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\textbf{Notebook:A set of questions and solutions.}

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1. Use mathematical analysis to show that:

e and $ \pi $ are irrational numbers;

Proof.

For a), For any positive integer $N$ we have
$$ \begin{aligned} e &=\sum_{n=0}^{\infty} \frac{1}{n !} \\ e_{N} &=\sum_{n=0}^{N} \frac{1}{n !}=\frac{p_{N}}{q_{N}}< e \end{aligned} $$$$ e-e_{N}<\frac{e}{(N+1) !}, \quad q_{N} \leqslant \frac{1}{N !} $$
Assume that $e$ is rational, then we can write $e=\frac{p}{q}$ for some $p, q \in \mathbb{N}$ with $q>0 .$ Choose $N$ sufficiently large so that $N>p,$ then we get $e<\frac{N+1}{q}$ thus $\frac{1}{q}>\frac{e}{N+1} .$ We have
$$ e-e_{N}=\frac{p}{q}-\frac{p_{N}}{q_{N}}=\frac{p q_{N}-q p_{N}}{q q_{N}} \geqslant \frac{1}{q q_{N}} \geqslant \frac{1}{q N !} \geqslant \frac{e}{(N+1) !} $$
which is a contradiction. So $e$ must be irrational.

2.$$ \lim_{n \rightarrow \infty}\sum_{k=1}^{n}\sum_{j=1}^{n} \sum_{m=1}^{n} \frac{\arctan \frac{j}{n}\arctan \frac{k^2}{n^2}}{kn^5+kj^2 n^3}m^3(\ln m- \ln n)^3(\ln(n+j)-\ln n).$$

Solution

$$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \sum_{j=1}^{n} \sum_{m=1}^{n} \frac{\arctan \frac{j}{n} \arctan \frac{k^{2}}{n^{2}} m^{3}(\ln m-\ln n)^{3}[\ln (j+n)-\ln n]}{k n^{5}+k j^{2} n^{3}} $$

$$ =\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \sum_{k=1}^{n} \sum_{j=1}^{n} \sum_{m=1}^{n} \frac{\arctan \frac{j}{n} \arctan \frac{k^{2}}{n^{2}}\left(\frac{m}{n}\right)^{3}\left(\ln \frac{m}{n}\right)^{3}\left[\ln \left(1+\frac{j}{n}\right)\right]}{\frac{k}{n}\left(1+\frac{j^{2}}{n^{2}}\right)} $$

$$ =\int_{0}^{1} \frac{\arctan z^{2}}{z} d z \int_{0}^{1}(y \ln y)^{3} d y \int_{0}^{1} \frac{\arctan x \ln (1+x)}{1+x^{2}} d x $$

$$ =\frac{G}{2} \int_{0}^{1}(y \ln y)^{3} d y \int_{0}^{1} \frac{\arctan x \ln (1+x)}{1+x^{2}} d x $$

$$ =\frac{G}{2}\cdot \frac{-3}{128} \int_{0}^{1} \frac{\arctan x \ln (1+x)}{1+x^{2}} d x $$

let
$$I_3= \int_{0}^{1} \frac{\arctan x \ln (1+x)}{1+x^{2}} d x $$

$$ I_{3}=\int_{0}^{\frac{\pi}{4}} x \ln (1+\tan x) d x=\int_{0}^{\frac{\pi}{4}} x \ln (\cos x+\sin x) d x -\int_{0}^{\frac{\pi}{4}} x \ln (\cos x) d x$$

$$ =\int_{0}^{\frac{\pi}{4}} x \ln \left(\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\right) d x-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(\frac{\pi}{2}-x\right) \ln (\sin x) d x$$

$$ =\frac{\pi^{2} \ln 2}{64} x-\frac{3 \pi}{4} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ln \sin x d x $$

$$ =\frac{\pi^{2} \ln 2}{64}+2 \psi_{1}-\frac{3 \pi}{4} \psi_{2}. $$
Try to compute $\psi_1,\psi_2$,

$$ \ln \sin x=-\ln 2-\sum_{n=1}^{\infty} \frac{\cos (2 n x)}{n}, 0< x<\pi $$

$$ \psi_{1}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} x \ln \sin x d x=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} x\left[-\ln 2-\sum_{n=1}^{\infty} \frac{\cos (2 n x)}{n}\right] d x$$

$$ =-\frac{3 \pi^{2} \ln 2}{32}-\sum_{n=1}^{\infty} \frac{n \pi \sin (n \pi)+\cos (n \pi)-\left(\frac{n \pi}{2}\right) \sin \left(\frac{n \pi}{2}\right)-\cos \left(\frac{n \pi}{2}\right)}{4 n^{3}} $$
$$ =-\frac{3 \pi^{2} \ln 2}{32}-\frac{1}{4} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{3}}+\frac{\pi}{8} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2 n-1)^{2}}-\frac{1}{32} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{3}} $$

$$ =\frac{21}{128} \zeta(3)-\frac{\pi}{8}-\frac{\pi^{2} \ln 2}{64}. $$

$$ \psi_{2}=\frac{G}{2}-\frac{\pi \ln 2}{4} $$ the answer is
$$ -\frac{3 G}{256}\left(\frac{21}{64} \zeta(3)-\frac{\pi}{8} G+\frac{\pi^{2} \ln 2}{64}\right). $$

3.
Find an explicit conformal transformation of an open set $U=\{|z|>$
1}$\backslash(-\infty,-1]$ to the unit disc.

Proof.
We will construct the mapping in several steps(*≧ω≦)

1. The mapping $w=-\frac{1}{z}$ maps $U$ onto $D_{1} \backslash[0,1]$

2. $w^{\prime}=\sqrt{w}$ maps $D_{1} \backslash[0,1]$ onto $\{z|| z \mid<1, \operatorname{Im}(z)>0\}$

3. The mapping $w^{\prime \prime}=\frac{w^{\prime}-1}{w^{\prime}+1}$ maps $\{z|| z \mid<1, \operatorname{Im}(z)>0\}$ onto
   $\{z \mid \operatorname{Re}(z)>0, \operatorname{Im}(z)>0\}$

4. $u=w^{\prime \prime 2} \operatorname{maps}\{z \mid \operatorname{Re}(z)>0, \operatorname{Im}(z)>0\}$ onto $\{z \mid \operatorname{Im}(z)>0\}$

5. $v=\frac{u-i}{u+i}$ maps the upper half plane onto the unit disc.

6. Composition of the mappings 1)-4) is $u=\left(\frac{\sqrt{-\frac{1}{2}}-1}{\sqrt{-\frac{1}{z}}+1}\right)^{2} \cdot$ Its composition with 5 ) gives the desired mapping.