分部求和法与积分中值定理
42. (分部积分法)设黎曼-斯提捷积分(Riemman-Stieltjes)积分$int_{a}^{b}alpha(x)df(x)$存在,则$int_{a}^{b}f(x)dalpha(x)$也存在并且有分部积分公式
$$int_{a}^{b}f(x)dalpha(x)=[f(x)alpha(x)]Big|_{a}^{b}-int_{a}^{b}alpha(x)df(x)$$
证明:一般的Riemman积分中的分部积分公式
$$int_{a}^{b}f(x)dalpha(x)=[f(x)alpha(x)]Big|_{a}^{b}-int_{a}^{b}alpha(x)df(x)$$
是与乘积的微分
$$d(falpha)=fdalpha+alpha df$$
对应的. 体现了微分与积分这一对矛盾. 但是要求$f,alpha$为可微函数.在$R-S$积分中,这一条件过强.$alpha(x)$实际上不必是可微函数甚至不必是连续函数.利用
$R-S$积分的定义,用离散与连续这对矛盾的眼光来看待分部积分.
作划分
$$pi: a=x_{0}<x_{1}<x_{2}<cdots<x_{n}=b$$
并且$|pi|=max ||Delta x_{i}|,Delta x_{i}=x_{i}-x_{i-1},x_{i-1}leq xi_{i}leq x_{i}$.
应用分部求和法
egin{align*}
sigma(f,pi,xi)&=sum_{k=1}^{n}f(xi_{k})[alpha(x_{k})-alpha(x_{k-1})]\
&=sum_{k=1}^{n}f(xi_{k})alpha(x_{k})-sum_{k=1}^{n}f(xi_{k})alpha(x_{k-1})\
&=sum_{k=1}^{n}f(xi_{k})alpha(x_{k})-sum_{k=0}^{n-1}f(xi_{k+1})alpha(x_{k})\
&=f(b)alpha(b)-f(a)alpha(a)+[f(xi_{n})-f(x_{n})]alpha(x_{n})+[f(x_{0})-f(xi_{1})]alpha(x_{0})\
&-sum_{1}^{n-1}[f(xi_{k+1})-f(xi_{k})]alpha(x_{k})\
&=f(b)alpha(b)-f(a)alpha(a)+[f(xi_{n})-f(x_{n})]alpha(x_{n})+[f(x_{0})-f(xi_{1})]alpha(x_{0})-sigma(alpha,pi',x)
end{align*}
序列
$$pi': a=xi_{1}<xi_{2}<cdots<xi_{n+1}=b,xi_{k}leq x_{k}leqxi_{k+1}$$
并且$|pi'|leq 2|pi|$,是积分$int_{a}^{b}alpha df$的一个划分.从而
$$lim_{|pi| o 0}sigma(f,pi,xi)=f(x)alpha(x)Big|_{a}^{b}-lim_{|pi'| o 0}sigma(alpha,pi',x)$$
即
$$int_{a}^{b}f(x)dalpha(x)=[f(x)alpha(x)]Big|_{a}^{b}-int_{a}^{b}alpha(x)df(x)$$
注意: 若$f$为连续函数而$alpha$为有界变差函数,则$int_{a}^{b}fdalpha$与$int_{a}^{b}alpha df$都存在.
43. (第一积分中值定理) 设$alpha(x)$为一单调函数而$f(x)$为实值连续函数则有中值公式
$$int_{a}^{b}f(x)dalpha(x)=f(xi)[alpha(b)-alpha(a)],\,\,(aleq xileq b)$$
证明: 设
$$m(f)=min_{xin[a,b]} f(x);M(f)=max_{xin[a,b]}f(x)$$
则有
$$m(f)leq frac{1}{alpha(b)-alpha(a)}int_{a}^{b}f(x)dalpha(x)leq M(f)$$
由实连续函数的介值定理知存在$xiin [a,b]$,使得
$$f(xi)=frac{1}{alpha(b)-alpha(a)}int_{a}^{b}f(x)dalpha(x)$$
44. 设$f(x)$连续而$psi(x)$为$[a,b]$上的勒贝格(H.Lebesge1875-1941)可积函数(简写作$psi in L$),并设$psi(x)geq 0$,则必有$xi,aleq xileq b$.使得
$$int_{a}^{b}f(x)psi(x)dx=f(xi)int_{a}^{b}psi(x)dx$$
证明: 设
$$m(f)=min_{xin[a,b]} f(x);M(f)=max_{xin[a,b]}f(x)$$
则有
$$m(f)leq frac{1}{int_{a}^{b}psi(x) dx}int_{a}^{b}f(x)psi(x)dxleq M(f)$$
由实连续函数的介值定理知存在$xiin [a,b]$,使得
$$int_{a}^{b}f(x)psi(x)dx=f(xi)int_{a}^{b}psi(x)dx$$
45. (长大不等式) 设在$[a,b]$上$f(x)$为连续函数而$alpha(x)$为有界变差函数,则
$$left|int_{a}^{b}f(x)dx
ight|leq M(f)cdot igvee_{a}^{b}(alpha)$$
这里$M(f)=max_{aleq xleq b}f(x)$,而 $igveelimits_{a}^{b}(alpha)$为$alpha$在$[a,b]$上的全变差.
