# B Applying L'Hospital's rule to Integration as the limit of a sum

Tags:
1. Mar 5, 2017

### Kumar8434

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?

Last edited: Mar 5, 2017
2. Mar 6, 2017

### Svein

Where have you read that? L'Hôpital's rule applies to quotients where nominator and denominator both tends to 0 (not ∞). Your expression is just a formula for the average value of f across the points you have chosen times the length of the interval - which is no great surprise.

3. Mar 6, 2017

### Staff: Mentor

L'Hopital's Rule also applies to quotients where both numerator and denominator are approaching infinity. See https://en.wikipedia.org/wiki/L'Hôpital's_rule, near the top of the page.

4. Mar 6, 2017

5. Mar 6, 2017

### Kumar8434

I remember reading in my book that all the other indeterminate forms can be reduced to $\frac{0}{0}$. So, $\frac{\infty}{\infty}$ is the same as $\frac{0}{0}$. And, I'm not saying that my last expression is a surprise. The series is not making any sense to me. How is it the average value of f multiplied by the length of the interval?

6. Mar 6, 2017

### Staff: Mentor

Your "..." depend on n in a non-continuous way. You cannot derive "..." with respect to n.

Counterexample: f(x)=1, a=0, b=1. Obviously the integral is 1, but f'(x)=0 for all x, hence your sum would give 0.

7. Mar 6, 2017

### Svein

OK. Let us take it step by step.
1. $\frac{(x_{1}+x_{2}+\dotso +x_{n})}{n}$ is by definition the mean (or average) value of the xi
2. Thus $\frac{f(a)+f\left(a+\frac{b-a}{n}\right)+.. ..+f\left(a+(n-2)\frac{b-a}{n}\right)+f\left(a+(n-1)\frac{b-a}{n}\right)}{n}$ is the mean (or average) value of the f values.
3. Your expression (before going to the limit) ($(b-a)\frac{f(a)+f\left(a+\frac{b-a}{n}\right)+.. ..+f\left(a+(n-2)\frac{b-a}{n}\right)+f\left(a+(n-1)\frac{b-a}{n}\right)}{n}$ is thus the mean (or average) value of the f values times the length of the interval (b - a)

8. Mar 6, 2017

### Staff: Mentor

In addition to the other comments, your (@Kumar8434) work requires f to be differentiable. For a definite integral $\int_a^b f(x)dx$, the only requirement on f is that it be continuous on (a, b). There is no guarantee that such a function will also be differentiable.

9. Mar 6, 2017

### Stephen Tashi

Yes.
But first, let's straighten out the terminology of "positive" vs "negative". You can't tell the sign of the terms because you don't know the sign of $f'(x)$. You are using the terminology "positive" to mean a term that has a "+" sign in front of it, but that does guarantee that the term is a positive number.

If you divide $[a,b]$ into 3 intervals the left endpoints of these intervals are $a,\ a+(1/3)(b-a),\ a+ (2/3)(b-a)$.
When you increase the number divisions to 4, the argument of $f(a + (b-a)/n)$ moves to the left from $a + (1/3)(b-a)$ to $a + (1/4)(b-a)$. The argument of $f(a + (n-1)(b-a)/n)$ moves to the right from $a + (2/3)(b-a)$ to $a + (3/4)(b-a)$.

So the "+" and "-" signs in front of the terms do correctly account for how the direction of the argument of a given function of is changing on the x-axis. What isn't obvious from the notation is that as $n$ increases, new terms are appearing as terms in the sum.

The general pattern is that you have a function of the form $G(n) = \sum_{i=1}^n w_i(n)$ and you are attempting to give an interpretation to the notation $G'(n)$ by doing a term-by-term differentiation of the sum, but to define the derivative, you need an interpretation of the difference quotient $\frac{ G(n+h) - G(n)}{h}$ and it isn't clear what you mean by $G(n+h)$ when $h$ is not an integer. For example if $n = 3$ and $h = 0.1$ then your interpretation of $G(n)$ appears to be to keep $n = 3$ as the upper index of the summation and use $n = 3.1$ when you evaluate the functions $w_i(n)$.

