h^{(n+1)}(t) \,\mathrm {d}t \\ \quad =(-1)^{n} \int _{a}^{h ^{-1}(b)}(t-a)^{n+1}h^{(n+1)}(t) \,\mathrm {d}t \\ \quad \ge (-1)^{n}\frac{ [h^{-1}(b)-a ]^{n+1}}{n+2} \bigl[h^{(n)} \bigl(h^{-1}(b) \bigr)-h^{(n)}(a) \bigr] \\ \quad =\frac{ [a-h^{-1}(b) ]^{n+1}}{n+2} \bigl[h^{(n)}(a)-h^{(n)} \bigl(h^{-1}(b) \bigr) \bigr]; \end{gathered} $$
(3.6)
By similar argument, we obtain that Substituting these inequalities into (3.4) and simplifying lead to inequality (3.5) and its reversed version for all cases. The proof of Theorem 3.3 is complete. □
1.
when \(a< h^{-1}(b)\), \(h^{(n+1)}(x)\) is decreasing, and \(n=2\ell \) for \(\ell \ge 0\), inequality (3.6) is reversed;
2.
when \(a< h^{-1}(b)\), \(h^{(n+1)}(x)\) is decreasing, and \(n=2\ell +1\) for \(\ell \ge 0\), inequality (3.6) is valid.
Theorem 3.4
Let
\(h(x)\in C^{n+1}[0,c]\)
such that
\(h(0)=0\)
and
\(h(x)\)
is strictly increasing on
\([0,c]\)
for
\(c>0\), let
\(h^{-1}\)
be the inverse function of
h, and let
\(a\in [0,c]\)
and
\(b\in [0,h(c)]\). If
\(h^{(n+1)}(x)\)
is convex on
\([\alpha ,\beta ]\), where
α, β
are defined as in (1.3), then
where
\(\ell \ge 0\)
is an integer. If
\(h^{(n+1)}(x)\)
is concave on
\([\alpha ,\beta ]\), all the above inequalities are reversed for all corresponding cases.
1.
when
\(h(a)>b\)
or when
\(h(a)< b\)
and
\(n=2\ell \), we have
$$\begin{aligned}& \frac{ [a-h^{-1}(b) ]^{n+2}}{n+2}h^{(n+1)} \biggl(\frac{a+(n+2)h ^{-1}(b)}{n+3} \biggr) \\& \quad \le \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab -\sum _{k=1} ^{n}h^{(k)} \bigl(h^{-1}(b) \bigr) \frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} \\& \quad \le \bigl[a-h^{-1}(b) \bigr]^{n+2}\frac{h^{(n+1)}(a)+(n+2)h^{(n+1)} (h^{-1}(b) )}{(n+3)!}; \end{aligned}$$
(3.7)
Proof
By substitution of variables, we have
Applying inequality (2.5) to the convex function \(h^{(n+1)}(x)\) yields that Substituting these inequalities into (3.4) and simplifying lead to the right inequality in (3.7) and its reversed version for all cases.
$$\begin{aligned}& \int _{h^{-1}(b)}^{a}(a-t)^{n+1}h^{(n+1)}(t) \,\mathrm {d}t \\& \quad = \bigl[a-h^{-1}(b) \bigr]^{n+2} \int _{0}^{1}(1-s)^{n+1}h^{(n+1)} \bigl(sa+(1-s)h ^{-1}(b) \bigr)\,\mathrm {d}s. \end{aligned}$$
1.
when \(a>h^{-1}(b)\), we have
$$\begin{aligned}& \int _{h^{-1}(b)}^{a}(a-t)^{n+1}h^{(n+1)}(t) \,\mathrm {d}t \\& \quad \le \bigl[a-h^{-1}(b) \bigr]^{n+2} \int _{0}^{1}(1-s)^{n+1} \bigl[sh^{(n+1)}(a)+(1-s)h^{(n+1)} \bigl(h^{-1}(b) \bigr) \bigr]\,\mathrm {d}s \\& \quad = \bigl[a-h^{-1}(b) \bigr]^{n+2}\frac{h^{(n+1)}(a)+(n+2)h^{(n+1)} (h^{-1}(b) )}{(n+2)(n+3)}; \end{aligned}$$
2.
when \(a< h^{-1}(b)\) and \(n=2\ell \), the above inequality is still valid;
3.
when \(a< h^{-1}(b)\) and \(n=2\ell +1\), the above inequality is reversed.
Applying inequality (2.6) to the convex function \(h^{(n+1)}(x)\) shows that Substituting these related inequalities into (3.4) and rearranging result in the left inequality in (3.7) and its reversed version for all cases.
