3

I am solving integrals using u-substitution. However, when I compile the code the box where it shows the substitution is not placed in the right place. And, when I continue to solve the problem another integral with a new boundary is typed inside the box.

\documentclass[12pt]{article}
\usepackage{authblk}
\usepackage{setspace}
\doublespacing% For double space
\usepackage{subeqnarray}
\usepackage{graphicx,epstopdf}
\usepackage[framed , numbered]{matlab-prettifier}% For MATLAB code
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{xcolor}
\usepackage{bigstrut}
\usepackage{tikz}

\renewcommand{\qedsymbol}{$\blacksquare$}


\usepackage{nccmath}% For adding Mathematical commands
\usepackage[english]{babel}
\usepackage{blindtext}
\usepackage[a4paper,margin=2.5cm]{geometry} % set page margins as needed
\newcommand{\diff}{\mathop{}!d}
\newcommand{\innerp}[2]{\left\langle #1 \vert #2 \right\rangle}
\usetikzlibrary{tikzmark,calc}
\begin{document}


\begin{proof}

$$ a_n\cos(nx)+b_n\sin(nx) $$
$$ = \Big[\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt)dt\Big]\cos(nx)+\Big[\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin(nt)dt\Big]\sin(nx) $$
$$ = \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\Big[\cos(nt)\cos(nx)+\sin(nt)\sin(nx)\Big]dt $$
$$ = \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos\big(n(t-x)\big)dt $$
$$ = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)dt+\frac{1}{\pi}\sum_{n=1}^{N}\int_{-\pi}^{\pi}f(t)\cos\big(n(t-x)\big)dt $$
$$ = \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\Big[\frac{1}{2}+\sum_{n=1}^{N}\cos\big(n(t-x)\big)\Big]dt $$
\begin{align}
 = \frac{1}{\pi}\displaystyle\int_{-\pi}^{\pi}f(t)D_N(t-x)dt 
\end{align}
\begin{tikzpicture}[overlay, remember picture]
\coordinate (x) at ($(pic cs:eqt)+(1,0)$);
\draw (x)--($(x)+(3,0)$);
\draw[red] (x) node[above right] {Substitution};
\draw ($(x)+(3,0)$) node[right,draw,fill=white!50] {% 
$\begin{aligned}
    u&=t-x,\
    du&= dt
\end{aligned}$};
\end{tikzpicture}
$$ = \frac{1}{\pi}\int_{-\pi-x}^{\pi-x}f(x+u)D_N(u)du = \frac{1}{\pi}\int_{-pi}^{pi}f(x+t)D_N(t)dt 
$$
$$ = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x+t)\frac{\sin\big((N+\frac{1}{2})t\big)}{\sin(\frac{1}{2}t)}dt $$


\end{proof}
\end{document}
Mico
  • 506,678
Riddle72
  • 107
  • 3

1 Answers1

8

I must confess that I don't see a need for the tikz machinery. I also wouldn't use $$ ... $$ in LaTeX documents. Please see Why is \[ ... \] preferable to $$ ... $$? for a longer discussion of this point.

Instead, I'd use an align* environment for the entire sequence of equations and align the individual equations on the & symbols. The "Substitution ..." interjection need not be encased in an \intertext directive.

enter image description here

\documentclass[12pt]{article}
%% Remark: I've streamlined your preamble as much as possible

\usepackage[nodisplayskipstretch]{setspace}
\doublespacing % For double space


\usepackage{amssymb,amsmath,amsthm}
\renewcommand{\qedsymbol}{$\blacksquare$}
\newcommand\intpp{\int_{-\pi}^{\pi}} % handy shortcut macro


\usepackage{xcolor}
\usepackage[english]{babel}
\usepackage[a4paper,margin=2.5cm]{geometry}


\begin{document}
\allowdisplaybreaks % allow page breaks in long displays
\begin{proof}

\begin{align}
a_n&\cos(nx)+b_n\sin(nx) \
&= \biggl[\frac{1}{\pi}\intpp f(t)\cos(nt),dt\biggr]\cos(nx)
  +\biggl[\frac{1}{\pi}\intpp f(t)\sin(nt),dt\biggr]\sin(nx) \
&= \frac{1}{\pi}\intpp f(t)\bigl[\cos(nt)\cos(nx)+\sin(nt)\sin(nx)\bigr]dt\
&= \frac{1}{\pi}\intpp f(t)\cos\bigl(n(t-x)\bigr),dt \
&= \frac{1}{2\pi}\intpp f(t),dt
  +\frac{1}{\pi}\sum_{n=1}^{N}\intpp f(t)\cos\bigl(n(t-x)\bigr),dt \
&= \frac{1}{\pi}\intpp f(t)
   \biggl[\frac{1}{2}+\sum_{n=1}^{N}\cos\bigl(n(t-x)\bigr)\biggr]dt \
&= \frac{1}{\pi}\intpp f(t)D_N(t-x),dt \
&  \qquad
   \textcolor{red}{\textup{Substitution}}
   \quad
   \boxed{\begin{aligned}
               u&=t-x,\
              du&=dt
          \end{aligned}} \
&= \frac{1}{\pi}\int_{-\pi-x}^{\pi-x}f(x+u)D_N(u),du \
&= \frac{1}{\pi}\intpp f(x+t)D_N(t),dt \
&= \frac{1}{2\pi}\intpp f(x+t),
   \frac{\sin\bigl((N+\frac{1}{2})t\bigr)}{\sin\bigl(\frac{1}{2}t\bigr)},dt 
   \qquad\qedhere
\end{align}
\end{proof}
\end{document}
Mico
  • 506,678