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The definite integral of a function $f(x)$ from $a$ to $b$ as the limit of a sum is: $$\int_a^bf(x)dx=\lim_{h\rightarrow 0}h(f(a)+f(a+h)+.. ..+f(a+(n-2)h)+f(a+(n-1)h))$$ where $h=\frac{b-a}{n}$. So, replacing $h$ with $\frac{b-a}{n}$ gives:

$$\lim_{n\rightarrow \infty}(b-a)\frac{f(a)+f\left(a+\frac{b-a}{n}\right)+.. ..+f\left(a+(n-2)\frac{b-a}{n}\right)+f\left(a+(n-1)\frac{b-a}{n}\right)}{n}$$

This seems to be of the form $\frac{\infty}{\infty}$ because the numerator is an infinite sum and the denominator also tends to infinity, so I think L'Hospital's rule can be applied. So, applying L'Hospital's rule gives: $$\int_a^bf(x)dx=\lim_{n\rightarrow \infty}\frac{(b-a)^2}{n^2}\left(-f'\left(a+\frac{b-a}{n}\right)-2f'\left(a+2\frac{(b-a)}{n}\right)-3f'\left(a+3\frac{(b-a)}{n}\right)-.....+3f'\left(a+(n-3)\frac{(b-a)}{n}\right)+2f'\left(a+(n-2)\frac{(b-a)}{n}\right)+f'\left(a+(n-1)\frac{(b-a)}{n}\right)\right)$$ Replacing $\frac{(b-a)}{n}$ with $h$ gives: $$\int_a^bf(x)dx=\lim_{h\rightarrow 0}h^2(-f'(a+h)-2f'(a+2h)-3f'(a+3h)-.....+3f'(a+(n-3)h)+2f'(a+(n-2)h)+f'(a+(n-1)h))$$ Replacing $f'(x)$ with $f(x)$ and hence $f(x)$ with $\int f(x)dx$ gives: $$\int_a^b\left(\int f(x)dx\right)dx=\lim_{h\rightarrow 0}h^2(-f(a+h)-2f(a+2h)-3f(a+3h)-.....+3f(a+(n-3)h)+2f(a+(n-2)h)+f(a+(n-1)h))$$ Does the above series make any sense to you? I'm asking this because the starting terms of the series are negative and the first term is multiplied by 1, the second term by 2, the third term by 3, etc,i.e the multipliers are increasing by 1, but the last terms of the series are positive and the multipliers are decreasing by 1 instead as we move forward in the series. Is it something like some transition from negative terms to positive terms takes place in the middle or something? Or Is applying L'Hospital's rule not allowed in this case?

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  • $\begingroup$ Do you have any conditions on f? Because if you take $f =1, \ [0,1]$. Then the integral is 1. But your series at the bottom is zero, since the derivative is zero. $\endgroup$ – Olba12 Mar 6 '17 at 4:06
  • $\begingroup$ @Olba12 In fact, I think my series would be not-defined in that case because it will be sum of an infinite no. of zeros, which is $0\cdot \infty$ which is not-defined. I don't think my series makes any sense, but what have I done wrong in the derivation? $\endgroup$ – Dove Mar 6 '17 at 4:16
  • $\begingroup$ The infinite sum of zeros is zero. But $\infty \cdot 0$ is undefined yes. The use of hospital's, it's not necessary the case that you have $\infty / \infty$. The numerator could be converging to a number x. $\endgroup$ – Olba12 Mar 6 '17 at 4:37
  • $\begingroup$ @Dove: I'll post an answer showing how you can fix this in a way. $\endgroup$ – RRL Mar 6 '17 at 4:52
  • $\begingroup$ @Dove, what you are trying to apply l'Hopital to is not the fundamental theorem of calculus but rather the definition of the Riemann integral. Perhaps the title should be changed. $\endgroup$ – Mikhail Katz Mar 7 '17 at 13:07
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Your argument is invalid but the question does suggest something that is valid. To keep the notation simple let $a = 0$ and $b=1$.

You are trying to apply L'Hospital's rule to find the limit of a Riemann sum as

$$\lim_{n \to \infty}\frac{F(n)}{G(n)} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n} \right),$$

where $F(n)$ is the sum and $G(n) = n$.

