R E S E A R C H
Open Access
Sharp inequalities for tangent function
with applications
Hui-Lin Lv
1, Zhen-Hang Yang
2, Tian-Qi Luo
1and Shen-Zhou Zheng
1**Correspondence:
1Department of Mathematics,
Beijing Jiaotong University, Beijing, 100044, China
Full list of author information is available at the end of the article
Abstract
In the article, we present new bounds for the functionetcot(t)–1on the interval (0,
π
/2) and find sharp estimations for the Sine integral and the Catalan constant based on a new monotonicity criterion for the quotient of power series, which refine the Redheffer and Becker-Stark type inequalities for tangent function.MSC: Primary 33B10; secondary 26D05
Keywords: trigonometric function; inequalities; Sine integral; Catalan constant
1 Introduction
The study of this paper is concerned with the following inequality:
sint
t ≥
π–t
π+t, t∈(,π), (.)
which was posted by Redheffer in [] and was proved by Williams []. Recently, Zhu and Sun [] extended the Redheffer inequality (.) to the tangent function, and they estab-lished the following inequalities:
π+ t
π– t
π/
<tant
t <
π+ t
π– t, t∈(,π/) (.)
with the best exponentsπ/ and . Zhu [] further refined the double inequality
√
π+ t
π– t
/
<tant
t <
√
π+ t
π– t
π/
, t∈(,π/). (.)
It is worth noting that Becker and Stark [] in showed the double inequality
π– t <
tant t <
π
π– t, t∈(,π/), (.)
where andπare the best constants. Later, Zhu and Hua [] gave a general refinement of the Becker-Stark inequalities (.) by the power series expansion of the tangent function in terms of the Bernoulli numbers. In particular, they proved that fort∈(,π/) the double
inequality
π+ (/π– )t
π– t <
tant t <
π+ (π/ – )t
π– t (.)
holds with the best constants (/π– ) and (π/ – ); also see []. Chen and Cheung
[] further presented an improvement of the left hand side inequality in (.), which states that
π
π– t
α <tant
t <
π
π– t
β
(.)
holds fort∈(,π/) with the best exponentsα=π/ andβ= (also cf. []). Another improvement involving the left hand side one in (.) was made in [] by Nishizawa. Very recently, Bhayo and Sándor [], Corollary , again proved the Becker-Stark inequal-ities (.) by using Redheffer inequality (.), which reveals the implicit relation between Redheffer’s and Becker-Stark’s inequalities. They in [], Corollaries , also stated that for
t∈(,π/) we have
π– t–πt/
π–t <
t
tant<
π– t
π–t . (.)
It is an important observation that Yanget al.[], (), in considered the bounds for functionetcott–and established a number of inequalities for trigonometric functions.
In particular, they in [], Corollary , showed that fort∈(,π/)
e–t/π<etcott–<e–t/, which can be written as
–t
π <
t
tant< – t
. (.)
Inspired by these results mentioned above, the aim of this paper is to determine the best bounds forY(t) =etcott–in terms of
Bp(t) =
( –pt)/(p) ifp∈(–∞, )∪(, /π],
e–t/ ifp= , (.)
on (,π/), that is to say, we will determine the best parametersp,q∈(–∞, /π] such that the double inequality
–pt/(p)<exp
t
tant–
< –qt/(q) (.)
holds for allt∈(,π/). Inequalities (.) also can be rewritten as
+ pln
–pt< t
tant < +
qln
–qt,
2 Some useful lemmas
In order to prove the main Theorem in the next section, we need some preliminary lemmas. To this end, we first introduce a useful auxiliary functionHf,g. For –∞ ≤a<b≤
∞, letf andg be differentiable on (a,b) and g= on (a,b). Then the functionHf,g is defined by
Hf,g:=
f
gg–f. (.)
The functionHf,g has been investigated with some well properties in [], Properties , , which plays an important role in the proof of a monotonicity criterion for the quotient of power series; also see [].