This situation is analogous to distinction between differentiating a function defined by an integral with constant limits of integration vs a function defined by an integral with variable limits of integration - e.g. finding the derivative of $P(x) = \int_0^{2x} xy^2 dy \$ vs $\$ finding the derivative of $Q(x) = \int_0^1 xy^2 dy$.

10. Mar 7, 2017

### Kumar8434

I think my sum would be undefined in that case. Because remember that it'll be a sum of an infinite number of 0's. I think that's an indeterminate form. Forget about adding the derivatives of $f(x)$. Instead in my last expression, it says that the double integral of f(x) from a to b is given by the series on the right, which doesn't involve derivatives. How's that possible? The double integral will be of the form F(x)+Cx+D. By taking definite integral the constant D will eliminate but there would still be a constant C. But my series on the right doesn't give any arbitrary constant. And, I don't see any way to evaluate such a series whose starting terms and ending terms are of opposite signs.

11. Mar 7, 2017

### Kumar8434

If it's differentiable, is the series true and does it make any sense?

12. Mar 7, 2017

### Kumar8434

Obviously the definite integral of a function from a to b is its average value between a and b multiplied by the length of the interval. Why are you talking about that? I thought you were saying that about my final expression.

13. Mar 7, 2017

### Stephen Tashi

but that does not define a unique process because there can be several different functions whose derivative is the given $f(x)$.

14. Mar 7, 2017

### Kumar8434

Oh, so it does give an arbitrary constant. Thanks.

15. Mar 7, 2017

### Stephen Tashi

Keep in mind that your notation only makes sense if you treat $n$ as a function of $h$.

There is a distinction between:

$A(h) = \sum_{i=1}^\infty w_i(h)$

versus

$B(h) = \sum_{i=1}^{g(h)} w_i(h)$

16. Mar 7, 2017

### Kumar8434

In this case, the limits are constants, aren't they?

17. Mar 7, 2017

### Kumar8434

Yeah, I know that. That's why I replaced $h$ in terms of $n$ to get an expression in a single variable before differentiating. However, if the function we're talking about is everywhere positive from a to b then certainly the starting and end terms of
$$\lim_{h\rightarrow 0}h^2(-f(a+h)-2f(a+2h)-3f(a+3h)-.....+3f(a+(n-3)h)+2f(a+(n-2)h)+f(a+(n-1)h))$$
differ in sign. How to interpret this change in sign?

18. Mar 7, 2017

### Stephen Tashi

By "this case", do you mean one of your own equations?

I'm drawing the analogy between an integral with a variable upper limit and your summations $\sum_{i=1}^n$ where $n$ depends on $h$.

In the usual sort of summation problem: $\sum_{i=1}^{n+1} w_i(h)$ can be computed as $\sum_{i=1}^{n} w_i(h) + w_{n+1}(h)$. $\$ However, you are dealing with summations where the functions $w_i(h)$ don't stay constant as the upper index of the summation increases. For example the second term involves $w_2(h)= f(a + (b-a)/n)$ where $n$ is a function of $h$ $\$ (i.e. $h = (b-a)/n$ so $n = (b-a)/h$.)

Using the notation "$h(n)$" to emphasize that $h$ is a function of $n$, Your sums are of the form $G(n) = \sum_{k=1}^n w_i(h(n))$ $\$ We cannot compute $G(n+1)$ as $G(n) + w_{n+1}(h(n))$.

So the sequence of sums $G_2, G_3,...$ is not a sequence of partial sums of the same infinite series and we can't apply convergence tests to it (e.g. the ratio test) that deal with partial sums of the same series. - not that you worry about such things ! :)

19. Mar 7, 2017

### Stephen Tashi

You've spoken of replacing $f'(x)$ by $f(x)$ , so speaking about "the function" being everywhere positive is ambiguous.

Even if $f(x)$ is a positive function and even if $f'(x)$ is a positive function, a function $w(h) = f(g(h))$ can have a negative derivative $w'(h)$ if $g'(h)$ is negative.