Application of Abelian distribution summation method (Iv.)

Partial summation method and integral mean value theorem

42. (Division integral method) Stite (Riemman-stieltjes) integral $\int_{a}^{b}\alpha (x) DF (x) $ present, then $\int_{a}^{b}f (x) d\alpha (x) $ Also exists and has a partial integral formula
$$\int_{a}^{b}f (x) d\alpha (x) =[f (x) \alpha (x)]\big|_{a}^{b}-\int_{a}^{b}\alpha (x) DF (x) $$
Proof: The partial integral formula in the general Riemman integral
$$\int_{a}^{b}f (x) d\alpha (x) =[f (x) \alpha (x)]\big|_{a}^{b}-\int_{a}^{b}\alpha (x) DF (x) $$
is the derivative of the product
$ $d (f\alpha) =fd\alpha+\alpha df$$
The corresponding. It embodies the contradiction between differential and integral. However, $f,\alpha$ is required to be a micro-function. In $r-s$ integral, this condition is too strong. $\alpha (x) $ actually does not have to be a micro function or even a continuous function. Using
$R the definition of-s$ integral, the division integral is viewed with the contradiction of discrete and continuous.
Make Division
$$\pi:a=x_{0}<x_{1}<x_{2}<\cdots<x_{n}=b$$
and $\|\pi\|=\max | | \delta X_{i}|,\delta X_{i}=x_{i}-x_{i-1},x_{i-1}\leq \xi_{i}\leq x_{i}$.
Application of Partial summation method
\begin{align*}
\sigma (F,\PI,\XI) &=\sum_{k=1}^{n}f (\xi_{k}) [\alpha (X_{k})-\alpha (x_{k-1})]\\
&=\sum_{k=1}^{n}f (\xi_{k}) \alpha (X_{k})-\sum_{k=1}^{n}f (\xi_{k}) \alpha (x_{k-1}) \ \
&=\sum_{k=1}^{n}f (\xi_{k}) \alpha (X_{k})-\sum_{k=0}^{n-1}f (\xi_{k+1}) \alpha (x_{k}) \ \
&=f (b) \alpha (b)-F (a) \alpha (a) +[f (\xi_{n})-F (X_{n})]\alpha (X_{n}) +[f (x_{0})-F (\xi_{1})]\alpha (x_{0}) \ \
&-\sum_{1}^{n-1}[f (\xi_{k+1})-F (\xi_{k})]\alpha (x_{k}) \ \
&=f (b) \alpha (b)-F (a) \alpha (a) +[f (\xi_{n})-F (X_{n})]\alpha (X_{n}) +[f (x_{0})-F (\xi_{1})]\alpha (x_{0})-\sigma (\ALPHA,\PI ', x)
\end{align*}
Sequence
$$\pi ': A=\xi_{1}<\xi_{2}<\cdots<\xi_{n+1}=b,\xi_{k}\leq x_{k}\leq\xi_{k+1}$$
and $\|\pi ' \|\leq 2\|\pi\|$, is a division of integral $\int_{a}^{b}\alpha df$.
$$\lim_{\|\pi\|\to 0}\sigma (F,\PI,\XI) =f (x) \alpha (x) \big|_{a}^{b}-\lim_{\|\pi ' \|\to 0}\sigma (\ALPHA,\PI ', x) $$
That
$$\int_{a}^{b}f (x) d\alpha (x) =[f (x) \alpha (x)]\big|_{a}^{b}-\int_{a}^{b}\alpha (x) DF (x) $$
Note: If $f$ is a continuous function and $\alpha$ is a bounded variation function, both $\int_{a}^{b}fd\alpha$ and $\int_{a}^{b}\alpha df$ exist.
43. (First integral mean value theorem) set $\alpha (x) $ for a monotone function and $f (x) $ for real value continuous function with median formula
$$\int_{a}^{b}f (x) d\alpha (x) =f (\XI) [\alpha (b)-\alpha (a)],\,\, (A\leq \xi\leq b) $$
Proof: Set
$ $m (f) =\min_{x\in[a,b]} f (x); M (f) =\max_{x\in[a,b]}f (x) $$
Then there are
$ $m (f) \leq \frac{1}{\alpha (b)-\alpha (a)}\int_{a}^{b}f (x) d\alpha (x) \leq m (f) $$
The existence of $\xi\in is known by the mean value theorem of the real continuous function [a,b]$, which makes
$ $f (\xi) =\frac{1}{\alpha (b)-\alpha (a)}\int_{a}^{b}f (x) d\alpha (x) $$
44. Set $f (x) $ continuous and $\PSI (x) $ for Lebesgue (h.lebesge1875-1941) integrable function on $[a,b]$ (Jane writes $\psi \in l$), and set $\psi (x) \geq 0$, then there must be $\xi,a\leq \xi\ Leq b$. Make
$$\int_{a}^{b}f (x) \psi (x) dx=f (\XI) \int_{a}^{b}\psi (x) dx$$
Proof: Set
$ $m (f) =\min_{x\in[a,b]} f (x); M (f) =\max_{x\in[a,b]}f (x) $$
Then there are
$ $m (f) \leq \frac{1}{\int_{a}^{b}\psi (x) dx}\int_{a}^{b}f (x) \psi (x) Dx\leq m (f) $$
The existence of $\xi\in is known by the mean value theorem of the real continuous function [a,b]$, which makes
$$\int_{a}^{b}f (x) \psi (x) dx=f (\XI) \int_{a}^{b}\psi (x) dx$$
45. (Growing inequality) is set on $[a,b]$ $f (x) $ is a continuous function and $\alpha (x) $ is a bounded variation function, then
$$\left|\int_{a}^{b}f (x) Dx\right|\leq M (f) \cdot \bigvee_{a}^{b} (\alpha) $$
Here $m (f) =\max_{a\leq X\leq b}f (x) $, while $\bigvee\limits_{a}^{b} (\alpha) $ is the total variation of $\alpha$ on $[a,b]$.