We can accept $G'(n) = 1$, but your argument breaks down trying to compute $F'(n)$ since you only applied the derivative to the terms of the sum. What about the $n$ appearing as the upper summation limit?

Following your steps you would claim

$$\frac{d}{dn} \sum_{k=1}^n \frac{k}{n} = -\frac{1}{n^2}\sum_{k=1}^n k = -\frac{1}{n^2} \frac{n^2 + n}{2}= -\frac{1}{2} - \frac{1}{2n},$$

when, in fact,

$$\frac{d}{dn} \sum_{k=1}^n \frac{k}{n} = \frac{d}{dn} \frac{n^2 + n}{2n}= \frac{1}{2}.$$

It is pointless to pursue this with L'Hospital's rule, but we can use the analog for sequences -- the Stolz-Cesaro theorem:

If for sequences $(a_n)$ and $(b_n)$ we have $b_n \uparrow \infty$ then

$$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n}, $$

if the limit on the right-hand side exists.

If $f$ is Riemann integrable, then the Riemann sum converges to the integral

$$\lim_{n \to \infty}\frac{F(n)}{G(n)} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n} \right) = \int_0^1 f(x) \, dx.$$

We now ask if it is also true that

$$\lim_{n \to \infty}\frac{F(n+1) - F(n)}{G(n+1)-G(n)} = \lim_{n \to \infty}\left( \sum_{k=1}^{n+1} f\left(\frac{k}{n+1} \right) - \sum_{k=1}^{n} f\left(\frac{k}{n} \right)\right) = \int_0^1 f(x) \, dx$$

The answer is yes if $f$ is continuously differentiable, $f \in C^1([0,1])$. However, it is not true if we only have $f$ continuous, $f \in C([0,1]) \setminus C^1([0,1]).$

The proof of this is not trivial. For example, see here.

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Some points. First, the numerator being an infinite sum, in the sense that the number of terms in the sum becomes unbounded, does not imply that the sum itself is infinite. That's precisely what the study of convergence of series is all about.

Moreover, L'Hôpital is used for calculating 'continuous' limits, not limits of sequences. This is not to say you can't use it to calculate things like

$$\lim_{n\to\infty}n\sin\left(\frac1n\right)=\lim_{n\to\infty}\frac{\sin\left(\frac1n\right)}{\frac1n},$$

by recognizing that as $\lim_{x\to0}\frac{\sin(x)}x$. But that's because a continuous a limit is stronger than the discrete one, in the sense that if $\lim_{x\to0}f(x)$ exists and equals $L$, then $\lim_{n\to\infty}f(a_n)=L$ for any sequence $a_n\to 0$. You can check this directly from the definition of these limits.

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  • $\begingroup$ The infinite series in my case is not something like 1/2+1/4+1/8+......., so that it will converge. Pick up ANY function and any number d such that 1/d is an integer and try to calculate the sum f(1)+f(1+d)+f(1+2d)+........f(2). Now, make d smaller and smaller. It will tend to infinity. $\endgroup$ – Dove Mar 6 '17 at 4:45
  • $\begingroup$ Not any function, no. If $f$ is continuous and, say, positive at $1$, then yes, you are right. $\endgroup$ – Fimpellizieri Mar 6 '17 at 4:46
  • $\begingroup$ Please let me know if you're able to find any continuous function (except f(x)=0) for which that sum converges. $\endgroup$ – Dove Mar 6 '17 at 4:47
  • $\begingroup$ I don't think that the derivation of L'Hospital's assumes that it can't be used for sequences. $\endgroup$ – Dove Mar 6 '17 at 4:48
  • $\begingroup$ Take $f(x)=x$ and $a=-b$. The sum will always be $0$. $\endgroup$ – Fimpellizieri Mar 6 '17 at 5:41
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As stated in my comment it fails for $f=1$. Given interval $[a,b]$, and a partition $\{a=x_1<...x_n=b\}$, it seems reasonable that there is a function such that $f(x_n) = 2^{-n}$. Thus your sum of infinite values divided by $n$ is not of the form $\infty / \infty$ necessarily.

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