How to write this equation in publication?

You should be using split not align inside equation. Introducing new variables for large repeating expressions and shifting your indexing by one will reduce the size of the equation. Finally I introduce an eqbreak command to shift a split expression:

\documentclass[twocolumn]{article}
\usepackage{amsmath, amsfonts, amssymb, textcomp}

\newcommand{\eqbreak}[1][2]{\\&\hskip#1em}

\begin{document}   

\begin{equation}
  \begin{split}  
    X(m+1)
    &= \frac{b-a}N \sum_{k=0}^{N-1} e^{-i2\pi km/N}\,  x(a_k) \\
    &= \frac{b-a}N \sum_{k=0}^{N-1} e^{-i2\pi a_km/(b-a)} \eqbreak[6]
    \times x(a_k)\,e^{i2\pi a m/(b-a)} \\
    &\xrightarrow{N\to\infty} e^{i2\pi a m/(b-a)} \,Qx\Bigl(\frac
    m{b-a}\Bigr),
  \end{split}
\end{equation}
where \( a_k = a + (b-a)k/N \).
\end{document}

Just for fun!

\documentclass{article}
\usepackage[a4paper,margin=2cm,twocolumn]{geometry}
\usepackage{mathtools}
\begin{document}
\begin{equation}
\begin{split}  
X(m) 
&=
\!
\begin{multlined}[t]
    \frac{b-a}{N} \sum_{k=1}^{N} \Bigg[  x\left(a + \frac{k-1}{N}(b-a)\right)\\
    \times e^{-\frac{2\pi i (k-1)(m-1)}{N}} \Bigg]
\end{multlined}\\ 
&=
\!
\begin{multlined}[t]
    \frac{b-a}{N} \sum_{k=1}^{N}\Bigg[ x\left(a + \frac{k-1}{N}(b-a)\right) \\
    \times e^{2\pi i a \frac{m-1}{b-a}}\\
    \times e^{-2\pi i \left(a + \frac{k-1}{N}(b-a)\right)\times \frac{m-1}{b-a}}\Bigg]
\end{multlined}\\
& \xrightarrow{ N\to\infty }
\!
\begin{multlined}[t] 
    e^{2\pi i a \frac{m-1}{b-a}} \\
    \times \int_{a}^{b} e^{2\pi i a \frac{m-1}{b-a}} Qx\left(\frac{m-1}{b-a}\right).
\end{multlined}
\end{split}
\end{equation}
\end{document}

I would propose you split the first two lines at a reasonably natural point. I would use explicit sizing instructions for the parentheses, and use \exp(...) expressions.

\documentclass[twocolumn]{article}
\usepackage{amsmath, amsfonts, amssymb, textcomp}
\begin{document}
\begin{align}
&X(m) \notag\\
&= \frac{b-a}{N} \sum_{k=1}^{N} \exp\bigl(-i2\pi (k-1)(m-1)/N\bigr) \notag\\
&\qquad  \times x\Bigl[a + (b-a)\frac{k-1}{N}\Bigr] \\
&= \frac{b-a}{N} \sum_{k=1}^{N} \exp\Bigl[-i2\pi \Bigl(a + (b-a)\frac{k-1}{N}\Bigr)
 \frac{m-1}{b-a}\Bigr] \notag \\
&\qquad  \times x\Bigl(a + (b-a)\frac{k-1}{N}\Bigr) 
        \exp\Bigl(i2\pi a \frac{m-1}{b-a}\Bigr) \\
& \xrightarrow{ N\to\infty } \exp\Bigl(i2\pi a \frac{m-1}{b-a}\Bigr) 
Qx\Bigl(\frac{m-1}{b-a}\Bigr)\,.
\end{align}
\end{document}