對齊問題？

Question

我試圖在以下程式碼的每一行開頭對齊 2 個等號和減號，但無論我做什麼，它都不起作用。

$C_{n}$  = $ \dfrac{1}{4\pi{i}} $ $ \displaystyle\int^\pi_{-\pi} x^2e^{ix({1-n})}\ dx $ $ - \dfrac{1}{4\pi{i}} $ $ \displaystyle\int^\pi_{-\pi} x^2e^{{-ix}({1+n})}\ dx $ 

\bigskip $  = \dfrac{1}{4\pi i}\Bigg[ \bigg[\dfrac {x^2ie^{-i(n-1)x}}{n-1} \bigg]_{-\pi}^\pi -  \displaystyle\int^\pi_{-\pi} \dfrac{2xie^{-i(n-1)x}}{n-1} \ dx \Bigg] $  

\bigskip $  - \dfrac{1}{4\pi i}\Bigg[ \bigg[\dfrac {x^2ie^{-i(n+1)x}}{n+1} \bigg]_{-\pi}^\pi -  \displaystyle\int^\pi_{-\pi} \dfrac {2xie^{-i(n+1)x}} {n+1} \ dx \Bigg] $

Answer 1

可能有更好的方法來做到這一點，透過根據align環境完全重構您的答案（見下文），但這個答案對您最初的嘗試的「影響」最小。本質上，我\phantom在第二行和第三行的開頭添加了 a 。

\documentclass{letter}
\usepackage{amsmath}
\begin{document}
$C_{n}$  = $ \dfrac{1}{4\pi{i}} $ $ \displaystyle\int^\pi_{-\pi} x^2e^{ix({1-n})}\ dx $ $ - \dfrac{1}{4\pi{i}} $ $ \displaystyle\int^\pi_{-\pi} x^2e^{{-ix}({1+n})}\ dx $ 

\bigskip $\phantom{C_{n}}  = \dfrac{1}{4\pi i}\Bigg[ \bigg[\dfrac {x^2ie^{-i(n-1)x}}{n-1} \bigg]_{-\pi}^\pi -  \displaystyle\int^\pi_{-\pi} \dfrac{2xie^{-i(n-1)x}}{n-1} \ dx \Bigg] $  

\bigskip $\phantom{C_{n}}  - \dfrac{1}{4\pi i}\Bigg[ \bigg[\dfrac {x^2ie^{-i(n+1)x}}{n+1} \bigg]_{-\pi}^\pi -  \displaystyle\int^\pi_{-\pi} \dfrac {2xie^{-i(n+1)x}} {n+1} \ dx \Bigg] $
\end{document}

---

這是一種方法align：

\documentclass{letter}
\usepackage{amsmath}
\begin{document}
\begin{align}
C_{n} &=  \dfrac{1}{4\pi{i}}  \displaystyle\int^\pi_{-\pi} x^2e^{ix({1-n})}\ dx  - \dfrac{1}{4\pi{i}}  \displaystyle\int^\pi_{-\pi} x^2e^{{-ix}({1+n})}\ dx 
\\[2ex]
&= \dfrac{1}{4\pi i}\Bigg[ \bigg[\dfrac {x^2ie^{-i(n-1)x}}{n-1} \bigg]_{-\pi}^\pi -  \displaystyle\int^\pi_{-\pi} \dfrac{2xie^{-i(n-1)x}}{n-1} \ dx \Bigg]  
\\[2ex]
&- \dfrac{1}{4\pi i}\Bigg[ \bigg[\dfrac {x^2ie^{-i(n+1)x}}{n+1} \bigg]_{-\pi}^\pi -  \displaystyle\int^\pi_{-\pi} \dfrac {2xie^{-i(n+1)x}} {n+1} \ dx \Bigg]
\end{align}
\end{document}