## 复分析作业

1.设$\displaystyle f_n\in O(D),f_n\rightrightarrows f$, $\displaystyle S_n=\cup_{k\geq n}f_k(D)$,令$\displaystyle S=\cap_{n=1}^\infty S_n$.问$S$的内部$\mathring{S}$和$f(D)$有什么关系?

2.Schwarz引理: $f:\triangle\to \triangle$解析, $f(0)=0$,则$|f'(0)|\leq 1$,并且$|f'(0)|= 1\Leftrightarrow f\in \mathrm{Aut} (\triangle)$.

(1)求出$\lambda$;

(2)若$f\in \mathscr{F}_{z_0}$,则$f\in \mathrm{Aut} (\triangle) \Leftrightarrow |f'(z_0)|=\lambda$.

3.设$f_n,g_n\in O(\triangle),n\geq 1$,设$|f_n(z)|\leq 2,|g_n(z)|\leq 2,|f_n(z)g_n(z)|\leq\frac1n,\forall z,\forall n$.设$f_n(0)=0,g_n(0)=1,\forall n$.设$\{z_n\}$是$\triangle$中子列,若满足$f_n(z_n)=1,g_n(z_n)=0,\forall n$,则$\lim\limits_{n\to\infty}|z_n|=1$.

4.对$r\in (0,1)$,记$A_r=\{z\in \mathbb{C}|r<|z|<1\}$

(1)对$\forall r\in (0,1)$,证明$A_r,\triangle^\ast,\mathbb{C}^\ast$不共形等价;

(2)设$r_1,r_2\in (0,1)$,若$A_{r_1}\cong A_{r_2}$,则$r_1=r_2$;

(3)计算$A_r$的自同构群$\mathrm{Aut}(A_r)$.

5.证明$\mathbb{C}^\ast,\triangle^\ast,A_r$同胚.

Let $q>0$ such that $[q[qn]]+1=[q^2n],n=1,2,\cdots$,solve $q$.

(1)计算$S^n(r)$上诱导度量$i^\ast g_0$在球极投影坐标下的表达式;

(2)计算$(S^n(r),i^\ast g_0)$的在一组正交标架下的联络形式;

(3)证明$(S^n(r),i^\ast g_0)$是常曲率黎曼流形.

(1) $f_{ij}=f_{ji}$,其中$i,j=1,2,\ldots,m$;

(2)$f_{ijk}-f_{ikj}=f_lR_{ijk}^l$,其中$i,j,k=1,2,\ldots,m$.

## 两个奇怪的积分

Evaluate integral
$$\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}$$
Well,I think we have
$$\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}=\frac{\pi}{e}$$

and

$$\int_{0}^{1}{x^{x}(1-x)^{1-x}\sin{\pi x}dx}=\frac{e\pi}{24}$$

With such nice result of these integral,why isn't worth to evaluate it?

I found a solution about the second one,but I wonder it will work for the first one
Note
$$S=\int_{0}^{1}{\sin{\pi x}x^{x}(1-x)^{1-x}dx}-\int_{0}^{1}{(1-x)e^{(i\pi+\ln{x}-\ln{(1-x)})x}dx}$$
Let $t=\ln{x}-\ln{(1-x)}$,$x=\frac{e^{t}}{1+e^{t}}$
Thus
\begin{align}S&=\int_{-\infty}^{+\infty}{\frac{1}{e^{t}+1}e^{(i\pi+t)\frac{e^{t}}{1+e^t}}\frac{e^{t}}{(1+e^{t})^{2}}dt}\\ &=\int_{-\infty+i\pi}^{-\infty-i\pi}{e^{\frac{te^{t}}{e^{t}-1} } \frac{e^{t}}{(e^{t}-1)^{3}}dt}\end{align}
Due to
$$f(z)=e^{\frac{te^{t}}{e^{t}-1} } \frac{e^{t}}{(e^{t}-1)^{3}},\qquad D=\{Z\in C|-\pi\leq Im(z) \leq \pi\}$$
Therefore
$res(f,0)=-\frac{e}{24}$when $z=0$
with $\zeta_{R}=\gamma_{R}+o_{R}+\tau_{R}$
$$\oint_{\zeta_{R}}{f(z)dz}=-2\pi i\cdot res(f,0)=\frac{2i\pi e}{24}$$
because
$$\{z_{n}\}\subset D,\qquad |z_{n}|\rightarrow\infty$$
Therefore
$$2S=2\lim_{R\rightarrow \infty}\int_{\gamma_{R}}{f(z)dz}$$
gives
$$\int_{0}^{1}{\sin{\pi x}x^{x}(1-x)^{1-x}dx}=Im(S)=\frac{e\pi}{24}$$

My friend tian_275461 told me he use a simliar method to deal with the first one to obtain the result $\frac{\pi}{e}$,but I am not figure it out.