证明:
egin{align*}
left|int_{a}^{b}f(x)dx
ight|&leq lim_{|pi| o 0}sum_{k=1}^{n}left|f(xi_{k})
ight|cdot |alpha(x_{k})-alpha(x_{k-1})|\
&leq M(f)cdot lim_{|pi| o 0}sum_{k=1}^{n}|alpha(x_{k})-alpha(x_{k-1})|\
&leq M(f)cdot igvee_{a}^{b}(alpha)
end{align*}
46. (第二积分中值定理) 设在$[a,b]$上$alpha(x)$为一实值连续函数而$f(x)$为一单调函数则必有$xi,aleq xileq b$使得
$$int_{a}^{b}f(x)dalpha(x)=f(a)int_{a}^{xi}dalpha(x)+f(b)int_{xi}^{b}dalpha(x)$$
证明: 利用分部积分和第一中值定理(涉及到端点值和中间值)
egin{align*}
int_{a}^{b}f(x)dalpha(x)&=f(x)alpha(x)Big|_{a}^{b}-int_{a}^{b}alpha(x)df(x)\
&=f(b)alpha(b)-f(a)alpha(a)-alpha(xi)int_{a}^{b}df(x)\
&=f(b)[alpha(b)-alpha(xi)]+f(a)[alpha(xi)-alpha(a)]\
&=f(a)int_{a}^{xi}dalpha(x)+f(b)int_{xi}^{b}dalpha(x)
end{align*}
47. (Bonnet) 设$varphi(x)in L$,又设$f(x)$单调. 则必存在$xi,aleq xileq b$
$$int_{a}^{b}f(x)varphi(x)dx=f(a)int_{a}^{xi}varphi(x)dx+f(b)int_{xi}^{b}varphi(x)dx$$
证明: 此命题是命题46的推论.
设$alpha(x)=int_{a}^{x}varphi(t)dt$.
48. (Bonnet)设在$[a,b]$上$alpha(x)$为实值连续函数而$f(x)geq 0$且$f(x)uparrow$,则必有$xi,aleq xileq b$,使得
$$int_{a}^{b}f(x)dalpha(x)=f(b)int_{xi}^{b}dalpha(x)$$
又若$f(x)leq 0$且$f(x)downarrow$,则必有$xi,aleq xileq b$,使得
$$int_{a}^{b}f(x)dalpha(x)=f(a)int_{a}^{xi}dalpha(x)$$
证明:假定$f(x)leq 0$且$f(x)downarrow$则
$$sigma(f,pi,xi)=sum_{k=1}^{n}f(xi_{k})[alpha(x_{k})-alpha(x_{k-1})]$$
关于$f(xi_{k})$
$$f(a)=f(xi_{1})geq f(xi_{2})geq cdots geq f(xi_{n})geq 0$$
而
$$inf_{aleq xleq b}int_{a}^{x}dalpha(t)leq sum_{k=1}^{n}[alpha(x_{k})-alpha(x_{k-1})]leq sup_{aleq xleq b}int_{a}^{x}dalpha(t)$$
由Abel引理即命题 4 有
$$f(a)inf_{aleq xleq b}int_{a}^{x}dalpha(t)leqsum_{k=1}^{n}f(xi_{k})[alpha(x_{k})-alpha(x_{k-1})]leq f(a)sup_{aleq xleq b}int_{a}^{x}dalpha(t)$$
取极限$|pi| o 0$
$$f(a)inf_{aleq xleq b}int_{a}^{x}dalpha(t)leqint_{a}^{b}f(x)dalpha(x)leq f(a)sup_{aleq xleq b}int_{a}^{x}dalpha(t)$$
由于$int_{a}^{x}dalpha(t)=alpha(x)-alpha(a)$为连续函数,由连续函数的介质定理知,存在$xi,aleq xileq b$
$$int_{a}^{b}f(x)dalpha(x)=f(a)int_{a}^{xi}dalpha(t)$$
若$alpha(x)=int_{a}^{x}varphi(t)dt$,结论为
$$int_{a}^{b}f(x)varphi(x)dx=f(a)int_{a}^{xi}varphi(x)dx$$
类似证明另一等式.