Lemma ([], Theorem ., [], Lemma ., and [], Lemma .) Let A(t) =∞k=aktk
and B(t) =∞k=bktk be two real power series converging on(–r,r)and bk> for all k.
Suppose that,for certain m∈N,the non-constant sequence{ak/bk}is increasing(resp.
de-creasing)for≤k≤m and decreasing (resp.increasing)for k≥m.Then the function A/B is strictly increasing(resp.decreasing)on(,r)if and only if HA,B(r–)≥(resp.≤).
Moreover,if HA,B(r–) <(resp.>),then there exists t∈(,r)such that the function A/B is
strictly increasing(resp.decreasing)on(,t)and strictly decreasing(resp.increasing)on
(t,r).
Lemma ([], p. ) Let <t<π.Then
cott=
t– ∞
n=
n (n)!|Bn|t
n–, (.)
sint=
t +
∞
n=
(n– )n (n)! |Bn|t
n–, (.)
where Bnis the Bernoulli number.
Lemma ([]) Let Bnand Bn–be the even-indexed Bernoulli numbers.Then
(π)
n(n– )(n–– ) n– <
|Bn|
|Bn–|
< (π)
n(n– )n–
n–– . (.)
Consequently,we have
n(n– ) (n+ )(n+ )
(n+– )(n–– ) n– <
|Bn|
|Bn–Bn+|
< n(n– ) (n+ )(n+ )
n– (n–– ).
Lemma Let the function g be defined on(,π)by
g(t) = t– costsint– tsin
t
t(t–sintcost) . (.)
Then g is strictly increasing from(,π/)onto(/, /(π))and decreasing from(π/,π)
Proof To avoid complicated calculations, we here make use of Lemmas , and to prove this lemma. For this purpose, we writeg(t) as
g(t) = t sint–
cost sint –
t
t
sint–tcostsint
:=g(t)
g(t)
,
then applying Lemma yields
g(t) = g(t)
g(t)
=
∞
n=
n+
(n+)!|Bn+|t n–
∞
n=
n
(n–)!|Bn|tn–
:=
∞
n=an+xn
∞
n=bn+xn
,
where
an= n+
(n+ )!|Bn+|, bn= n
(n– )!|Bn|, x=t
∈,π.
We now prove that the sequence{an/bn}is increasing for ≤n≤ and decreasing for
n≥. A simple check yields
a
b
= >
a
b
= <
a
b
= <
a
b
= ,
and it remains to show thatan–/bn–>an/bnforn≥. Indeed, we have
an–
bn–
an
bn
=
(n– )(n– )
|Bn|
|Bn–|
n(n+ )
|Bn+|
|Bn|
= n(n+ ) (n– )(n– )
|Bn|
|Bn–|
|Bn|
|Bn+|
.
Then by Lemma , we get
an–
bn–
an
bn
– > n(n+ ) (n– )(n– )
n(n– ) (n+ )(n+ )
(n+– )(n–– )
n– –
= n
n–
(n+– )(n–– ) n– –
=(
n+– n)n+ n
×n(n– )(n+ ) > forn≥,
where the inequality holds due to n+– n> forn≥. This proves the piecewise
monotonicity of{an/bn}n≥.
According to Lemma , we also have to check thatHg,g(π
–) < andH
g,g(π/) = . In
fact, we have
Hg,g( √
x) = g
(
√ x)
g(√x)g(
√ x) –g(
√ x)
=( t
sint –costsint –t)
(sintt–tcostsint)
t
sint–t
cost
sint
–
t
sint–
cost
sint –
t
= sint
tcostsint+ tcost+t(sint– )sintcost+ costsint
tcost– tsint+costsint
→–∞ ast→π–,
then Lemma leads to the result that there is a uniquet∈(,π) such thatgis increasing
on (,t) and decreasing on (t,π). Note that
g(t) = g
(t)
g(t)
Hg,g(t) = g(t)
g(t)
g(t)
g(t)g(t) –g(t)
→ ast→π/,
we clearly see that the uniquet=π/. A simple computation yields
g+= , g
π
=
π, g
π–= π,
which completes the proof.
