5

I've been stuck on this problem for a while and can't find anything similar that is able to help me. I am attempting to write an equation where there is an integral, with limits, of a fraction, where the denominator has a nested fraction in it. I am using the bigints package to use larger integrands, but the vinculum of the outer fraction is lined up with the vertical centre of the integrand so that it towers over the numerator and doesn't go low enough around the denominator.

\begin{align}
\theta_f &= \pi - 2 {\displaystyle \bigints_{r_\text{min}}^\infty}
\cfrac{b}{r^2 \left[1 - \bigg(\cfrac{b}{r}\bigg)^2 - \cfrac{2 V(r)}
{m v_0^2}\right]^{\frac{1}{2}}} dr \nonumber \\
\end{align}

enter image description here

Any ideas about how I can, preferably get the integrand to rubber band around fractions in the same way that fractions can, or how to manually change the vertical alignment/spacing to shift the whole fraction up until I'm happy with it?

Let me know if I need to give any more information. I've been stuck on this for about an hour now!

Thanks

Steve

  • 2
    Hello to TeX.SE! Well, the idea is not to change the size of the integral symbol. You simply keep it smaller. It's the standard thing to do, it's completely ok and enlarging it looks ugly and inconsistent. The only thing that's then necessary (advised) is getting rid of \left...\right and tweaking the braces' size and spacing manually. – yo' Mar 21 '18 at 16:43
7

Just get rid of the unnecessarily large fractions and the large integral symbol, and you'll be fine:

enter code here

\documentclass{article}
\usepackage{mathtools}
\begin{document}

\begin{align}
\theta_f &= \pi - 2 \int_{r_\mathrm{min}}^\infty
\frac{b}{r^2 \Bigl[1 - (\frac{b}{r})^2 - \frac{2 V(r)}
{m v_0^2}\Bigr]^{1/2}}\, \mathrm{d}r \nonumber
\end{align}

\end{document}

Note especially the usage of 1/2 in the exponent, removal of the spacing-killing group and meaningless \displaystyle, removal of all illogical \cfrac, and addition of the thin space before \mathrm{d}r. And finally, the change of \text to \mathrm in r_{min} as it should not follow the text font, but rather be always upright rm.

Also note that I personally would not use [] at all and stick to () (or even \sqrt), but that's a personal choice.

  • @campa I owe you something for this; today is not the best day for my LaTeX parser. Thanks once more! – yo' Mar 21 '18 at 16:56
  • The rendering of that (\frac{b}{r})^2 in the denominator looks a little weird to me. I think that {(\frac b r)}^2 (to make the superscript respect the height of the fraction) looks better (but, of course, to each his or her own). – LSpice Mar 21 '18 at 23:28
  • Thank you @yo' very helpful. Can you explain to me why I should use frac rather than cfrac or dfrac? I'm not really sure when to use them. Also, what does mathrm do that text doesn't? – Steven Thomas Mar 22 '18 at 13:10
5

You could express the integrand as a product rather than a fraction by giving negative exponents to $r^2$ and square-bracketed expression:

\begin{align}
\theta_f &= \pi - 2 {\displaystyle \bigints_{r_\text{min}}^\infty}
{b}{r^{-2} \left[1 - \bigg(\cfrac{b}{r}\bigg)^2 - \cfrac{2 V(r)}
{m v_0^2}\right]^{-\frac{1}{2}}} dr \nonumber \\
\end{align}
  • 2
    Well, I would stick to \frac{b}{r^2} [...]^{-1/2} instead of r^{-2}. Other than that, a great idea. – yo' Mar 21 '18 at 16:50
  • Do you still need \bigints? – Mico Mar 21 '18 at 20:25
4

Would using \mfrac (medium-sized fractions) from nccmath be an acceptable solution for you?

\documentclass{article}
\usepackage{mathtools, nccmath}
\usepackage{bigints}

\begin{document}

\begin{align}
\theta_f &= \pi - 2 {\displaystyle \bigints_{r_\text{min}}^\infty}
\cfrac{b}{r^2 \Bigl[1 - \Bigl(\mfrac{b}{r}\Bigr)^{\mkern-5mu 2} - \frac{2 V(r)}
{m v_0^2}\Bigr]^{\!\frac{1}{2}}}\, dr \nonumber
\end{align}

\end{document} 

enter image description here

2

Here's a solution that dispenses with any need for large integral symbols.

enter image description here

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\theta_f = \pi - 2\int_{r_{\min}}^\infty \frac{b}{r^2} 
\biggl[1 - \biggl(\frac{b}{r}\biggr)^{\!2}
         - \frac{2 V(r)} {m v_0^2}\biggr]^{-1/2} dr
\]
\end{document}

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.