Today in second semester calculus, I found myself stumbling a bit to provide a natural-sounding explanation for all the letters involved in the expression $$ \lim_{t \rightarrow \infty} \int_1^t \frac{1}{x^p} dx, $$ where $p$ is an arbitrary real number. There are some subtle things going on.

  • The value of $p$ is fixed throughout the evaluation, but is arbitrarily chosen at the beginning.
  • The value of $t$ is fixed as regards the definite integral, but is later pushed toward infinity when the limit is evaluated.
  • The value of $x$ is not like either of $p$ or $t$, but rather is the variable of integration. It ranges over the domain $[1,t]$, which again is finite for any fixed $t$.

I think these distinctions arise frequently in analysis, so I'd be interested in advice on making them more concrete to students.

One thing I think might help is to separate the limit and the integral. That is, I would first ask them to consider $\int_1^t \frac{1}{x^p} dx$ for arbitrary (but fixed) $p$ and $t > 1$. In this case, the only thing varying is the variable of integration. Now we have a function $A(t)$ that gives the area under $\frac{1}{x^p}$ on $[1,t]$, and we can ask how this function behaves as $t$ tends to infinity. While this viewpoint is helpful to me, I wonder if it is still too abstract for a typical Calculus student.

  • 3
    $\begingroup$ Being able to pull apart a problem to determine what type of object is each component part is a hurdle in everyone's math education. There is a lot happening in this problem to keep track of. $\endgroup$
    – Chris C
    Feb 4, 2015 at 1:47

1 Answer 1


The logicians' words "bound" and "free" may help.

  • The variable $x$ is bound in $\int_1^t dx/x^p$, and so also bound in the expression for the limit.
  • The variable $t$ is free in $\int_1^t dx/x^p$, but bound in the expression for the limit.
  • The variable $p$ is free in the expressions for both the integral and the limit.
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    $\begingroup$ I use this in class, without stressing the formalism too much. Usually I'll explain in an aside, "Are we free to plug in any value for $x$, say $x = 2$? Does $d2/2^p$ make sense here?" (+1) $\endgroup$
    – user1815
    Feb 4, 2015 at 3:34
  • $\begingroup$ Liked explanation. Going to use that, thanks. $\endgroup$
    – Karl
    Feb 4, 2015 at 18:16
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    $\begingroup$ And when the expression appears in an equation, $p$ is bound to an implicit $\forall p$. $\endgroup$
    – Jordan
    Feb 19, 2020 at 15:58

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