49. 试由命题48导出命题46.
证明: 设$f(x)downarrow$,命题 48要求$f(x)$非负,命题46无此要求只对单调性有要求.因此$f(x)-f(b)geq 0,\,(aleq xleq b)$.
$$int_{a}^{b}[f(x)-f(b)]dalpha(x)=[f(a)-f(b)]int_{a}^{xi}dalpha(x)=f(a)int_{a}^{xi}dalpha+f(b)int_{xi}^{b}dalpha-f(b)int_{a}^{b}dalpha$$
两边消去$f(b)int_{a}^{b}dalpha$即得命题46.
50. 设在$[a,b]$上$alpha(x)$为一有界变差函数而$f(x)$为非负连续函数.若$f(x)uparrow$,则
$$int_{a}^{b}f(x)dalpha(x)=Af(b)$$
此处
$$inf_{aleq xleq b}int_{x}^{b}dalpha(x)leq Aleq sup_{aleq xleq b}int_{x}^{b}dalpha(x)$$
若$f(x)downarrow$,则
$$int_{a}^{b}f(x)dalpha(x)=Bf(a)$$
此处
$$inf_{aleq xleq b}int_{a}^{x}dalpha(x)leq Bleq sup_{aleq xleq b}int_{a}^{x}dalpha(x)$$
证明:若$f(x)geq 0$且$f(x)uparrow$则
$$f(b)inf_{aleq xleq b}int_{x}^{b}dalpha(t)leq sum_{k=1}^{n}f(xi_{k})[alpha(x_{k})-alpha(x_{k-1})]leq f(b)sup_{aleq xleq b}int_{x}^{b}dalpha(t) $$
取极限$|pi| o 0$
$$f(b)inf_{aleq xleq b}int_{x}^{b}dalpha(t)leqint_{a}^{b}f(x)dalpha(x)leq f(b)sup_{aleq xleq b}int_{x}^{b}dalpha(t)$$
所以
$$inf_{aleq xleq b}int_{x}^{b}dalpha(t)leq A=frac{1}{f(b)}int_{a}^{b}fdalphaleq sup_{aleq xleq b}int_{x}^{b}dalpha(t)$$
同理可证$f(x)downarrow$时
$$inf_{aleq xleq b}int_{a}^{x}dalpha(t)leq B=frac{1}{f(a)}int_{a}^{b}fdalphaleq sup_{aleq xleq b}int_{a}^{x}dalpha(t)$$
51. (陈建功) 设$alpha>0,A>0,0leq a<b$.试证
$$left|int_{a}^{b}cosleft(nt-frac{A}{t^{alpha}}
ight)dt
ight|<frac{2}{n}$$
$$left|int_{a}^{b}sinleft(nt-frac{A}{t^{alpha}}
ight)dt
ight|<frac{2}{n}$$
证明: 作变量替换
$$omega=t-frac{A}{nt^{alpha}}$$
则
$$frac{domega}{dt}=1+frac{Aalpha}{nt^{alpha+1}}>0,\, tin [a,b]$$
$$frac{d^{2}(t)}{domega^{2}}=left(1+frac{Aalpha}{nt^{alpha+1}}
ight)^{-2}frac{alpha(alpha+1)A}{nt^{alpha+2}}frac{dt}{domega}>0$$
则由积分第二中值定理
egin{align*}
left|int_{a}^{b}cosleft(nt-frac{A}{t^{alpha}}
ight)dt
ight|&=left|int_{omega(a)}^{omega(b)}cos (nomega)frac{dt}{domega}domega
ight|\
&=left(1+frac{Aalpha}{nb^{alpha+1}}
ight)^{-1}left|int_{xi}^{b}cos(nomega)domega
ight|\
&leq left|frac{sin (nxi)-sin(nomega(b))}{n}
ight|\
&leq frac{2}{n}
end{align*}
第二个不等式的证明完全类似.
