如何对LaTeX⽂档中的所作的修改进⾏⾼亮显⽰
\[\int_0^1 \dfrac{dx}{\sqrt{1-x} \sqrt[4]{2x-x^2 \sqrt{3}}} = \dfrac{2 \sqrt{2} \pi }{3 \sqrt[8]{3}}\]
For $\alpha, \beta, \gamma \in (0,1)$ satisfying $\alpha+\beta+\gamma = 1$ and
$\mu \in \mathbb{C} \setminus [1,\infty)$, define
$$
F_{\alpha\beta}(\mu) = \int_0^1\frac{dx}{x^\alpha(1-x)^\beta(1-\mu x)^\gamma}
\quad\text{ and }\quad
\Delta = \frac{\Gamma(1-\alpha)\Gamma(1-\beta)}{\Gamma(1+\gamma)}
$$
When $|\mu| < 1$, we can rewrite the integral $F_{\alpha\beta}(\mu)$ as
$$\begin{align}
F_{\alpha\beta}(\mu)
= & \int_0^1 \frac{1}{x^\alpha(1-x)^{\beta}}\left(\sum_{n=0}^{\infty}\frac{(\gamma)_n}{n!}\mu^n x^n\right) dx
= \sum_{n=0}^{\infty}\frac{(\gamma)_n}{n!}\frac{\Gamma(n+1-\alpha)\Gamma(1-\beta)}{\Gamma(n+1+\gamma)}\mu^n\\
= & \Delta\sum_{n=0}^{\infty}\frac{(\gamma)_n (1-\alpha)_n}{n!(\gamma+1)_n}\mu^n
= \Delta\gamma \sum_{n=0}^{\infty}\frac{(1-\alpha)_n}{n!(\gamma+n)}\mu^n
\end{align}$$
This implies
$$\mu^{-\gamma} \left(\mu\frac{\partial}{\partial \mu}\right) \mu^{\gamma} F_{\alpha\beta}(\mu) =
\Delta\gamma \sum_{n=0}^{\infty}\frac{(1-\alpha)_n}{n!}\mu^n
= \Delta\gamma\frac{1}{(1-\mu)^{1-\alpha}}
$$
and hence
$$F_{\alpha\beta}(\mu)
= \Delta\gamma \mu^{-\gamma} \int_0^\mu \frac{\nu^{\gamma-1}d\nu}{(1-\nu)^{1-\alpha}}
= \Delta\gamma \int_0^1 \frac{t^{\gamma-1} dt}{(1-\mu t)^{1-\alpha}}
= \Delta \int_0^1 \frac{dt}{(1 - \mu t^{1/\gamma})^{1-\alpha}}$$
Notice if we substitute $x$ by $y = 1-x$, we have
$$F_{\alpha\beta}(\mu) = \int_0^1 \frac{dy}{y^\beta(1-y)^\alpha(1-\mu - \mu y)^{\gamma}}
= \frac{1}{(1-\mu)^\gamma} F_{\beta\alpha}(-\frac{\mu}{1-\mu})$$
Combine these two representations of $F_{\alpha\beta}(\mu)$ and let $\omega = \left(\frac{\mu}{1-\mu}\right)^{\gamma}$, we obtain
$$F_{\alpha\beta}(\mu) = \frac{\Delta}{(1-\mu)^{\gamma}}\int_0^1 \frac{dt}{( 1 + \omega^{1/\gamma} t^{1/\gamma})^{1-\beta}} = \frac{\Delta}{\mu^\gamma}\int_0^ Let $(\alpha,\beta,\gamma) = (\frac14,\frac12,\frac14)$ and $\mu = \frac{\sqrt{3}}{2}$, the identity we want to check becomes
$$\frac{\Gamma(\frac34)\Gamma(\frac12)}{\Gamma(\frac54) (\sqrt{3})^{1/4}}\int_0^\omega \frac{dt}{\sqrt{1+t^4}} \stackrel{?}{=} \frac{2\sqrt{2}}{3\sqrt[8]{3}} \pi\ta Let $K(m)$ be the complete elliptic integral of the first kind associated with modulus $m$. i.e.
$$K(m) = \int_0^1 \frac{dx}{\sqrt{(1-x^2)(1-mx^2)}}$$
It is known that $\displaystyle K(\frac12) = \frac{8\pi^{3/2}}{\Gamma(-\frac14)^2}$. In term of $K(\frac12)$, it is easy to check $(*1)$ is equivalent to
$$\int_0^\omega \frac{dt}{\sqrt{1+t^4}} \stackrel{?}{=} \frac23 K(\frac12)\tag{*2}$$
To see whether this is the case, let $\varphi(u)$ be the inverse function of above integral.