1.
when \(a>h^{-1}(b)\), we have
$$\begin{aligned}& \int _{h^{-1}(b)}^{a}(a-t)^{n+1}h^{(n+1)}(t) \,\mathrm {d}t \\& \quad \ge \biggl[ \int _{h^{-1}(b)}^{a}(a-t)^{n+1}\,\mathrm {d}t \biggr]h^{(n+1)} \biggl(\frac{\int _{h^{-1}(b)}^{a}(a-t)^{n+1}t\,\mathrm {d}t}{\int _{h^{-1}(b)} ^{a}(a-t)^{n+1}\,\mathrm {d}t} \biggr) \\& \quad =\frac{ [a-h^{-1}(b) ]^{n+2}}{n+2}h^{(n+1)} \biggl(\frac{a+(n+2)h ^{-1}(b)}{n+3} \biggr); \end{aligned}$$
2.
when \(a< h^{-1}(b)\) and \(n=2\ell \), the above inequality is still valid;
3.
when \(a< h^{-1}(b)\) and \(n=2\ell \), the above inequality is reversed.
For the concave function \(h^{(n+1)}(x)\), one can derive everything similarly. The proof of Theorem 3.4 is complete. □
Theorem 3.5
Let
\(n\ge 0\)
and
\(h(x)\in C^{n+1}[0,c]\)
such that
\(h(0)=0\)
and
\(h(x)\)
is strictly increasing on
\([0,c]\)
for
\(c>0\), let
\(h^{-1}\)
be the inverse function of
h, let
\(a\in [0,c]\)
and
\(b\in [0,h(c)]\), and let
\(h^{(n+1)}(x)\)
be nonnegative and convex on
\([\alpha ,\beta ]\), where
α, β
are defined as in (1.3). If
\(h(a)>b\), then
If
\(h(a)< b\)
and
\(n=2\ell \)
for
\(\ell \ge 0\), then
If
\(a< h^{-1}(b)\)
and
\(n=2\ell +1\)
for
\(\ell \ge 0\), the double inequality (3.9) is reversed.
$$\begin{aligned}& \frac{ [a-h^{-1}(b) ]^{n+2}}{(n+1)!} \biggl[\frac{1}{2^{n}}h ^{(n+1)} \biggl(\frac{a+h^{-1}(b)}{2} \biggr) -\frac{2h^{(n+1)}(a)+h ^{(n+1)} (h^{-1}(b) )}{6} \biggr] \\& \quad \le \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab-\sum _{k=1} ^{n}h^{(k)} \bigl(h^{-1}(b) \bigr) \frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} \\& \quad \le \frac{ [a-h^{-1}(b) ]^{n+2}}{(n+1)!}\frac{h^{(n+1)}(a)+2h ^{(n+1)} (h^{-1}(b) )}{6}. \end{aligned}$$
(3.8)
$$\begin{aligned}& \frac{ [h^{-1}(b)-a ]^{n+2}}{(n+1)!} \biggl[\frac{1}{2^{n}}h ^{(n+1)} \biggl(\frac{a+h^{-1}(b)}{2} \biggr) -\frac{2h^{(n+1)}(a)+h ^{(n+1)} (h^{-1}(b) )}{6} \biggr] \\& \quad \le \int _{0}^{a}h(x)\,\mathrm {d}x+ \int _{0}^{b}h^{-1}(x)\,\mathrm {d}x-ab-\sum _{k=1} ^{n}h^{(k)} \bigl(h^{-1}(b) \bigr) \frac{ [a-h^{-1}(b) ]^{k+1}}{(k+1)!} \\& \quad \le \frac{ [h^{-1}(b)-a ]^{n+2}}{(n+1)!}\frac{h^{(n+1)}(a)+2h ^{(n+1)} (h^{-1}(b) )}{6}. \end{aligned}$$
(3.9)
Proof
When \(a>h^{-1}(b)\), employing the double inequality (2.7) gives
$$\begin{aligned}& 2 \biggl[\frac{a-h^{-1}(b)}{2} \biggr]^{n+1}h^{(n+1)} \biggl( \frac{a+h ^{-1}(b)}{2} \biggr) \\& \qquad {}-\frac{ [a-h^{-1}(b) ]^{n+1}h^{(n+1)} (h^{-1}(b) )}{6}-\frac{ [a-h^{-1}(b) ]^{n+1}h^{(n+1)}(a)}{3} \\& \quad \le \frac{1}{a-h ^{-1}(b)} \int _{h^{-1}(b)}^{a}(a-t)^{n+1}h^{(n+1)}(t) \,\mathrm {d}t \\& \quad \le \frac{ [a-h^{-1}(b) ]^{n+1}h^{(n+1)} (h^{-1}(b) )}{3}+\frac{ [a-h^{-1}(b) ]^{n+1}h^{(n+1)}(a)}{6}. \end{aligned}$$
When \(a< h^{-1}(b)\), it is clear that
If \(a< h^{-1}(b)\) and \(n=2\ell \), utilizing the double inequality (2.7) gives
If \(a< h^{-1}(b)\) and \(n=2\ell +1\), the above double inequality is reversed.