52. (Dirichlet-Jordan)设$f(x)$为以$2pi$为周期的的可积函数,那么采用命题29的证明中的记法时,$f(x)$的傅里叶级数的部分和可表达成
$$S_{n}(x)=frac{1}{pi}int_{-pi}^{pi}f(x+ heta)D_{n}( heta)d heta$$
现设$f(x)$在$[-pi,pi]$上为连续的有界变差函数,试证$S_{n}(x)$一致收敛于$f(x)$.
证明:先论证$f(x)$的傅里叶级数的部分和可表达成
$$S_{n}(x)=frac{1}{pi}int_{-pi}^{pi}f(x+ heta)D_{n}( heta)d heta$$
其中
$$D_{n}(x)=frac{sin (n+frac{1}{2}) heta}{2sinfrac{ heta}{2}}$$
考察
$$sum_{-N}^{N}e^{in heta}=sum_{-N}^{N}(cos n heta+sin n heta)=2D_{N}( heta)$$
利用周期性,奇偶性,函数$f(x)$的傅里叶级数的部分和
egin{align*}
S_{N}(f)(x)&=sum_{-N}^{N}hat{f}(n)e^{inx}\
&=sum_{-N}^{N}left(frac{1}{2pi}int_{-pi}^{pi}f( heta)e^{-in heta}d heta
ight)e^{inx}\
&=frac{1}{pi}int_{-pi}^{pi}f( heta)D_{N}(x- heta)d heta\
&=frac{1}{pi}int_{-pi}^{pi}f(x- heta)D_{N}( heta)d heta\
&=frac{1}{pi}int_{-pi}^{pi}f(x+ heta)D_{N}( heta)d heta
end{align*}
经计算
$$frac{1}{pi}int_{-pi}^{pi}D_{N}( heta)d heta=1$$
拟合法
egin{align*}
S_{N}(f)(x)-f(x)&=frac{1}{pi}int_{-pi}^{pi}[f(x+ heta)-f(x)]D_{N}( heta)\
&=frac{1}{pi}int_{0}^{pi}[f(x+ heta)+f(x- heta)-2f(x)]D_{N}( heta)d heta\
&=frac{1}{pi}int_{0}^{pi}Delta_{ heta}f(x)frac{sin n heta}{ heta}d heta+frac{1}{pi}int_{0}^{pi}Delta_{ heta}f(x)left(frac{cot frac{ heta}{2}}{2}-frac{1}{ heta}
ight)d heta\
&+frac{1}{2pi}int_{0}^{pi}Delta_{ heta}f(x)cos n heta d heta\
&=I_{1}+I_{2}+I_{3}
end{align*}
由黎曼-勒贝格定理知$I_{2} o 0, I_{3} o 0$且一致收敛到0. $I_{1}$主要是在零点附近会出现问题
$$I_{1}=int_{0}^{frac{pi}{n}}+int_{frac{pi}{n}}^{eta}+int_{eta}^{pi}=I_{11}+I_{12}+I_{13}$$
分别估计
$$|I_{11}|leqmax_{0leq hetaleq frac{pi}{n}}|Delta_{ heta}f(x)|int_{0}^{pi}frac{sin heta}{ heta}d hetaleq max_{0leq hetaleq frac{pi}{n}}|Delta_{ heta}f(x)|cdot pi $$
$$|I_{12}|=frac{n}{pi}int_{frac{pi}{n}}^{xi}Delta_{ heta}f(x)sin n heta d hetaleq frac{2}{pi}max_{0leq hetaleq eta}|Delta_{ heta}f(x)|+frac{1}{pi}igvee_{0}^{eta}(Delta_{ heta}f)$$
于是$I_{11},I_{12}$一致收敛到0. $I_{13}$可使用黎曼—勒贝格定理.