Prove:
\begin{align*}
\left|\int_{a}^{b}f (x) Dx\right|&\leq \lim_{\|\pi\|\to 0}\sum_{k=1}^{n}\left|f (\xi_{k}) \right|\cdot |\alpha (x_ {k}) -\alpha (x_{k-1}) |\\
&\leq M (f) \cdot \lim_{\|\pi\|\to 0}\sum_{k=1}^{n}|\alpha (X_{k})-\alpha (x_{k-1}) |\\
&\leq M (f) \cdot \bigvee_{a}^{b} (\alpha)
\end{align*}
46. (The second integral median theorem) is set on $[a,b]$ $\alpha (x) $ for a real value continuous function and $f (x) $ for a monotone function must have $\xi,a\leq \xi\leq b$ make
$$\int_{a}^{b}f (x) d\alpha (x) =f (a) \int_{a}^{\xi}d\alpha (x) +f (b) \int_{\xi}^{b}d\alpha (x) $$
Proof: The use of partial integrals and the first median theorem (involving end-point values and intermediate values)
\begin{align*}
\int_{a}^{b}f (x) d\alpha (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}d\alpha (x) +f (b) \int_{\xi}^{b}d\alpha (x)
\end{align*}
(Bonnet) Set $\varphi (x) \in l$, also set $f (x) $ monotonic. Then there must be $\xi,a\leq \xi\leq b$
$$\int_{a}^{b}f (x) \varphi (x) dx=f (a) \int_{a}^{\xi}\varphi (x) dx+f (b) \int_{\xi}^{b}\varphi (x) dx$$
Proof: This proposition is the inference of Proposition 46.
Set $\alpha (x) =\int_{a}^{x}\varphi (t) dt$.
(Bonnet) set on $[a,b]$ $\alpha (x) $ is a real continuous function and $f (x) \geq 0$ and $f (x) \uparrow$, there must be $\xi,a\leq \xi\leq b$, making
$$\int_{a}^{b}f (x) d\alpha (x) =f (b) \int_{\xi}^{b}d\alpha (x) $$
And if $f (x) \leq 0$ and $f (x) \downarrow$, there must be $\xi,a\leq \xi\leq b$, making
$$\int_{a}^{b}f (x) d\alpha (x) =f (a) \int_{a}^{\xi}d\alpha (x) $$
Proof: Assume $f (x) \leq 0$ and $f (x) \downarrow$
$$\sigma (F,\PI,\XI) =\sum_{k=1}^{n}f (\xi_{k}) [\alpha (X_{k})-\alpha (x_{k-1})]$$
About $f (\xi_{k}) $
$ $f (a) =f (\xi_{1}) \geq f (\xi_{2}) \geq \cdots \geq f (\xi_{n}) \geq 0$$
and
$$\inf_{a\leq X\leq B}\int_{a}^{x}d\alpha (t) \leq \sum_{k=1}^{n}[\alpha (X_{k})-\alpha (x_{k-1})]\leq \sup_{a\leq x\ Leq B}\int_{a}^{x}d\alpha (t) $$
By the Abel lemma that Proposition 4 has
$ $f (a) \inf_{a\leq X\leq B}\int_{a}^{x}d\alpha (t) \leq\sum_{k=1}^{n}f (\xi_{k}) [\alpha (X_{k})-\alpha (x_{k-1})]\leq F (a) \sup_{a\leq X\leq B}\int_{a}^{x}d\alpha (t) $$
Take limit $\|\pi\|\to 0$
$ $f (a) \inf_{a\leq X\leq B}\int_{a}^{x}d\alpha (t) \leq\int_{a}^{b}f (x) d\alpha (x) \leq F (a) \sup_{a\leq X\leq b}\int_{a} ^{x}d\alpha (t) $$
Since $\int_{a}^{x}d\alpha (t) =\alpha (x)-\alpha (a) $ is a continuous function, the media theorem of the continuous function is known to exist $\xi,a\leq \xi\leq b$
$$\int_{a}^{b}f (x) d\alpha (x) =f (a) \int_{a}^{\xi}d\alpha (t) $$
If $\alpha (x) =\int_{a}^{x}\varphi (t) dt$, the conclusion is
$$\int_{a}^{b}f (x) \varphi (x) dx=f (a) \int_{a}^{\xi}\varphi (x) dx$$
Similar proof another type.