Remark If we use an ordinary method to prove the piecewise monotonicity ofg, then it is very troublesome. For example, a direct computation yields
g(t) =
t– sint+ tcost t(t–sint)
u=t∈(,π)
=
u– sinu+ucosu
u(u–sinu) :=g(u),
then differentiatingg(u) gives
g(u) = –
usinu+ ( +cosu)u–u( +cosu)sinu+ sinu
u(u–sinu) .
As a result, there are various approaches to showing the piecewise monotonicity ofgon
(, π), but it seems to be difficult. It thus can be seen that our method used previously is relatively easy.
Lemma For t∈(,π/), let p−→Bp(t) and p−→αp(t) be respectively defined on (–∞, /π]by(.)and
αp(t) =
exp(tcott– )
( –pt)/(p) if p= , and α(t) =exp
tcott– +t
. (.)
Then p −→ Bp(t) and p −→ αp(π/)Bp(t) are strictly decreasing and increasing on (–∞, /π],respectively.Moreover,there is a unique p≈.such thatαp(π/) <
for p∈(–∞,p)andαp(π/) > for p∈(p, /π),where pis the unique solution of the
equationαp(π/) = on(–∞, /π).
Proof Letp= . Logarithmic differentiation yields
∂lnBp(t) ∂p = –
pr(t),
∂ln[αp(π/)Bp(t)]
∂p =
p
r
π
–r(t)
where
r(t) =ln –pt+ –pt– .
Differentiation again leads to
r(t) = p
t
( –pt)>
fort∈(,π/), which means that =r(+) <r(t) <r(π/). These together withB (t) =
limp→Bp(t) andα(t) =limp→αp(t) show thatp−→Bp(t) andp−→αp((π/))Bp(t) are strictly decreasing and increasing on (–∞, /π], respectively.
Note the increasing property ofp−→lnαp(π/) = – –lnBp(π/) on (–∞, /π] and
lnα(π/) =lim
p→
– – pln
–pπ
=π
– < ,
lnα/(π/) = – –ln
–π
> ,
which implies that there is a uniquep∈(, /) such thatlnαp< forp∈(–∞,p) and
αp> forp∈(p, /π). Solving the equationlnαp(π/) = forpgivesp=p≈..
The proof is finished.
3 Main results
This section is devoted to stating and proving the main results concerning some inequal-ities for the tangent function. More precisely, we have the following.
Theorem For p∈(–∞, /π],let Y(t) =exp(tcott– )and Bp(t)be defined on(,π/)
by(.).
(i) Ifp≤/≈.,then the functiont−→Y(t)/Bp(t)is strictly decreasing on the interval(,π/).Consequently,for allt∈(,π/)
αp(π/)
–pt/(p)<exp
t
tant–
< –pt/(p) (.)
with the best coefficientsandαp(π/)defined by(.). (ii) Ifp≥/(π)≈.,then the functiont−→Y(t)/B
p(t)is strictly increasing on (,π/),and therefore,for allt∈(,π/),
–pt/(p)<exp
t
tant–
<αp(π/)
–pt/(p) (.)
with the best coefficientsandαp(π/)defined by(.).
(iii) If/ <p< /(π),then there is at∈(,π/)such that the function
t−→Y(t)/Bp(t)is strictly increasing on(,t)and decreasing on(t,π/),and
hence,for allt∈(,π/),
minαp(π/),
–pt/(p)<exp
t
tant–
<βp
with
βp=
tcott–
( –pt)/(p)
,
wheretis the unique solution of the equation
cost
sint – t
sint+
t
( –pt)= (.)
on(,π/).
When p∈[p, /(π)),the double inequality(.)still holds for all t∈(,π/).In
partic-ular,when p=p,we have
–pt
/(p)
<exp
t
tant–
<βp
–pt
/(p)
(.)
with the best constantsandβp≈..
Proof Let
f(t) =lnY(t) –lnBp(t) =
t
tant– –
pln( –pt) ifp= , t
tant– + t
ifp= .
(.)