53. 设$f(x)$是$[a,b]$上的可积函数,
$$F(x)=int_{a}^{x}f(t)dt,\,|F(x)|leq M(x-a) (aleq x leq b)$$
又设$g(x)$为是$[a,b]$上的非负并且非增的可积函数,则
$$left|int_{a+}^{b}f(x)g(x)dx
ight|leq Mint_{a}^{b}g(x)dx$$
证明: 设$a<alpha<etaleq b$. 则由分部积分法
egin{align*}
&int_{alpha}^{eta}f(x)g(x)dx-g(eta)int_{alpha}^{eta}f(x)dx\
=&int_{alpha}^{eta}f(x)[g(x)-g(eta)]dx\
=&-F(alpha)[g(alpha)-g(eta)]-int_{alpha}^{eta}F(x)dg(x)
end{align*}
把关于$|F(x)|$的条件用到上式的末端可知其绝对值不超过
egin{align*}
&M(alpha-a)[g(alpha)-g(eta)]-int_{alpha}^{eta}M(x-a)dg(x)\
=&M(alpha-a)[g(alpha)-g(eta)]-[M(x-a)g(x)]_{alpha}^{eta}+Mint_{alpha}^{eta}g(x)dx\
=&Mint_{alpha}^{eta}g(x)dx+2M(alpha-a)g(alpha)-M(alpha+eta-2a)g(eta)
end{align*}
当$alpha,eta o a+$时,右端两项是无穷小量,所以$int_{a+}^{b}f(x)g(x)dx$收敛. 置$eta=b,alpha o a+$, 由上式得
$$left|int_{a+}^{b}f(x)g(x)dx
ight|leq Mint_{a}^{b}g(x)dx-M(b-a)g(b)+g(b)left|int_{a}^{b}f(x)dx
ight|leq Mint_{a}^{b}g(x)dx$$
54. （Steffensen） 设$f(x)$和$g(x)$是$[a,b]$上的两个可积函数,$f(x)$不增(注:不增的意思就是递减),$g(x)$满足$0leq g(x) leq 1$,则
$$int_{b-lambda}^{b}f(t)dtleq int_{a}^{b}f(t)g(t)dtleq int_{a}^{a+lambda}f(t)dt$$
证明: 非常有意思的不等式. 不妨设$f(t)$非负否则置$f(x)-m$代入原不等式.
egin{align*}
&int_{a}^{a+lambda}f(x)dx-int_{a}^{b}f(x)g(x)dx\
=&int_{a}^{a+lambda}f(x)[1-g(x)]dx-int_{a+lambda}^{b}f(x)g(x)dx\
geq&f(a+lambda)int_{a}^{a+lambda}[1-g(x)]dx-f(a+lambda)int_{a}^{xi}g(x)dx\
=&f(a+lambda)int_{xi}^{b}g(x)dx\
geq&0
end{align*}
另一方面
egin{align*}
&int_{a}^{b}f(x)g(x)dx-int_{b-lambda}^{b}f(x)dx\
=&int_{a}^{b-lambda}f(x)g(x)dx+int_{b-lambda}^{b}f(x)[g(x)-1]dx\
geq&f(b-lambda)int_{a}^{b-lambda}g(x)dx+f(b-lambda)int_{b-lambda}^{b}[g(x)-1]dx\
=&f(b-lambda)int_{a}^{b}g(x)dx-f(b-lambda)int_{b-lambda}^{b}dx\
=&0
end{align*}
另证: 只需要注意到恒等变形
egin{align*}
&int_{a}^{a+lambda}f(x)dx-int_{a}^{b}f(x)g(x)dx\
=&int_{a}^{a+lambda}[f(x)-f(a+lambda)]cdot [1-g(x)]dx
+int_{a+lambda}^{b}[f(a+lambda)-f(x)]cdot g(x)dx
end{align*}
55. (Stefensen) 设$g_{1}(x)$与$g_{2}(x)$满足
$$int_{a}^{x}g_{1}(x)leq int_{a}^{x}g_{2}(x)\,\,(aleq xleq b)$$
$$int_{a}^{b}g_{1}(x)=int_{a}^{b}g_{2}(x)$$
又设$f(x)$非增,则
$$int_{a}^{x}g_{1}(x)leq int_{a}^{x}f(x)g_{2}(x)$$
证明: 使用第二积分中值定理,$f(x)-f(b)geq 0$且非增.
$$int_{a}^{b}f(x)cdot[g_{2}(x)-g_{1}(x)]dx=int_{a}^{b}[f(x)-f(b)]cdot[g_{2}(x)-g_{1}(x)]dx
=[f(a)-f(b)]int_{a}^{xi}[g_{2}(x)-g_{1}(x)]dxgeq 0$$
56. 试由命题55推导出命题44.