$$ \int _{h^{-1}(b)}^{a}(a-t)^{n+1}h^{(n+1)}(t) \,\mathrm {d}t =(-1)^{n} \int _{a} ^{h^{-1}(b)}(t-a)^{n+1}h^{(n+1)}(t) \,\mathrm {d}t. $$
$$\begin{aligned}& 2 \biggl[\frac{h^{-1}(b)-a}{2} \biggr]^{n+1}h^{(n+1)} \biggl( \frac{a+h ^{-1}(b)}{2} \biggr) \\& \qquad {}-\frac{ [h^{-1}(b)-a ]^{n+1}h^{(n+1)} (h^{-1}(b) )}{6}-\frac{ [h^{-1}(b)-a ]^{n+1}h^{(n+1)}(a)}{3} \\& \quad \le \frac{1}{h ^{-1}(b)-a} \int _{a}^{h^{-1}(b)}(t-a)^{n+1}h^{(n+1)}(t) \,\mathrm {d}t \\& \quad \le \frac{ [h^{-1}(b)-a ]^{n+1}h^{(n+1)} (h^{-1}(b) )}{3}+\frac{ [h^{-1}(b)-a ]^{n+1}h^{(n+1)}(a)}{6}. \end{aligned}$$
4 Applications and examples
As applications of Theorems 3.1 to 3.5, we now estimate several concrete definite integrals, including a definite integral of \(e^{-1/x}\), the exponential integral \(\operatorname{Ei}(x)\), and the logarithmic integral \(\operatorname{li}(x)\).
Example 4.1
By taking \(h(x)=\sqrt[4]{x^{4}+1} -1\), \(a=3\), and \(b=2\) in Theorem 1.2, it was obtained in [4] that
The difference between the upper and lower bounds is 0.00000000490079353… .
$$\begin{aligned} 9+\frac{4\sqrt[4]{125} }{27} \bigl[3-2\sqrt[4]{5} \bigr]^{2} =&9.000042866 \ldots \\ < & \int ^{3}_{0}\sqrt[4]{x^{4}+1} \,\mathrm {d}x + \int ^{3}_{1}\sqrt[4]{x^{4}-1} \,\mathrm {d}x \\ < &9+\frac{27}{2\sqrt[4]{82^{3}} } \bigl(3-2\sqrt[4]{5} \bigr)^{2}=9.000042871 \ldots . \end{aligned}$$
In [5, 6], by Theorems 1.3 and 1.4, the above double inequality was improved as
$$ 9.00004286805\ldots < \int ^{3}_{0}\sqrt[4]{x^{4}+1} \,\mathrm {d}x + \int ^{3} _{1}\sqrt[4]{x^{4}-1} \,\mathrm {d}x< 9.000042868057\ldots . $$
It is easy to see that \(h^{-1}(x)= [(x+1)^{4}-1 ]^{1/4}\). Straightforward computation gives
Since \(h^{-1}(2)=80^{1/4}=2.990\ldots <3\), then \((\alpha ,\beta )= (80^{1/4},3 )\). The third derivative \(h'''(x)\) has two real nonzero zeros \(\pm (\frac{2}{5} )^{1/4}=\pm 0.79527\ldots \) . This implies that the third derivative \(h'''(x)\) is negative and \(h''(x)\) is concave on the closed interval \([\alpha ,\beta ]\). Substituting these data into the reversed version of inequality (3.7) yields
which can be rewritten as
This lower estimation is better than all before.
