General ⇒ Size of the Integral sign and alignment

[curiouslearn]

Hello,

I have a the following math equation in the display environment.

Code: Select all

 \begin{displaymath}
                        m_{R}(m_{N},x) =
                        \begin{array}{l}
                                \int_{r_{L}}^{r_{H}}p(x)f(x+r)g(r)\md r - \int_{r_{L}}^{r_{H}}p'(x)[1-F(x+r)]g(r)\md r \\
                                \qquad \qquad+ p'(x)(1-F(x))(1-m_{N})-p(x)f(x)(1-m_{N}) \\
                                \hline 
                                \int_{r_{L}}^{r_{H}}p(x)f(x+r)g(r)\md r - \int_{r_{L}}^{r_{H}}p'(x)[1-F(x+r)]g(r)\md r]
                        \end{array}                
   \end{displaymath}

(1) The size of the Integral sign seems small. Is there a way to get a larger integral sign?
(2) Is there a way to align the " m_{R}(m_{N},x) = " part of the expression with the horizontal line? The way the code appears above this part of the expression occurs at the same height as the second line in the numerator.

Thanks in advance for your help.

[curiouslearn]

A search on the internet revealed the mathtools package which allows me to take care of the second problem (the alignment problem) using the following code instead:

Code: Select all

               \begin{equation}
                        m_{R}(m_{N},x) =
                        \dfrac{
                                \splitdfrac{\int_{r_{L}}^{r_{H}}p(x)f(x+r)g(r)\md r - \int_{r_{L}}^{r_{H}}p'(x)[1-F(x+r)]g(r)\md r}
                                {+ p'(x)(1-F(x))(1-m_{N})-p(x)f(x)(1-m_{N})}}
                                {\int_{r_{L}}^{r_{H}}p(x)f(x+r)g(r)\md r - \int_{r_{L}}^{r_{H}}p'(x)[1-F(x+r)]g(r)\md r}
                \end{equation}

However, the integral signs are still small. Any suggestions for making them larger would be helpful.

Thanks.

[gmedina]

One possibility is to use the \mathlarger command provided by the relsize package, as the following example suggests:

Code: Select all

\documentclass{article}
\usepackage{amsmath}
\usepackage{mathtools}
\usepackage{relsize}
\newcommand\md{\ }

\begin{document}
\begin{equation}
  m_{R}(m_{N},x) =
  \dfrac{
    \splitdfrac{\mathlarger{\int}_{r_{L}}^{r_{H}}p(x)f(x+r)g(r)\md r - 
        \mathlarger{\int}_{r_{L}}^{r_{H}}p'(x)[1-F(x+r)]g(r)\md r}
        {+ p'(x)(1-F(x))(1-m_{N})-p(x)f(x)(1-m_{N})}}
    {\mathlarger{\int}_{r_{L}}^{r_{H}}p(x)f(x+r)g(r)\md r - 
        \mathlarger{\int}_{r_{L}}^{r_{H}}p'(x)[1-F(x+r)]g(r)\md r}
\end{equation}

\end{document}

I had to define the \md command since I do not know how you originally defined it. Change it back to the original definition.

[curiouslearn]

Thanks very much gmedina. That worked very well. I was using the command \md as a shortform for \mathrm{d}.

It would have been perfect had the space between integrand and the integral sign been a little less. I tried using \mathclap available in mathtools, which reduces that space but creates the problem of the integral limits running into the integral sign (since they are not above or below) as in the summation sign. I suppose I can live with that.

[balf]

To reduce the space, apart from adding a \! before the integrand, you could try (it modify the very last integral from gmedina's post):

Code: Select all

\mathlarger{\int}_{r_{L}}^{\mathrlap{r_{H}}}p'(x)[1-F(x+r)]g(r)\md r}

Btw, the spacing would be better, I find, if you defined \md as \,\mathrm{d}.

B.A.

[curiouslearn]

Thanks very much Balf. That worked great.