Exactly the same method works for the other case.
$$\int_0^1 x^{-x} (1-x)^{x-1}\sin{\pi x} dx = \mathrm{Im}\left[\int_0^1 \frac{e^{(i\pi+\ln(1-x)-\ln x)x}}{1-x}dx\right]$$
Write $t=\ln((1-x)/x)$ and $z=t+i\pi$ as you did above to get
$$S = \int_0^1 \frac{e^{(i\pi+\ln(1-x)-\ln x)x}}{1-x}dx=\int_{-\infty+i\pi}^{\infty+i\pi} \frac{e^{\frac{z}{1-e^z}}}{1-e^z}dz$$

Then with $$f(z)=\frac{e^{\frac{z}{1-e^z}}}{1-e^z}$$
the only pole is at $z=0$, $res(f,0)=-\frac{1}{e}$ and in the limit $2S = \oint f(z)dz=-2\pi i \cdot res(f,0) = 2\pi i/e$ and your answer follows.

This one can be done with "residue at infinity" calculation. This method is shown in the Example VI of http://en.wikipedia.org/wiki/Methods_of_contour_integration

First, we use $z^z = \exp ( z \log z )$ where $\log z$ is defined for $-\pi\leq \arg z < \pi$.

For $(1-z)^{1-z} = \exp ( (1-z)\log (1-z) )$, we use $\log (1-z)$ defined for $0\leq \arg(1-z) <2\pi$.

Then, let  $f(z)= \exp( i\pi z + z \log z + (1-z) \log (1-z) )$.

As shown in the Ex VI in the wikipedia link, we can prove that $f$ is continuous on $(-\infty, 0)$ and $(1,\infty)$, so that the cut of $f(z)$ is $[0,1]$.

We use the contour: (consisted of upper segment: slightly above $[0,1]$, lower segment: slightly below $[0,1]$, circle of small radius enclosing $0$, and circle of small radius enclosing $1$, that looks like a dumbbell having knobs at $0$ and $1$, can someone edit this and include a picture of it please? In fact, this is also the same contour as in Ex VI, with different endpoints.)

On the upper segment, the function $f$ gives, for $0\leq r \leq 1$,
$$\exp(i\pi r) r^r (1-r)^{1-r} \exp( (1-r) 2\pi i ).$$

On the lower segment, the function $f$ gives, for $0\leq r \leq 1$,
$$\exp(i\pi r) r^r(1-r)^{1-r}.$$

Since the functions are bounded, the integrals over circles vanishes when the radius tend to zero.

Thus, the integral of $f(z)$ over the contour, is the integral over the upper and lower segments, which contribute to

$$\int_0^1 \exp(i\pi r) r^r (1-r)^{1-r} dr - \int_0^1 \exp(-i\pi r) r^r(1-r)^{1-r} dr$$

which is
$$2i \int_0^1 \sin(\pi r) r^r (1-r)^{1-r} dr.$$

By the Cauchy residue theorem, the integral over the contour is
$$-2\pi i \textrm{Res}_{z=\infty} f(z) = 2\pi i \textrm{Res}_{z=0} \frac{1}{z^2} f(\frac 1 z).$$

From a long and tedious calculation of residue, it turns out that the value on the right is
$$2i \frac{\pi e}{24}.$$
Then we have the result:
$$\int_0^1 \sin(\pi r) r^r (1-r)^{1-r} dr = \frac{\pi e}{24}.$$

## i的i次方等于多少?

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$，这是我们从初中就开始熟悉的概念。

$i= (-1)^{1/2} = ((-1)(-1)(-1))^{1/2} =(-1)^{1/2}(-1)^{1/2}(-1)^{1/2}=iii=-i$，

$-1=(-1)^1 = ((-1)^2)^{1/2}=1^{1/2}=1$。

$i^i= \exp(i Ln i) = \exp(i (i\pi/2 + i2k\pi)) = \exp(-\pi/2 - 2k\pi)=\exp(-\pi/2) \exp(-2k\pi)$

