poorly sized square brackets due to equation over two lines -- how to equate their sizes?

Question

Consider the following:

\documentclass[12pt,a4paper,oneside]{book}
\usepackage{amsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs,amsmath}
\begin{document}
\begin{align*}
M_3&=\frac{(\mu(k)+\rho(k)+\xi(k))\left(\sigma_1^2(k)+2\int_M D_1^2(k,y)\nu(dy)\right)}{2M_1}S_k^{*^2}\\[1ex]    &+\frac{(\mu(k)+\alpha(k)+\theta(k)+\gamma(k))\left(\sigma_2^2(k)+2\int_M D_2^2(k,y)\nu(dy)\right)}{2M_2}I_k^{*^2}\\[1ex]
&+\frac{2\mu(k)+\rho(k)+\xi(k)+\alpha(k)+\theta(k)+\gamma(k)}{2\beta(k)(1-\rho(k))(1-\xi(k))}\left[\sigma_2^2(k)\right.\\
&\left.+2\int_M (D_2(k,y)-\log(1+D_2(k,y))\nu(dy)) \right]I_k^*
\end{align*}
\end{document}

How to i get the \left[ -- the "big" square bracket in line 3 -- to match the \right] big bracket in line 4?

Answer 1

To preserve the pairwise contents of the numerator terms of the three additive components of M_3, I would like to suggest you employ the \splitdfrac macro of the mathtools package for the numerator on line 3.

A separate issue: I think that the terms S_k^{*^2} and I_k^{*^2} in lines 1 and 2 are wrong, as the exponent 2 is too small. I believe {S_k^*}^2 and {I_k^*}^2 is better, as the exponent 2 will be rendered in \scriptstyle math mode instead of in \scriptscriptstyle math mode.

Note that there are no instances of \left and \right anywhere. That's quite deliberate.

\documentclass[12pt,a4paper,oneside]{book}
\usepackage{%amsfonts, %% amsfonts is loaded automatically by amssymb
      graphicx, verbatim,
      mathtools, % for '\splitdfrac' macro    
      amssymb, amsthm, mathrsfs,
      %amsmath %% amsmath is loaded automatically by mathtools
      }
\begin{document}
\begin{align*}
M_3
&=\frac{\bigl[\mu(k)+\rho(k)+\xi(k)\bigr]
        \bigl[\sigma_1^2(k)+2\int_M D_1^2(k,y)\nu(dy)\bigr]}{%
        2M_1} \, {S_k^*}^2 \\[1ex]
&\quad+
  \frac{\bigl[\mu(k)+\alpha(k)+\theta(k)+\gamma(k)\bigr]
        \bigl[\sigma_2^2(k)+2\int_M D_2^2(k,y)\nu(dy)\bigr]}{%
        2M_2} \, {I_k^*}^2 \\[1ex]
&\quad+
  \frac{\splitdfrac{\bigl[2\mu(k)+\rho(k)+\xi(k)
            +\alpha(k)+\theta(k)+\gamma(k)\bigr]}
        {\cdot\bigl[\sigma_2^2(k) +2\int_M
            \bigl(D_2(k,y)-\log(1+D_2(k,y))\nu(dy) \bigr) \bigr]}}%
     {2\beta(k)(1-\rho(k))(1-\xi(k))} \, I_k^* \\
\end{align*}
\end{document}

Answer 2

Mico suggestion is the good one, but if you still want to use your coding as is, then try with the below:

Using left and right

\documentclass[12pt,a4paper,oneside]{book}
\usepackage{amsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs,amsmath}
\begin{document}
\begin{align*}
M_3&=\frac{(\mu(k)+\rho(k)+\xi(k))\left(\sigma_1^2(k)+2\int_M D_1^2(k,y)\nu(dy)\right)}{2M_1}S_k^{*^2}\\[1ex]
&\quad +\frac{(\mu(k)+\alpha(k)+\theta(k)+\gamma(k))\left(\sigma_2^2(k)+2\int_M D_2^2(k,y)\nu(dy)\right)}{2M_2}I_k^{*^2}\\[1ex]
&\quad+\frac{2\mu(k)+\rho(k)+\xi(k)+\alpha(k)+\theta(k)+\gamma(k)}{2\beta(k)(1-\rho(k))(1-\xi(k))}\left[\vphantom{+2\int_M (D_2(k,y)-\log(1+D_2(k,y))\nu(dy))}\sigma_2^2(k)\right.\\
&\quad\left.+2\int_M (D_2(k,y)-\log(1+D_2(k,y))\nu(dy)) \right]I_k^*
\end{align*}

Using biggl and biggr

\begin{align*}
M_3&=\frac{(\mu(k)+\rho(k)+\xi(k))\left(\sigma_1^2(k)+2\int_M D_1^2(k,y)\nu(dy)\right)}{2M_1}S_k^{*^2}\\[1ex]
&\quad +\frac{(\mu(k)+\alpha(k)+\theta(k)+\gamma(k))\left(\sigma_2^2(k)+2\int_M D_2^2(k,y)\nu(dy)\right)}{2M_2}I_k^{*^2}\\[1ex]
&\quad+\frac{2\mu(k)+\rho(k)+\xi(k)+\alpha(k)+\theta(k)+\gamma(k)}{2\beta(k)(1-\rho(k))(1-\xi(k))}\biggl[\sigma_2^2(k)\biggr.\\
&\quad\biggl.+2\int_M (D_2(k,y)-\log(1+D_2(k,y))\nu(dy)) \biggr]I_k^*
\end{align*}