More precisely, define $\varphi(u)$ by following relation:
$$u = \int_0^{\varphi(u)} \frac{dt}{\sqrt{1+t^4}}$$
Let $\psi(u)$ be $\frac{1}{\sqrt{2}}(\varphi(u) + \varphi(u)^{-1})$. It is easy to check/verify
$$
$$
\varphi'(u)^2 = 1 + \varphi(u)^4
\implies
\psi'(u)^2 = 4 (1 - \psi(u)^2)(1 - \frac12 \psi(u)^2)
$$
Compare the ODE of $\psi(u)$ with that of a Jacobi elliptic functions with modulus $m = \frac12$, we find
系统变量path修改了怎么恢复$$\psi(u) = \text{sn}(2u + \text{constant} | \frac12 )\tag{*3}$$
Since we are going to deal with elliptic functions/integrals with $m = \frac12$ only,
we will simplify our notations and drop all reference to modulus, i.e
$\text{sn}(u)$ now means $\text{sn}(u|m=\frac12)$ and $K$ means $K(m = \frac12)$.
Over the complex plane, it is known that $\text{sn}(u)$ is doubly periodic with
fundamental period $4 K$ and $2i K$. It has two poles at $i K$ and $(2 + i)K$ in the fundamental domain.
When $u = 0$, we want $\varphi(u) = 0$ and hence $\psi(u) = \infty$. So the constant
in $(*3)$ has to be one of the pole. For small and positive $u$, we want $\varphi(u)$ and hence $\psi(u)$ to be positive. This fixes the constant to $i K$. i.e. $$\psi(u) = \text{sn}(2u + iK )$$
and the condition $(*2)$ becomes whether following equality is true or not.
$$\frac{1}{\sqrt{2}} (\omega + \omega^{-1}) \stackrel{?}{=} \text{sn}( \frac43 K + i K)\tag{*4}$$
Notice $ 3( \frac43 K + i K) = 4 K + 3 i K $ is a pole of $\text{sn}(u)$. if one repeat
apply the addition formula for $\text{sn}(u+v)$
$$\text{sn}(u+v) = \frac{\text{sn}(u)\text{cn}(v)\text{dn}(v)+\text{sn}(v)\text{cn}(u)\text{dn}(u)}{1-m\,\text{
sn}(u)^2 \text{sn}(v)^2}$$
One find in order for $\text{sn}(3u)$ to blow up, $\text{sn}(u)$ will be a root of
following polynomial equation:
$$3 m^2 s^8-4 m^2 s^6-4 m s^6+6 m s^4-1 = 0$$
Substitute $m = \frac12$ and $s = \frac{1}{\sqrt{2}}(t+\frac{1}{t})$ into this, the equation $\omega$ need to satisfy is given by:
$$(t^8 - 6 t^4 - 3)(3 t^8 + 6 t^4 - 1 ) = 0$$
One can check that $\omega = \sqrt[4]{\frac{\sqrt{3}}{2-\sqrt{3}}}$ is indeed a root of this polynomial. As a result, the original equality is valid.
论⽂投出之后，如果好消息，通常是minor/major revision难免; 很少有⽂不加点直接接受的.
修改的时候为了⽅便编辑和审稿⼈,⾼亮显⽰⽂档中改动的部分是专业的.
MS word的comment 审阅修订中"对⽐compare⽂档"功能是很⽅便的.
但是在LaTeX中如何实现? -- 实际上是对tex源代码的两个版本的对⽐,
如果安装了CTeX，只需要两点:
(1) 确保等⽂件所在的bin⽂件夹也被添加到系统环境变量path中(可能要⼿动⾃⼰添加，否则只能在该⽂件夹下使⽤
<命令⾏是郁闷的)
(2)再安装ActivePerl，安装的时候选择把ActivePerl添加到系统环境变量path中即可;
W windows下activeperl 5.16的下载地址，可以从这⾥进⼊: ：或直接⽤下⾯的下载链接:
命令⾏中输⼊:>> latexdiff-so old_ new_ >
然后⽤同样⽅法编译 即可
(⽐如LaTeX 1次> BibTeX ⼀次>> LaTeX 2次>> dvi2ps 1次>> ps2pdf 完成;)
尝试之后，发现对使⽤了gnuplot绘图的命令⽀持不好，如果有相应的改动，可能⽆法通过编译. 但其它的则鲜有例外.
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HEN  several authors are editing a paper, it is sometimes
necessary to quickly highlight what changes have been made in a
particular version of the paper compared to previous versions. Word processors have long had revision tracking features for this but some LaTeX authors are unaware that there is a handy utility called latexdiff that can also highlight changes between two revisions by comparing two LaTeX documents:
This page briefly describes this program. While LaTeX authors can use revision control software like software developers to track changes,there are cases where this is not convenient. For example, grad students may revise papers, get feedback from their advisor and then revise the paper again (I know I did). The advisor just wants to see quickly and with formatting what changes were made on the manus
cript without learning how to use revision control software. latexdiff provides a way to do this. It performs a “diff” (like the Unix utility) on the two documents and then creates a new LaTeX document where the changes are made clear.
Obviously you need to have two versions of your document to compare. latexdiff can be run with the following command:
latexdiff-so >
The file can be run through LaTeX to produce the version with revisions highlighted.
It has been pointed out to me that the file names should not have spaces or else you will have to enclose them in quotes.
Sometimes latexdiff produces wrong output that will give an error when you run LaTeX on the file. In the cases I have seen, you simply need to go to the place where the error occurs and make sure that the various commands that latexdiff adds are properly closed. For example, you may have to add some \DIFaddend to unclosed \DIFadd commands.

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