This is probably very simple but I don't see the solution now. Let us say I have a variable e defined as


ReplaceAll can subsitute n with value 3 using


The problem is when e is given by an integral of some function f[{n_}, r_] with n as a parameter

e=Integrate[f[{n}, r],{r, rmin, rmax}]

Then the use of


will take too long if the integral in question is hard to evaluate and so, better would be FIRST to replace n with its numerical value and THEN evaluate the integral.

If e was defined as a function

e[n_]:=Integrate[f[{n}, r],{r, rmin, rmax}]

there would be no problem. But my function f depends on quite many definitions (using "n") I made earlier. It works ok but then in the end of my code there is a complicated f which makes the integration (i.e. evaluation of "e") too long - and I don't want to change all the definitions I made before.

What to do if I want to keep "e" as above, i.e. I don't want to define it as a function?

  • $\begingroup$ Oh, sure, you're right, I used another symbol, I forgot about that. Thanks, edited. $\endgroup$ – wondering Jan 28 '15 at 23:51
  • $\begingroup$ If you are not making functions, and n is eventually replaced by a number, why not simply define n beforehand? E.g. n=3;e=n. $\endgroup$ – 2012rcampion Jan 29 '15 at 1:28
  • $\begingroup$ @2012rcampion Right, I should have noted that. I cannot use your solution because I will need to evaluate e for several values of n, (e.g. using Table). $\endgroup$ – wondering Jan 29 '15 at 1:35
  • $\begingroup$ This question may be a duplicate of: (3864) $\endgroup$ – Mr.Wizard Jan 29 '15 at 2:05


The first method I recommended does not work properly. I fooled myself because I have a $PrePrint definition by default. However following your update I think you do not need ReplaceAll. Instead try Block:

e := Integrate[1/(x^3 + 1), {x, 0, n}];

Block[{n = 2.0}, e]

Not the use of SetDelayed to keep the definition from evaluating early.

It happens that Table uses a Block-like mechanism(1) therefore you can also use:

Table[e, {n, 1`, 4`, 1`}]
{0.835649, 1.09, 1.15445, 1.17814}

I recommend that you also read:

  • $\begingroup$ +1. But, two or more levels of Unevaluated usually signal (to me), that Unevaluated is generally the wrong tool for the job (not in this particular case). $\endgroup$ – Leonid Shifrin Jan 29 '15 at 0:25
  • $\begingroup$ Thanks. I corrected the $E$ and added some motivation on why I do not want to define a function. To your answer - for some reason I am not gettin "1.09" as output but rather $\text{Unevaluated}\left[\int_0^{2.} \frac{1}{x^3+1} \, dx\right]$ Why is that? $\endgroup$ – wondering Jan 29 '15 at 1:13
  • $\begingroup$ @wondering You're right, my result only appears because I am using a $PrePrint definition. I forgot to test this in a default configuration. Sorry. Let me see what I can do. $\endgroup$ – Mr.Wizard Jan 29 '15 at 1:49
  • $\begingroup$ @wondering Please see the update. $\endgroup$ – Mr.Wizard Jan 29 '15 at 1:59
  • $\begingroup$ @Mr.Wizard Thanks, this works. Actually this solution (with SetDelayed + Table came to my mind but because I have two parameters m and n from which only n one takes several values, m only one, I would have had anyway to define m beforehand as 2012rcampion suggested (or to make a Table[e,{n,nmin,nmax},{m,m0,m0}], but that is not very good). I think the best solution for this situation is a combination of your two suggestions, with Block and Table, i.e.: Block[{m=1}, Table[e, {n, 1, 5}]] +1 $\endgroup$ – wondering Jan 29 '15 at 3:07

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