Differentiation yields
f(t) =cost
sint – t
sint+
t
( –pt)
=t
(t–costsint)
( –pt)sint
p–t– costsint– tsin
t
t(t–sintcost)
=t
(t–costsint)
( –pt)sint
p–g(t),
whereg(t) is defined by (.).
Noticing that (t–costsint) = [t–sin(t)]/ > fort∈(,π/) and ( –pt) > for
p∈(–∞, /π] andt∈(,π/), we easily see that, for allt∈(,π/),
sgnf(t) =sgnp–g(t). (.)
As shown in Lemma , the function g is strictly increasing from (,π/) onto (/, /(π)). We are now in a position to distinguish three cases to prove the required
result.
Case :p≤mint∈(,π/)g(t) = /. Then we obtainf(t)≤ fort∈(,π/), which means
thatf is strictly decreasing on (,π/). Consequently, we can deduce the following obser-vation:
lnαp
π
=f
π
–
which is equivalent to the double inequality (.) holding for allt∈(,π/) with the best coefficients andαp.
Case :p≥maxt∈(,π/)g(t) = /(π). Similarly, we havef(t) > fort∈(,π/), which
implies that the double inequality (.) holds for allt∈(,π/) with the best coefficients andαp(π/).
Case : / <p< /(π). Sincet−→p–g(t) :=g(t) is strictly decreasing on (,π/)
with
g
+=p–
> and g
π
–
=p– π < ,
we find that there ist∈(,π/) such that g(t) > fort∈(,t) andg(t) < fort∈
(t,π/). This indicates thatf is strictly increasing on (,t) and decreasing on (t,π/).
Therefore, we deduce that
min,lnαp(π/)
=min
f+,f
π
–
<f(t)≤f(t) =lnβp
for allt∈(,π/), that is, (.) holds for allt∈(,π/), whereβp=Y(t)/Bp(t).
Whenp∈[p, /(π)), by Lemma we haveαp≥, and it follows that
<f(t)≤f(t) =lnβp,
that is, the double inequality (.) still holds fort∈(,π/).
In particular, forp=p≈., solving equation (.) fortyieldst≈., and
henceβp≈.. Thus we complete the proof.
Takingp= /≈., /, /, , and →–∞, respectively. Then by part (i) of The-orem and the monotonicity ofp−→Bp(t) andp−→αp(π/)Bp(t) given in Lemma , we immediately obtain the following conclusion.
Corollary For t∈(,π/),the inequalities
e–<αe–t
/
<α/
–t
<α/
–t
/
(.)
<α/
–t
/
<exp
t
tant–
<
–t
/
<
–t
/
<
–t
<e–t/<
holds with the best coefficients
α/=e–
–π
–/
≈., α/=e–
–π
–/
≈.,
α/=e–
–π
–
≈., α=eπ
/–
Likewise, takingp= /(π)≈., /, /, /, and /π, respectively, we have the
following.
Corollary For t∈(,π/),the inequalities
–t
π
π/
<
–t
<
–t
<
–t
/
<
– t
π
π/
<exp
t
tant–
<α/(π)
– t
π
π/
<α/
–t
/
<α/
–t
<α/
–t
(.)
hold with the best coefficients
α/(π)=e–
π/
≈., α/=e–
–π
–/
≈.,
α/=e–
–π
–
≈., α/=e–
–π
–
≈..
Theorem Let p,q∈(–∞, /π].Then the double inequality
–pt/(p)<exp
t
tant–
< –qt/(q) (.)
holds for all t∈(,π/)if and only if p≥p≈.and q≤/≈.,where p
is defined in Lemma.
Proof Clearly, the sufficiency easily follows by Theorem . The necessary condition for the right hand side inequality in (.) to hold fort∈(,π/) follows from the limit relation
lim
t→+
lnY(t) –lnBq(t)
t =
(q– )≤.
The necessary condition for the left hand side inequality in (.) to hold fort∈(,π/) can be obtained from the inequality
lim
t→π/
Y(t)
Bp(t) =e–
–pπ
–/(p)
=αp≥.