证明:取截断函数即可
egin{equation*}
g_{1}(x)=left{egin{array}{ll}
1& ext{$b-lambdaleq xleq b$}\
0& ext{其他}
end{array}
ight.
end{equation*}
egin{equation*}
g_{3}(x)=left{egin{array}{ll}
1& ext{$aleq xleq a+lambda$}\
0& ext{其他}
end{array}
ight.
end{equation*}
则当$b-lambda leq xleq $且$0leq g(x)leq 1$时有
$$int_{a}^{x}g_{1}(x)dx=int_{a}^{b}g(x)dx+x-bleq int_{a}^{x}g(x)dx$$
且
$$int_{a}^{b}g_{1}(x)dx=int_{a}^{b}g(x)dx=int_{a}^{b}g_{3}(x)$$
利用把$g_{1}(x),g_{2}(x),g_{3}(x)$代入命题55即得命题54.
57. (Hayashi)以$0leq g(t)leq A$代替命题54中的$0leq g(x)leq 1$,则有
$$Aint_{b-lambda}^{b}f(t)dtleq int_{a}^{b}f(t)g(t)dtleq Aint_{a}^{a+lambda}f(t)dt$$
这里$A$是正的常数,$lambda=frac{1}{A}int_{a}^{b}g(t)dt$.
证明:归一化处理
$$ ilde{g}(x)=frac{g(x)}{A}\,\,(aleq xleq b)$$
则$0leq g(x)leq 1$,
$$lambda=int_{a}^{b} ilde{g}(x)dx=frac{1}{A}int_{a}^{b}g(x)dx$$
由命题54得
$$int_{b-lambda}^{b}f(t)dtleq int_{a}^{b}f(t) ilde{g}(t)dtleq int_{a}^{a+lambda}f(t)dt$$
即
$$Aint_{b-lambda}^{b}f(t)dtleq int_{a}^{b}f(t)g(t)dtleq Aint_{a}^{a+lambda}f(t)dt$$
58. 不等式
$$int_{b-lambda}^{b}f(t)dtleq int_{a}^{b}f(t)g(t)dtleq int_{a}^{a+lambda}f(t)dt$$
式中$lambda=int_{a}^{b}g(t)dt$对 extbf{每个}不增的函数$f(t)$都成立的充分且必要条件是函数$g(t)$对所有的$xin [a,b]$满足
$$0leqint_{x}^{b}g(t)dtleq b-x ext{且}0leqint_{a}^{x}g(t)dtleq x-a$$
证明: 先证明必要性，取
egin{equation*}
f(t)=left{egin{array}{ll}
1& ext{$aleq tleq x$}\
0& ext{$x< t$}
end{array}
ight.
end{equation*}
则$f(x)$为不增函数
$$int_{a}^{b}f(t)g(t)dt=int_{a}^{x}g(t)dtleq int_{a}^{x}dx=x-a$$
取
egin{equation*}
f(t)=left{egin{array}{ll}
1& ext{$xleq tleq b$}\
0& ext{$t< x$}
end{array}
ight.
