In the answer to this question Eric Näslund showed that logarithms can be written as the following limit of a sum:

$$\displaystyle \log(x) = \lim_{k\to \infty } \, \sum\limits_{n=k}^{x k} \frac{1}{n}$$

Changing the symbol for the sum $\sum$ into an integral $\int$ with the same limits as in the sum, one still gets logarithms:

$$\displaystyle \log(x) = \lim_{k\to \infty } \, \int\limits_k^{x k} \frac{1}{n} \, dn$$

or at least in Mathematica, so I guess it is true.

Are there other sums and integrals such that the function to be integrated or summed is the same, as well as the integration and summation limits, that give the same result? This provided that integrals or sums giving zero as result are not considered.

As a Mathematica program this is:

Clear[n, k, x];
Table[Limit[Integrate[1/n, {n, k, x*k}], k -> Infinity], {x, 1, 12}]
Table[Limit[Sum[1/n, {n, k, x*k}], k -> Infinity], {x, 1, 12}]

which gives the output:

{0, Log[2], Log[3], Log[4], Log[5], Log[6], Log[7], Log[8], Log[9], Log[10], Log[11], Log[12]}

in both cases.

Edit 7.4.2013: I now realized that the integral is not dependent on that the limit goes to infinity. Actually any value of k will give the same logarithm.

A small Mathematica program to illustrate this:

Table[Integrate[1/n, {n, k, 2*k}], {k, 1, 12}]

which gives:

{Log[2], Log[2], Log[2], Log[2], Log[2], Log[2], Log[2],
 Log[2], Log[2], Log[2], Log[2], Log[2]}
  • $\begingroup$ Näslund? $ $ $ $ $\endgroup$ – Did Nov 15 '12 at 16:59
  • $\begingroup$ I assumed that is his real name with swedish umlaut. But if he has grown up in an english speaking country the spelling might be Naslund. $\endgroup$ – Mats Granvik Nov 15 '12 at 17:02
  • $\begingroup$ You might look at this question and my answer which has functions that sum, integrate, sum squared, and integrate squared to the same answer $\endgroup$ – Ross Millikan Nov 15 '12 at 17:09
  • $\begingroup$ I will have a look at that. $\endgroup$ – Mats Granvik Nov 15 '12 at 17:11

Your Answer

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

Browse other questions tagged or ask your own question.