49. Try to derive proposition 46 from Proposition 48.
Proof: Set $f (x) \downarrow$, Proposition 48 requires $f (x) $ non-negative, Proposition 46 no this requirement is only required for monotonicity. Therefore $f (x)-F (b) \geq 0,\, (A\leq X\leq b) $.
$$\int_{a}^{b}[f (x)-F (b)]d\alpha (x) =[f (a)-F (b)]\int_{a}^{\xi}d\alpha (x) =f (a) \int_{a}^{\xi}d\alpha+f (b) \int_{\ Xi}^{b}d\alpha-f (b) \int_{a}^{b}d\alpha$$
Both sides eliminate $f (b) \int_{a}^{b}d\alpha$ is the Proposition 46.
50. Set on $[a,b]$ $\alpha (x) $ is a bounded variation function and $f (x) $ is a nonnegative continuous function. If $f (x) \uparrow$, then
$$\int_{a}^{b}f (x) d\alpha (x) =af (b) $$
Here
$$\inf_{a\leq X\leq b}\int_{x}^{b}d\alpha (x) \leq a\leq \sup_{a\leq x\leq b}\int_{x}^{b}d\alpha (x) $$
If $f (x) \downarrow$, then
$$\int_{a}^{b}f (x) d\alpha (x) =BF (a) $$
Here
$$\inf_{a\leq X\leq b}\int_{a}^{x}d\alpha (x) \leq b\leq \sup_{a\leq x\leq b}\int_{a}^{x}d\alpha (x) $$
Proof: If $f (x) \geq 0$ and $f (x) \uparrow$
$ $f (b) \inf_{a\leq X\leq B}\int_{x}^{b}d\alpha (t) \leq \sum_{k=1}^{n}f (\xi_{k}) [\alpha (X_{k})-\alpha (x_{k-1})]\leq F (b) \sup_{a\leq X\leq B}\int_{x}^{b}d\alpha (t) $$
Take limit $\|\pi\|\to 0$
$ $f (b) \inf_{a\leq X\leq B}\int_{x}^{b}d\alpha (t) \leq\int_{a}^{b}f (x) d\alpha (x) \leq f (b) \sup_{a\leq X\leq b}\int_{x} ^{b}d\alpha (t) $$
So
$$\inf_{a\leq X\leq B}\int_{x}^{b}d\alpha (t) \leq a=\frac{1}{f (b)}\int_{a}^{b}fd\alpha\leq \sup_{a\leq X\leq b}\int_{ X}^{b}d\alpha (t) $$
The same can be $f (x) \downarrow$
$$\inf_{a\leq X\leq B}\int_{a}^{x}d\alpha (t) \leq b=\frac{1}{f (a)}\int_{a}^{b}fd\alpha\leq \sup_{a\leq X\leq b}\int_{ A}^{x}d\alpha (t) $$
51. (Chen Jiangong) set $\alpha>0,a>0,0\leq a<b$. Test Certificate
$$\left|\int_{a}^{b}\cos\left (nt-\frac{a}{t^{\alpha}}\right) dt\right|<\frac{2}{n}$$
$$\left|\int_{a}^{b}\sin\left (nt-\frac{a}{t^{\alpha}}\right) dt\right|<\frac{2}{n}$$
Proof: Make variable substitution
$$\omega=t-\frac{a}{nt^{\alpha}}$$
The
$$\frac{d\omega}{dt}=1+\frac{a\alpha}{nt^{\alpha+1}}>0,\, T\in [a,b]$$
$$\frac{d^{2} (t)}{d\omega^{2}}=\left (1+\frac{a\alpha}{nt^{\alpha+1}}\right) ^{-2}\frac{\alpha (\alpha+1) A}{nt^{\ alpha+2}}\frac{dt}{d\omega}>0$$
Then the second mean value theorem of integrals
\begin{align*}
\left|\int_{a}^{b}\cos\left (Nt-\frac{a}{t^{\alpha}}\right) Dt\right|&=\left|\int_{\omega (a)}^{\omega (b)}\ cos (n\omega) \frac{dt}{d\omega}d\omega\right|\\
&=\left (1+\frac{a\alpha}{nb^{\alpha+1}}\right) ^{-1}\left|\int_{\xi}^{b}\cos (N\omega) d\omega\right|\\
&\leq \left|\frac{\sin (n\xi)-\sin (N\omega (b))}{n}\right|\\
&\leq \frac{2}{n}
\end{align*}
The proof of the second inequality is exactly the same.

Application of Abelian distribution summation method (Iv.)