end{equation*}
类似可得
$$0leqint_{x}^{b}g(t)dtleq b-x$$
再证充分性
egin{align*}
&int_{a}^{a+lambda}f(t)dt-int_{a}^{b}f(t)g(t)dt\
=&int_{a}^{a+lambda}f(t)cdot[1-g(t)]dt-int_{a+lambda}^{b}f(t)g(t)dt\
=&int_{a}^{a+lambda}[f(t)-f(a+lambda)]cdot [1-g(t)]dt+int_{a+lambda}^{b}[f(a+lambda)-f(t)]g(t)dt\
=&I_{1}+I_{2}
end{align*}
利用分部积分
egin{align*}
I_{1}&=int_{a}^{a+lambda}[f(t)-f(a+lambda)]dleft[t-int_{a}^{t}g(x)
ight]\
&=[f(t)-f(a+lambda)]left[t-int_{a}^{t}g(x)
ight]Bigg|_{a}^{a+lambda}+int_{a}^{a+lambda}left[t-int_{a}^{t}g(x)
ight]d(-f)\
&=[f(a)-f(a+lambda)]a+int_{a}^{a+lambda}left[t-int_{a}^{t}g(x)
ight]d(-f)\
&geq [f(a)-f(a+lambda)]a+int_{a}^{a+lambda}a d(-f)\
&geq 0
end{align*}
对$I_{2}$估计
egin{align*}
I_{2}&=int_{a+lambda}^{b}[f(a+lambda)-f(t)]g(t)dt\
&=int_{a+lambda}^{b}[f(a+lambda)-f(t)]dint_{a}^{t}g(x)dx\
&=lim_{|pi| o 0}sum_{k=1}^{n}[f(a+lambda)-f(xi_{k})]cdot [alpha(t_{k})-alpha(t_{k-1})]\
&geq 0
end{align*}
其中$alpha(t)=int_{a}^{t}g(x)dx$.
所以
$$int_{a}^{b}f(t)g(t)dtleq int_{a}^{a+lambda}f(t)dt$$
同理可证
$$int_{b-lambda}^{b}f(t)dtleq int_{a}^{b}f(t)g(t)dt$$
59. (Abel)设无穷积分$int_{a}^{infty}varphi(x)dx$为收敛. 又设$psi(x)$为一单调的有界函数.则积分$int_{a}^{infty}varphi(x)psi(x)dx$必收敛.
证明:由收敛的Cauchy准则知任意$varepsilon>0$,存在$X$,$A>Bgeq X$时
$$left|int_{A}^{B}varphi(x)
ight|leq varepsilon$$
利用积分第二中值定理
$$left|int_{A}^{B}varphi(x)psi(x)
ight|=left|psi(A)int_{A}^{xi}varphi(x)dx+psi(B)int_{xi}^{B}varphi(x)dx
ight|leq 2Mcdot varepsilon$$
其中$aleq xileq b,M=suplimits_{aleq xleq infty}{psi(x)}$. 由Cauchy收敛准则积分$int_{a}^{infty}varphi(x)psi(x)dx$必收敛.
60. (Dirichlet)设函数$alpha(x)=int_{a}^{x}$为有界$(aleq xleq infty)$. 又设当$x oinfty$时$psi(x)downarrow 0$.则积分$int_{a}^{infty}varphi(x)psi(x)dx$必收敛.
证明: 由$psi(x)downarrow 0,x o infty$知任意$varepsilon>0$存在$X$,当$A>B>X$时,$|psi(x)|<varepsilon$.
由积分第二中值定理
$$left|int_{A}^{B}varphi(x)psi(x)dx
ight|=left|psi(B)int_{A}^{xi}varphi(x)dx
ight|leq 2Mcdot varepsilon$$
其中$M=sup_{aleq infty}left|int_{a}^{x}varphi(x)
ight|$.由Cauchy收敛准则积分$int_{a}^{infty}varphi(x)psi(x)dx$必收敛.
61. 设$varphi(x)downarrow 0 (x oinfty)$试证下列二积分必收敛
$$int_{a}^{infty}varphi(x)sin xdx,\,int_{a}^{infty}varphi(x)cos xdx$$
证明:
$$left|int_{a}^{x}sin x
ight|leq 2;\,left|int_{a}^{x}cos x
ight|leq 2$$
又$varphi(x)downarrow 0 (x oinfty)$,由狄利克雷判别法即命题60知$int_{a}^{infty}varphi(x)sin xdx,\,int_{a}^{infty}varphi(x)cos xdx$收敛.
62. 设$a$是正的常数,试证不等式
$$left|int_{a}^{infty}cos (x^{2})dx
ight|leq frac{1}{a}$$
证明:
$$left|int_{a}^{A}cos (x^{2})dx
ight|=left|int_{a^{2}}^{A^{2}}frac{cos t}{2sqrt{t}}dt
ight|=frac{1}{2a}left|int_{a^{2}}^{xi}cos t
ight|leq frac{2}{a}$$
令$A oinfty$.由狄利克雷判别法知左边级数收敛.