The problem is to find: $$\int\limits_0^1 \lim\limits_{n\rightarrow\infty}(nz^{n-1})\mathrm{d}z$$

I started by finding $\lim\limits_{n\rightarrow\infty}(nz^{n-1})$. Naturally it converges to zero on $[0,1)$. However at $z = 1$, the limit diverges. So in essence on the contour of integration we have a zero function except for the endpoint, which is $\infty$. My question now is how do I integrate this?

I thought I could break it up into two integrals: $$\lim\limits_{\beta\rightarrow1^-}\left(\int\limits_0^\beta 0\mathrm{ d}z +\int\limits_\beta^1\lim\limits_{n\rightarrow\infty}(nz^{n-1})\mathrm{d}z\right)$$

My question now is does the integral of an infinite point equal zero? In the limit we get, essentially, $\int\limits_\beta^1\infty\mathrm{ d}z$. Usually a point integral is zero, but does that hold when the point itself is divergent to infinity?

I know the answer is supposed to be $0$, and that makes sense. I'm having trouble getting there rigorously, though.

Any help is greatly appreciated.

PS: This is for complex variables, but I realize the contour of integration is on the real line. I assume that would be the best way to do it. However, if you know of some fancy way to do it that strays from the real line feel free to do so, as this is complex variables after all.

  • $\begingroup$ en.wikipedia.org/wiki/… is helpful (Halil: calm down. If Greg is incorrect then downvote, leave one comment on his answer, and then move on. It's not that important.) $\endgroup$ Commented Feb 19, 2013 at 23:22

1 Answer 1


The question you are asking is: define $f(x)=0$ for $0\le x<1$, and define $f(1)=\infty$. (I don't like treating $\infty$ like a real number, but in this problem the integrand does not exist at $1$ unless we do.) Then, what is $\int_0^1 f(x)\,dx$?

So, what definition of integral are you using? The Riemann integral? The Lebesgue integral? For example, if it's a Riemann integral, then we can choose to take $1$ as one of our sample points (getting $\infty$ for the Riemann sum) or not (getting $0$), and so the limit of Riemann sums doesn't exist. If it's a Lebesgue integral, then you can change the integrand's value on a set of measure 0 without affecting the integral (important lemma in the subject - valid even if values $\pm\infty$ are allowed!), so the answer is $0$.

  • $\begingroup$ So is this function Lebesgue integrable? The question does not specify (and the text never specifies) what integral is being used. If you were given the question as it is ($\int\limits_0^1 \lim\limits_{n\rightarrow\infty}(nz^{n−1})dz$), what would you do? $\endgroup$ Commented Feb 19, 2013 at 19:16
  • $\begingroup$ Is it okay to say $\int\limits_0^1 \lim\limits_{n\rightarrow\infty}(nz^{n-1})dz = \lim\limits_{b\rightarrow 1}\int\limits_0^b \lim\limits_{n\rightarrow\infty}(nz^{n-1})dz = 0$? $\endgroup$ Commented Feb 19, 2013 at 19:18
  • $\begingroup$ @Halil Any justification? $\endgroup$ Commented Feb 19, 2013 at 20:42
  • $\begingroup$ Halil proposed an edit to this post that I voted to reject because it deleted the Riemann integral stuff but with the edit comment "improved formatting" which was clearly not what it was doing. $\endgroup$ Commented Feb 19, 2013 at 23:12
  • $\begingroup$ ...I'm actually inclined to think that Halil is right, though. The only way to interpret the integral in the question as a Riemann integral is if it is an improper one, in which case the limit is correct. $\endgroup$ Commented Feb 19, 2013 at 23:21

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