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# 数学代写|实分析代写Real Analysis代考|Riemann’s Lemma

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## 数学代写|实分析代写Real Analysis代考|Riemann’s Lemma

We shall first show that
$$\int_a^{\pi / 2} \frac{\sin [(2 n+1) u]}{\sin u} F(x+2 u) d u, \quad 0<a<\pi / 2,$$
can be made arbitrarily close to zero by taking $n$ sufficiently large. To do this, we shall concentrate on a single interval $(a, b), 0<a<b \leq \pi / 2$, where $F(x+2 u) / \sin u$ is continuous and bounded. Since our whole interval from $a$ to $\pi / 2$ is made up of a finite number of such pieces, it is enough to show that the integral over each such piece can be forced to be arbitrarily close to zero.

Working with the integral over $[a, b]$, we do not change the value of the integral if we change the value of our function just at $x=a$ or $b$. It is convenient when working on this particular piece to redefine $F$, if necessary, so that $F(x+2 a)=F(x+2 a+0)$ (the limit from the right) and $F(x+2 b)=F(x+2 b-0)$ (the limit from the left) so that we can consider $F(x+2 u) / \sin u$ to be continuous over the closed interval $[a, b]$. Since we are now working over a closed interval, the restriction that $F$ be bounded is implied by the continuity (see Theorem 3.6). For simplicity of notation, we write $g(u)$ for $F(x+2 u) / \sin u$. The only restriction on $g$ is that it be continuous over $[a, b]$ where we mean continuous from the right at $a$, continuous from the left at $b$, and continuous in the usual sense at all points between $a$ and $b$. We also write $M$ in place of $2 n+1$. For this lemma, it is not necessary for this multiplier to be an odd integer.

Lemma 6.2 (Riemann’s Lemma). If $g(u)$ is a continuous function over the interval $[a, b], 0<a<b \leq \pi / 2$, then
$$\lim _{M \rightarrow \infty} \int_a^b \sin (M u) g(u) d u=0$$

## 数学代写|实分析代写Real Analysis代考|The Integral of the Dirichlet Kernel

Lemma 6.4 (Integral of Dirichlet Kernel). For any positive integer $n$,
$$\int_0^{\pi / 2} \frac{\sin [(2 n+1) u]}{\sin u} d u=\frac{\pi}{2}$$
Proof: From equation (6.11), we know that
$$\frac{\sin [(2 n+1) u]}{\sin u}=1+2 \cos (2 u)+2 \cos (4 u)+\cdots+2 \cos (2 n u) .$$
Substituting this into our integral and integrating each summand, we get exactly $\pi / 2$.

Bonnet’s Mean Value Theorem
The final lemma that we need is Bonnet’s form of the mean value theorem, a version that he discovered and proved in 1849 specifically to simplify the proof of Dirichlet’s theorem. As he pointed out, it also has many other applications. We shall postpone the proof of this lemma until the next section. Here, for the first time, we shall need to be very careful about exactly what we mean by an integral.

Lemma 6.5 (Bonnet’s Mean Value Theorem). Let $f$ be integrable and let $g$ be a nonnegative, increasing function on $[\alpha, \beta]$. There is at least one value $\zeta$ strictly between $\alpha$ and $\beta$ for which
$$\int_\alpha^\beta f(t) g(t) d t=g(\beta) \int_\zeta^\beta f(t) d t .$$
As an example, let $f(t)=\sin t$ and $g(t)=t^2$ on the interval $[0,2 \pi]$. This lemma promises us a number $\zeta$ for which
$$\int_0^{2 \pi} t^2 \sin t d t=4 \pi^2 \int_\zeta^{2 \pi} \sin t d t$$

## 数学代写|实分析代写Real Analysis代考|Riemann’s Lemma

$$\int_a^{\pi / 2} \frac{\sin [(2 n+1) u]}{\sin u} F(x+2 u) d u, \quad 0<a<\pi / 2$$

$$\lim _{M \rightarrow \infty} \int_a^b \sin (M u) g(u) d u=0$$

## 数学代写|实分析代写Real Analysis代考|The Integral of the Dirichlet Kernel

$$\int_0^{\pi / 2} \frac{\sin [(2 n+1) u]}{\sin u} d u=\frac{\pi}{2}$$

$$\frac{\sin [(2 n+1) u]}{\sin u}=1+2 \cos (2 u)+2 \cos (4 u)+\cdots+2 \cos (2 n u) .$$

Bonnet 的中值定理

$$\int_\alpha^\beta f(t) g(t) d t=g(\beta) \int_\zeta^\beta f(t) d t$$

$$\int_0^{2 \pi} t^2 \sin t d t=4 \pi^2 \int_\zeta^{2 \pi} \sin t d t$$

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