It follows from Lemma thatp≥p, which completes the proof.
4 Comparisons and remarks By Theorem , we have
+ p
ln –pt
< t
tant< +
ln
– t
, (.)
We denote the lower bounds fort/tantgiven in the inequalities (.), (.), (.), (.), (.), and (.), respectively, by
LY(t) = + p
ln –pt
, ZS(t) =π
– t
π+ t, Z(t) =
π– t
√
π+ t
π/
,
BS(t) = –
πt
, ZH(t) = π– t
π+ (π/ – )t, BS(t) =
π– t–πt/
π–t .
Proposition The comparison inequalities
LY(t) >ZH(t) >BS(t) >max
ZS(t),Z(t),BS(t)
(.)
hold for t∈(,π/).Moreover,ZS(t),Z(t)and BS(t)are not comparable with each other
for all t∈(,π/).
Proof (i) We first prove
D(x) =LY(
√
x) –ZH(√x) = + p
ln( –px) –
π– x
π+ (π/ – )x>
forx∈(,π/). Differentiation yields
D(x) = – ( –π
)x
( –xp)(π+ (π/ – )x)
x–π
– π– πp
( –π)
,
which shows thatDis increasing on (,x) and decreasing on (x,π/), where
x=
π– π– πp
( –π) ≈..
Then we conclude thatD(x) >min(D(),D(π/)) = withD(π/) = due top
sat-isfyingαp(π/) = shown in Lemma .
(ii) The second inequality directly follows from
ZH(t) –BS(t) =
π– t
π+ (π/ – )t –
– πt
=
t( –π)(π– t)
π(π+ (π/ – )t)>
fort∈(,π/).
(iii) The third one is deduced by
BS(t) –ZS(t) =
– πt
–π
– t
π+ t =
t(π– t)
π(π+ t) > ,
BS(t) –Z(t)
π– t =
π –
(π– t)π/–
(√π+ t)π/ >
π –
(π)π/–
(√π)π/ = , BS(t) –BS(t) =
– πt
–π
– t–πt/
π–t =
π
t(t+π– π)
(π–t) > ,
(iv) Finally, we prove thatZS(t),Z(t) andBS(t) are not comparable with each other for
allt∈(,π/). Simple computations yield
lim
t→+
ZS(t) –Z(t)
t =tlim→+t
–
π– t
π+ t –
π– t
√
π+ t
π/
=
π–
π < ,
lim
t→(π/)–
ZS(t) –Z(t)
π– t =t→lim(π/)–
π+ t –
(π– t)π/–
(√π+ t)π/
= π > ,
lim
t→+
Z(t) –BS(t)
t =tlim→+t
–
π– t
√
π+ t
π/
–π
– t–πt/
π–t
= –π
–
π < ,
lim
t→(π/)–
Z(t) –BS(t)
= lim
t→(π/)–
π– t
√
π+ t
π/
–π
– t–πt/
π–t
= π
> ,
ZS(t) –BS(t) =
π– t
π+ t–
π– t–πt/ π–t
= t
(π+ )
(π–t)(π+ t)
t–π
( –π)
( +π)
⎧ ⎨ ⎩
< if <t<π
–π
+π, > if π
–π
+π <t<
π
.
This completes the proof.
Remark From the above proposition we see that the sharp lower bound in (.) is su-perior to those ones given in (.), (.), (.), (.), and (.).
Remark Analogously, by comparing the limits att= andt=π/, we find the sharp upper bound in (.) is not comparable with those ones given in (.), (.), (.), (.), and (.). Here we omit all the details.
Remark We claim that the result stated in Theorem is stronger than the inequality (.), that is, fort∈(,π/), we have the inequalities
– πt
< +
p
ln –pt
< t
tant< +
ln
– t
< –t
. (.)
Indeed, the right hand side for this inequality in (.) follows from Corollary , while the left hand side one is the inequality connecting the first and third bounds in (.).
Remark Lemma tells us that
<
t– costsint– tsint
t(t–sintcost) <
π fort∈(,π/), (.)