$$ h'(x)=\frac{x^{3}}{(x^{4}+1)^{3/4}},\qquad h''(x)= \frac{3 x^{2}}{(x ^{4}+1)^{7/4}}, \qquad h'''(x)= \frac{x (6-15x^{4} )}{(x^{4}+1)^{11/4}}. $$
$$\begin{aligned}& \frac{ (3-80^{1/4} )^{3}}{3}h'' \biggl(\frac{3+380^{1/4}}{4} \biggr) \\& \quad \ge \int ^{3}_{0}\sqrt[4]{x^{4}+1} \,\mathrm {d}x-3+ \int ^{3}_{1}\sqrt[4]{x ^{4}-1} \,\mathrm {d}x-6 -h' \bigl(80^{1/4} \bigr) \frac{ (3-80^{1/4} )^{2}}{2!} \\& \quad \ge \bigl(3-80^{1/4} \bigr)^{3} \frac{h''(3)+3h'' (80^{1/4} )}{4!}, \end{aligned}$$
$$\begin{aligned}& \frac{ (3-80^{1/4} )^{3}}{3}\frac{3072 (\sqrt[4]{95} \sqrt{2} +3 )^{2}}{ [ (\sqrt[4]{95} \sqrt{2} +3 )^{4}+256 ]^{7/4}} +9+\frac{8\times 5^{3/4}}{27}\frac{ (3-80^{1/4} )^{2}}{2!} \\& \quad =9.0000428983186013\ldots \ge \int ^{3}_{0}\sqrt[4]{x^{4}+1} \,\mathrm {d}x+ \int ^{3}_{1}\sqrt[4]{x^{4}-1} \,\mathrm {d}x \\& \quad \ge \frac{ (3-80^{1/4} )^{3}}{4!} \biggl(\frac{27}{82^{7/4}}+3 \times \frac{4\sqrt{5} }{729} \biggr) +9+\frac{8\times 5^{3/4}}{27}\frac{ (3-80^{1/4} )^{2}}{2!} \\& \quad =9.0000428680640760\ldots . \end{aligned}$$
Example 4.2
Let \(h(x)=e^{-1/x}\) for \(x>0\) and \(h(0)=0\). Then \(h^{-1}(x)=-\frac{1}{ \ln x}\) for \(x\in (0,1)\) and \(h^{-1}(0)=0\). The function \(h(x)=e^{-1/x}\) plays an indispensable role in the proof of the existence of partitions of unity in differential geometry [30] and is a generating function of the Lah numbers in combinatorics [13‐18, 20, 22].
Direct computation gives
This means that the function \(h'(x)\) is increasing on \((0, \frac{1}{2} ]\) and decreasing on \([\frac{1}{2},\infty )\).
$$ h'(x)=\frac{e^{-1/x}}{x^{2}} \quad \text{and}\quad h''(x)= \frac{e^{-1/x} (1-2 x)}{x^{4}}. $$
Choosing \(a=b=\frac{1}{2}\) means that \(a< h^{-1} (\frac{1}{2} )=\frac{1}{\ln 2}=1.44\ldots \) and \((\alpha ,\beta )= ( \frac{1}{2},\frac{1}{\ln 2} )\). Then it follows from the double inequality (1.4) that
where
are respectively called the exponential integral and the logarithmic integral.
$$\begin{aligned}& \frac{1}{4}+ \biggl(\frac{1}{2}-\frac{1}{\ln 2} \biggr) \biggl[ \frac{1}{e ^{\frac{1}{2}(\frac{1}{2}+\frac{1}{\ln 2})}}-\frac{1}{2} \biggr] \\& \quad =0.364469045537996606 \ldots \\& \quad \le \int _{0}^{1/2}\frac{1}{e^{1/x}}\,\mathrm {d}x- \int _{0}^{1/2}\frac{1}{ \ln x}\,\mathrm {d}x =\operatorname{Ei}(-2)-\operatorname{li}\biggl(\frac{1}{2} \biggr)+\frac{1}{2 e^{2}} \\& \quad \le \frac{1}{4}+\frac{1}{2} \biggl(\frac{1}{2}- \frac{1}{\ln 2} \biggr) \biggl(\frac{1}{e^{2}}-\frac{1}{2} \biggr) =0.421883810040011829\ldots , \end{aligned}$$
$$ \operatorname{Ei}(x)=- \int _{-x}^{\infty }\frac{e^{-t}}{t}\,\mathrm {d}t \quad \text{and}\quad \operatorname{li}(x)= \int _{0}^{x}\frac{\mathrm{d} t}{\ln t} $$
Example 4.3
Let \(h(x)=e^{x^{2}}-1\) for \(x\ge 0\). Then \(h^{-1}(x)= \sqrt{\ln (1+x)} \) for \(x\ge 0\).