$(-2)^{1/3}, \hspace{5 mm}, (-1)^i$

（把你们的解答放在评论里，我的答案后天附在下面）

【答案】

$(-2)^{1/3} = \exp(Ln(-2)/3) = \{-\sqrt[3]{2}, \;\;\; \sqrt[3]{2}(1/2 -i\sqrt{3}/2), \;\;\; \sqrt[3]{2}(1/2 + i\sqrt{3}/2) \}$

$(-1)^i = \exp(iLn(-1))= e^{-\pi+2k\pi} = 0.043213918 \times 535.4916555^k$

$x_0 = -\sqrt[5]{2}$

$x_1 = -\sqrt[5]{2}((\sqrt{5}-1)/4+ i \sqrt{10+2\sqrt{5}}/4), \;\;\; x_2 =-\sqrt[5]{2}(-(\sqrt{5}+1)/4 + i \sqrt{10-2\sqrt{5}}/4)$

$x_3 = -\sqrt[5]{2}(-(\sqrt{5}+1)/4 - i \sqrt{10-2\sqrt{5}}/4),\;\;\; x_4 = -\sqrt[5]{2}((\sqrt{5}-1)/4- i \sqrt{10+2\sqrt{5}}/4),$

$x_n = -\sqrt[5]{2}(\cos(2n\pi/5) +i\sin(2n\pi/5)), \;\;\; n=0,1,2,3,4$

$\cos(2\pi/5)=(\sqrt{5}-1)/4= 0.309016994...$，

$\sin(2\pi/5)=\sqrt{10+2\sqrt{5}}/4= 0.951056516...$，

$\cos(\pi/5)=(\sqrt{5}+1)/4= 0.809016994…$

$\sin(\pi/5)=\sqrt{10-2\sqrt{5}}/4= 0.587785252...$

## 涉及到一个有意思的方程的积分

$$f(z) = {1\over\log{(i(1-e^{i2z}))}},$$
the logarithm being the principal branch. As can be deduced from the comments, the original integral is equal to
$$\int_0^\pi \operatorname{Re}{f(x)}\,dx.$$
Next consider, for $R>\epsilon > 0$, the following contour:

It is straightforward to check that for each fixed $\epsilon$ and $R$, the function $f$ is analytic on an open set containing this contour (just consider where $i(1-e^{i2z})\leq0$; this can only happen when $\operatorname{Im}{z}<0$ or when $\operatorname{Im}{z} = 0$ and $\operatorname{Re}{z}$ is an integer multiple of $\pi$). It then follows from Cauchy's theorem that $f$ integrates to zero over it. First let's see that the integrals over the quarter-circular portions of the contour vanish in the limit $\epsilon \to 0$. I'll look at the quarter circle near $\pi$ (near the bottom right corner of the contour) but the one near zero is similar, if not easier. Writing $i(1-e^{i2z}) = i(1-e^{i2(z-\pi)})$, it is clear that $i(1-e^{i2z}) = O(z-\pi)$ as $z\to\pi$. It follows from this that $f(z) = O(1/\log{|z-\pi|})$ as $z\to0$, and therefore that the integral of $f$ over the bottom right quarter circle is $O(\epsilon/\log{\epsilon})$ as $\epsilon \to0$, hence it vanishes in the limit as claimed.

Thus for fixed $R$, we can let $\epsilon \to 0$ to see that $f$ integrates to zero over the rectangle with corners $0,\pi, \pi +iR,$ and $iR$. Now the vertical sides of this rectangle give the contribution
\begin{align*}-\int_0^R f(iy)\,idy + \int_0^R f(iy+\pi)\,idy = 0,\end{align*}
since $f(iy) = f(iy+\pi)$. It follows at once that for each $R$, we can set the contribution from the horizontal sides equal to zero, giving
\begin{align*}\int_0^{\pi} f(x)\,dx - \int_0^\pi f(x+iR)\,dx=0, \qquad R>0. \tag{1}\end{align*}
(Note that the above equation implies that the integral in $x$ of $f(x+iR)$ over the interval $[0,\pi]$ is constant as a function of $R$; another way to evaluate the integral is to prove this directly, which I'll add below in a moment.) Now $f(x+iR) \to 1/\log{i} = 2/\pi$ as $R\to\infty$, uniformly for $x\in[0,\pi]$. Thus
\begin{align*}\lim_{R\to\infty} \int_0^\pi f(x+iR)\,dx = \pi\cdot {2\over \pi} = 2,\end{align*}
and it follows from $(1)$ that the original integral is equal to $2$ as well.
(By the way, the above idea is basically a replication of the technique I used here and here, and which I originally got from Ahlfors' book on complex analysis.)