Question

Is there a way to adjust the size integral sign automatically depending on "size" of the integrand?

Answer 1

Like, is there a way for the integral sign to change size automatically, kind of like \left( and \right) automatically adjust the size of your parentheses? Or can you only make it bigger manually?

For example, if you have this made-up integral like:
\[\int_{0}^{\infty}\frac{\frac{2^nb^2}{\alpha^{n-1}-1}-\sum\limits_{i=1}^nx_i^2}{(n^{2-\frac{\gamma}{2}}\alpha)\left(\frac{2^x}{\sum_{i=1}^nx_i^2}\right)^{n^{2-\frac{\gamma}{2}}}} \, db\], the integral sign looks rather small compared to the integrand..

Answer 2

\[\huge\int_{0}^{\infty}\small
\frac{\frac{2^nb^2}{\alpha^{n-1}-1}-\sum\limits_{i=1}^nx_i^2}{(n^{2-\frac{\gamma}{2}}\alpha)\left(\frac{2^x}{\sum_{i=1}^nx_i^2}\right)^{n^{2-\frac{\gamma}{2}}}} \, db\]
This is one way, but it doesn't look very good...

The \left( and \right) are called delimiters. As of right now, I cannot find a way to make an integral or sum into a delimiter.

Answer 3

\[\huge\int\limits_{0}^{\infty}\small
\frac{\frac{2^nb^2}{\alpha^{n-1}-1}-\sum\limits_{i=1}^nx_i^2}{(n^{2-\frac{\gamma}{2}}\alpha)\left(\frac{2^x}{\sum_{i=1}^nx_i^2}\right)^{n^{2-\frac{\gamma}{2}}}} \, db\]
This looks slightly better. I used \int\limits to put the integral's limits above and below the integral.

Answer 4

@BTaylor ... hm yeah it does look strange too when it's bigger... maybe I should just stick with the regular-sized integral...

@TheOcean: What I meant was if there was a way the integral sign could "stretch out" vertically according to the "height" of the integrand, so that you don't have a tiny integral sign next to a large/high integrand. But I was curious if \(\LaTeX\) had an input that would this this "stretching out" automatically (kind of like how \left( and \right) stretch out the parentheses vertically to match the size of its argument inside). For example:
\[ (\frac{3z-2}{2y+x}) \,\,\, \text{vs.} \,\,\, \left(\frac{3z-2}{2y+x}\right) \] But BTaylor showed me that the result doesn't necessarily look better (but they adjusted the integral sign height by manually putting in the command "\huge", which is not size-dependent on the size of the integral's argument, but it is a fixed size )

Answer 5

\[\large \int\limits_0^\infty\left(
\frac{\frac{2^nb^2}{\alpha^{n-1}-1}-\sum\limits_{i=1}^nx_i^2}
{(n^{2-\gamma/2}\alpha)
\left(\frac{2^x}{\sum\limits_{i=1}^nx_i^2}\right)^{n^{2-\gamma/2}}}
\right)\, \mathrm db\]

Answer 6

Right click what UnkleRhaukus did, go to Show Math As > and select TeX commands.

Answer 7

Another thing is is you use `\[ \]` to enclose it instead of `\( \)` it does different sizing.

\[\int\limits_0^\infty\left(
\frac{\frac{2^nb^2}{\alpha^{n-1}-1}-\sum\limits_{i=1}^nx_i^2}
{(n^{2-\gamma/2}\alpha)
\left(\frac{2^x}{\sum\limits_{i=1}^nx_i^2}\right)^{n^{2-\gamma/2}}}
\right)\, \mathrm {db} \]

The thing that gets me is that you can do a one side one for the | bar when evaluating things.

\[ \int\limits^{2}_{1} x\;dx = \left. \dfrac{x^{2}}{2}\right|^{2}_{1} = 4 - \frac{1}{2} = \frac{7}{2} \]

Answer 8

Oh, and if you were going to do it in print, they have packages for it.
http://tex.stackexchange.com/questions/39181/big-integral-sign

I just do not see MathJax support for these online... though you can make only a few characters larger if you want.

\[\huge\int\limits_0^\infty\normalsize \left(
\frac{\frac{2^nb^2}{\alpha^{n-1}-1}-\sum\limits_{i=1}^nx_i^2}
{(n^{2-\gamma/2}\alpha)
\left(\frac{2^x}{\sum\limits_{i=1}^nx_i^2}\right)^{n^{2-\gamma/2}}}
\right)\, \mathrm {db} \]

Answer 9

Thanks for the info

Answer 10

np. And just think, \(\int\) started out life as an S that printers played with. So formatting it different has a long tradition.