π <
t– costsint– tsint
t(t–sintcost) <
Then from equation (.) we find that for f(t) < fort∈(,π) whenp= /π, and so
f(t) <f(+) = . This gives the following inequality:
tcost
sint – –
π
ln
– t
π
<
for allt∈(,π), which can be stated as the following proposition.
Proposition For all t∈(,π),we have
t
tant< +
π
ln
– t
π
.
Remark The inequality
sint t <
+cost
, t∈
,π
,
is true due to Cusa and Huygens’ paper (see, e.g.[]), which is now known as Cusa’s inequality(seee.g.[, –]). Some refinements and generalizations ofCusa’s inequality can be found in[, , , –]. Now by lettingt=x/ and simplifying, inequalities (.) and (.) can be written as
x(x–sinx) <
+cosx
–
sinx x <
x(x–sinx)
π forx∈(,π),
x(x–sinx) π <
+cosx
–
sinx x <
x(x–sinx)
π forx∈(π, π),
which give stronger versions of Cusa’s inequality.
Proposition We have
+cosx
–
x(x–sinx) π <
sinx x <
+cosx
–
x(x–sinx)
for x∈(,π), (.)
+cosx
–
x(x–sinx) π <
sinx x <
+cosx
–
x(x–sinx)
π for x∈(π, π). (.)
Moreover,the two double inequalities are sharp.
Remark In [], Corollary , Yanget al.proved that, fort∈(,π/),
exp
t
tant–
<sint
t <exp
t
tan(t/)–
.
Then by inequalities (.) forp= /(π) andq= /, we obtain
– t
π
π/
<
exp
t
tant–
<sint
t <exp
t/
tan(t/)–
<
– t
Proposition The inequalities
ρr
–rt/(r)<λsexp
st
tan(st)–
<sint
t <exp
st
tan(st)–
< –rt/(r) (.)
hold for t∈(,π/)with the best constants s= /√,r= /and
λs=
πexp(√πcot(√π/) – )≈.,
ρr=
π( –π/)/ ≈..
Proof Let
h(t) =stcos(st)
sin(st) – –ln
sint t ,
wheres= /√. Differentiation yields
h(t) =
t –
cost
sint +s
cosst
sinst –s
t
sinst.
Expanding in power series leads to
h(t) =
∞
n=
– nsn
n (n)!|Bn|t
n–=
∞
n=
n– n
n
(n– )!|Bn|t
n–> .
This indicates thath(π/) >h(t) >h(+) = fort∈(,π/), which proves the second and
third inequalities of (.). Considering the limit
lim
t→
h(t)
t =limt→
stcos(st)
sin(st) – –ln
sint t
t =
–s
,
it is seen thats= /√ andλsare the best possible constants.
The first and fourth ones are derived from the decreasing property off(st)≡f(u) for
u∈(,sπ/)⊂(,π/) proved in Theorem forp=r/s= /, and thenr= / andρ
r
are also the best. This completes the proof.
5 Applications
In this section, we give some precise estimations for the Sine integral and Catalan constant. The Sine integral is defined by
Si(x) =
x
sint t dt.
Proposition For x∈(,π],we have
x+sinx
–
x+ xcosx– sinx
π <Si(x) <
x+sinx
–
x+ xcosx– sinx
. (.)
In particular,we have
.≈ π +
π–π+ π
√
<Si
π
<π( –π
)
+ –π
√
≈.,
.≈π
+ π+
π <Si
π
< + π–π
≈.,
.≈π
+
π <Si(π) <
π( –π)
≈..
Proof Indeed, integrating both sides over [,x] for double inequality (.) easily yields (.). Direct computations give the approximation values ofSi(x) forx=π/,π/,π.
Note that
π/
ln(sinx)dx= –π
ln≈–.
and
x
t
tantdt=xln(sinx) – x
ln(sint)dt. (.)
We are now in the position to evaluate the integralxln(sint)dtforx∈(,π/).
