# Lower expression in mathmode

I am using the mtpro2 math font and I want to lower the expression inside the big parenthesis in the following equation.

\documentclass{article}
\usepackage{mtpro2}

\begin{document}
$y(x)=c_1y_1(x)\XL\int{\PARENS{\left( \frac{y_2}{y_1}\right)'(x)\XL\int{\frac{1}{\left[\left( \frac{y_2}{y_1}\right)'(x)\right]^2\cdot y_1^3(x)\cdot e^{\frac{a_2}{a_3}x} }dx}} dx}+c_2y_2(x)+c_3y_1(x)\ ,\ x\in I$
\end{document}


I tryied to use \raisebox but it didn't work

• \PARENS will center its contents with respect to the math axis, so the same problem appears whenever the contents is a fraction with different heights in the numerator and the denominator. – egreg Aug 18 '16 at 14:04
• You can add \vphantom{\left[\left( \frac{y_2}{y_1}\right)'(x)\right]^2\cdot y_1^3(x)\cdot e^{\frac{a_2}{a_3}x}} to the first argument of \frac. – Manuel Aug 18 '16 at 14:05

What \PARENS{<subformula>} does (apart from a different choice of parentheses) is

\left(
\vcenter{\hbox{$\displaystyle <subformula>$}}
\right)


This means that the output of something as easy as

$\frac{a}{b}\PARENS{\frac{a}{\dbinom{b}{c}}}$


is the completely wrong

Of course nobody would want to write that subformula, but it's a good example of what happens.

You get a better output with

\documentclass{article}
\usepackage[lite]{mtpro2}
\usepackage{amsmath}

\begin{document}

$y(x)=c_1y_1(x)\XL\int \Biggl( \left(\frac{y_2}{y_1}\right)^{\!\prime}(x) \XL\int\frac{1}{\Bigl[\left( \frac{y_2}{y_1}\right)'(x)\Bigr]^2} \cdot y_1^3(x)\cdot e^{\frac{a_2}{a_3}x}\,dx \Biggr) \,dx+c_2y_2(x)+c_3y_1(x)\ ,\ x\in I$

\end{document}


Alternatively, build the big parentheses separately:

\documentclass{article}
\usepackage[lite]{mtpro2}
\usepackage{amsmath}

\begin{document}

$y(x)=c_1y_1(x)\XL\int \LEFTRIGHT(.{\vphantom{\XL\int}}\kern-2\nulldelimiterspace \left(\frac{y_2}{y_1}\right)^{\!\prime}(x) \XL\int\frac{1}{\Bigl[\left( \frac{y_2}{y_1}\right)'(x)\Bigr]^2} \cdot y_1^3(x)\cdot e^{\frac{a_2}{a_3}x}\,dx \kern-2\nulldelimiterspace\LEFTRIGHT.){\vphantom{\XL\int}} \,dx+c_2y_2(x)+c_3y_1(x)\ ,\ x\in I$

\end{document}