In [21, Sect. 1.4], it was computed that
This implies that the function \(h(x)\) is absolutely monotonic, that is, \(h^{(n)}(x)\ge 0\) for all \(n\ge 0\), on \((0,\infty )\). For more information on absolutely monotonic functions, please refer to [3] and closely related references therein.
$$ h^{(n)}(x)= \bigl(e^{x^{2}} \bigr)^{(n)} =e^{x^{2}}\frac{n!}{(2x)^{n}} \sum_{k=0}^{n} \binom{k}{n-k}\frac{(2x)^{2k}}{k!},\quad n\ge 1. $$
Choosing \(a=x\) and \(b=x^{2}\) for \(x>0\) implies that \(h^{-1} (x ^{2} )=\ln ^{1/2} (1+x^{2} )< x\) for all \(x>0\). Applying these data to the double inequality (3.8), we arrive at
for \(n\ge 0\) and \(x>0\). In particular, letting \(n=1\) derives
Further taking \(x=1\) results in
which can be numerically computed as
$$\begin{aligned}& \frac{ [x-\ln ^{1/2} (1+x^{2} ) ]^{n+2}}{(n+1)!} \biggl[\frac{1}{2^{n}}h^{(n+1)} \biggl( \frac{x+\ln ^{1/2} (1+x^{2} )}{2} \biggr) \\& \qquad {}-\frac{2h^{(n+1)}(x)+h^{(n+1)} (\ln ^{1/2} (1+x^{2} ) )}{6} \biggr] \\& \quad \le \int _{0}^{x}e^{t^{2}}\,\mathrm {d}t+ \int _{0}^{x^{2}}\sqrt{\ln (1+t)} \,\mathrm {d}t -x-x^{3} \\& \qquad {}-\sum_{k=1}^{n}h^{(k)} \bigl(\ln ^{1/2} \bigl(1+x^{2} \bigr) \bigr)\frac{ [x-\ln ^{1/2} (1+x^{2} ) ]^{k+1}}{(k+1)!} \\& \quad \le \frac{ [x-\ln ^{1/2} (1+x^{2} ) ]^{n+2}}{(n+1)!}\frac{h ^{(n+1)}(x)+2h^{(n+1)} (\ln ^{1/2} (1+x^{2} ) )}{6} \end{aligned}$$
$$\begin{aligned}& \frac{ [x-\ln ^{1/2} (1+x^{2} ) ]^{3}}{2} \biggl[ \frac{1}{2}h'' \biggl(\frac{x+\ln ^{1/2} (1+x^{2} )}{2} \biggr) \\& \qquad {}-\frac{2h''(x)+h'' (\ln ^{1/2} (1+x^{2} ) )}{6} \biggr] \\& \quad \le \int _{0}^{x}e^{t^{2}}\,\mathrm {d}t+ \int _{0}^{x^{2}}\sqrt{\ln (1+t)} \,\mathrm {d}t -x-x^{3} \\& \qquad {}-h' \bigl(\ln ^{1/2} \bigl(1+x^{2} \bigr) \bigr) \frac{ [x-\ln ^{1/2} (1+x^{2} ) ]^{2}}{2} \\& \quad \le \frac{ [x-\ln ^{1/2} (1+x^{2} ) ]^{3}}{2}\frac{h''(x)+2h'' (\ln ^{1/2} (1+x^{2} ) )}{6}. \end{aligned}$$
$$\begin{aligned}& \frac{ (1-\ln ^{1/2}2 )^{3}}{2} \biggl[\frac{1}{2}h'' \biggl(\frac{1+ \ln ^{1/2}2}{2} \biggr) -\frac{2h''(1)+h'' (\ln ^{1/2}2 )}{6} \biggr]+h' \bigl(\ln ^{1/2}2 \bigr) \frac{ (1-\ln ^{1/2}2 )^{2}}{2}+2 \\& \quad \le \int _{0}^{1}e^{t^{2}}\,\mathrm {d}t+ \int _{0}^{1}\sqrt{\ln (1+t)} \,\mathrm {d}t \\& \quad \le \frac{ (1-\ln ^{1/2}2 )^{3}}{2}\frac{h''(1)+2h'' ( \ln ^{1/2}2 )}{6} +h' \bigl(\ln ^{1/2}2 \bigr) \frac{ (1-\ln ^{1/2}2 )^{2}}{2}+2, \end{aligned}$$
$$ 2.044751320\ldots \le \int _{0}^{1}e^{t^{2}}\,\mathrm {d}t+ \int _{0}^{1}\sqrt{ \ln (1+t)} \,\mathrm {d}t\le 2.060536019 \ldots . $$
Acknowledgements
The authors thank Di-San Chen (Lijiang College of Guangxi Normal University), Jing-Feng Tian (North China Electric Power University), and Bo-Yan Xi (Inner Mongolia University for Nationalities) for their supplying electronic versions of several references in this paper. The authors appreciate anonymous referees for their careful corrections to and valuable comments on the original version of this paper.
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