On Mon, 21 Jan 2008, Peter I. Hansen wrote:

Hi, I'm sending you this off-list because of the attachment...

Your patch works nicely for what I want it do do (The displaylimits), but I'm a bit confused about the behaviour of the intext math in with the third option (intlimitcode = 2). Did I mess things up?

Best, Peter

I had made a mistake in the code. Here is a corrected version, which works for all three cases Aditya \chardef\intlimitcode\zerocount %0 nolimits 1 displaylimits 2 limits \def\intlimits {\ifcase\intlimitcode \nolimits \or \displaylimits \or \limits \fi} \definemathcommand [int] {\intop \intlimits} \definemathcommand [oint] {\ointop\intlimits} \def\repeatintegral#1% {\scratchtoks\emptytoks \let\dointlimits\donothing \let\dodointlimits\intlimits \dorecurse{#1}{\appendtoks \intop \dointkern \to \scratchtoks} \appendtoks \intop \dointlimits \dodointlimits \to \scratchtoks \edef\dodorepeatintegral{\the\scratchtoks}% \futurelet\next\dorepeatintegral} \def\dorepeatintegral {\ifx\next\limits \dointlimitcorrection \else \ifx\next\displaylimits \dointlimitcorrection \else \ifx\next\nolimits \donothing \else \ifcase\intlimitcode\else \dointlimitcorrection \fi\fi\fi\fi \dodorepeatintegral} \def\dointlimitcorrection {\mkern-7mu\mathchoice{\mkern-2mu}{}{}{}% \mathop\bgroup \mkern7mu\mathchoice{\mkern2mu}{}{}{}% \let\dointlimits\egroup} \starttext \startbuffer $\int_a^b f(x) dx$ and also $\iint_a^b f(x,y) dxdy$, $\iiint_a^b f(x,y) dxdy$, $\iiiint_a^b f(x) dx$ \startformula \int_a^b f(x) dx \quad \iint_a^b f(x) dx \quad \iiint_a^b f(x) dx \quad \iiiint_a^b f(x) dx \quad \stopformula $\iint\limits_a^b$ $\iint\nolimits_a^b$ $\iint\displaylimits_a^b$ \startformula \iint\limits_a^b \iint\nolimits_a^b \iint\displaylimits_a^b \stopformula \stopbuffer Default: \getbuffer Displaylimits: \chardef\intlimitcode\plusone \getbuffer Limits: \chardef\intlimitcode\plustwo \getbuffer \stoptext