Proposition Let x∈(,π/).Then,for p∈(, /),we have
Lp(x) <
x
ln(sint)dt<Up(x), (.)
where
Lp(x) =xln(sinx) – p–
p x–
pxln
–px– p/ln
√px+
–√px,
Up(x) =xln(sinx) –
x
tanx+
px–
p/ln
√ px+ –√px.
The double inequality(.)is reversed for p∈[/(π), /π].In particular,we have
xln(sinx) –x+ x
<
x
ln(sint)dt<xln(sinx) – x
tanx–
x
. (.)
Proof By the proof of Theorem we see that the function
t→lnY(t) –lnBp(t) =
t
tant– –
pln
is strictly decreasing on (,π/) ifp≤/ and increasing on (,π/) if /(π)≤p≤
/π. Then, fort∈(,x]⊂(,π/), we have, forp≤/,
x
tanx– –
pln
–px≤ t
tant– –
pln
–pt< ,
that is,
x
tanx–
pln
–px+ pln
–pt≤ t
tant< +
pln
–pt,
which is the reverse for /(π)≤p≤/π.
Integrating both sides over [,x] gives, forp∈(, /),
x
tanx– x
pln
–px+ p
xln –px+√
pln √
px+ –√px– x
≤ x
t
tantdt<x+
p
xln –px+√
pln √
px+ –√px– x
.
Combining with equation (.) gives the double inequality (.) forp∈(, /]. Moreover, it is clear that the double inequality (.) is reversed if /(π)≤p≤/π.
Lettingp→+in (.) yields (.), which completes the proof.
Remark Takingp= /(π) andx=π/,π/ in the double inequality (.) and
com-puting give
L/(π)
π
= –ππ
ln –πln +√πln( +√) – π+
≈–.,
U/(π)
π
= –π
√
ln(√ + ) –
≈–.,
L/(π)
π
= –π
√
πln( + √) –πln –π(√ – )ln – π+ ln +
≈–.,
U/(π)
π
= –π
√
πln(√ + ) – √πln – π+π+ ln
≈–..
Then we obtain
–.≈U/(π)
π
<
π/
ln(sint)dt<L/(π)
π
≈–., (.)
–.≈U/(π)
π
<
π/
ln(sint)dt<L/(π)
π
≈–.. (.)
It is well known that the Catalan constant appearing in [–]
G=
∞
n=
(–)n
(n+ ) = . . . .
is a famous mysterious constant appearing in many places in mathematics and physics. Its integral representations [] include the following:
G=
arctanx
x dx=
π/
x
sinxdx
= –
π/
ln(sinx)dx=π
– π ln +
π/
x
sinxdx.
Now, by using the third integral representation for Gand (.), we easily obtain a very accurate approximation forG, the absolute error of which is less than ..
Proposition We have
.≈–π
ln – L/(π)
π
<G< –π
ln – U/(π)
π
≈..
Remark Clearly, the above estimate forGis superior to Yang’s presentation in [], Proposition , [], Remark ..
6 Conclusions
Rather than using classical approaches, we in this paper presented the new upper and lower bounds of tant
t on the interval (,π/) by way of the monotonicity criterion for the quotient of power series. Our conclusions have not only refined the Redheffer and Becker-Stark type inequalities concerning the tangent function, but they also showed some more precise estimations to the Sine integral and the Catalan constant. More precisely, our con-clusion is that the sharp lower bound of tantt is superior to all given results as showed by Proposition in Section , although its sharp upper bound is not comparable with those given ones. In addition, we also derived a stronger version of Cusa’s inequality, and a very accurate approximation of the Catalan constant with the absolute error being less than ..
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed to each part of this work equally, and they all read and approved the final manuscript.
Author details
1Department of Mathematics, Beijing Jiaotong University, Beijing, 100044, China.2Department of Science and
Technology, State Grid Zhejiang Electric Power Company Research Institute, Hangzhou, Zhejiang 310014, China.
Acknowledgements
This paper is supported by the National Science Foundation of China grant No. 11371050 and the National Training Program of Innovation and Entrepreneurship for Undergraduates Project grant No. 160170009.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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