# Something weird with the fundamental theorem of calculus

1. May 17, 2012

### AxiomOfChoice

Suppose I know my function G is infinitely differentiable on the closed interval [a,b] and that all derivatives of G (including G itself) vanish at b. For any z in [a,b], I have by the FTC that

$$\int_z^b G'(w) dw = G(b) - G(z).$$

Or, switching limits,

$$\int_b^z G'(w) dw = G(z) - G(b).$$

One can integrate by parts on the left-hand side and obtain (this is basically what the integral remainder form of Taylor's theorem tells you)

$$G(z) - G(b) = G'(b)(z-b) + \int_b^z (z-w)G''(w)dw,$$

or $G(z) = \int_b^z (z-w)G''(w)dw$ when you drop the terms that are zero (i.e., G(b) and G'(b)). Now, I want to take absolute values of both sides and do some estimating. I get

$$|G(z)| \leq \int_b^z |z-w||G''(w)|dw \leq C \int_b^z |z-w|dw,$$

since we are assuming G is $C^\infty$ and therefore has bounded second derivative on $[z,b]$. But on this interval, w > z, so |z-w| = w-z, so we have

$$|G(z)| \leq C \int_b^z (w-z) dw,$$

which evaluates to

$$|G(z)| \leq - \frac 12 C (z-b)^2,$$

which is telling me that a positive quantity is less than, or equal to, a negative number! What on earth have I done wrong here?

2. May 17, 2012

### theorem4.5.9

I don't see this. I come up with
$G'(z)z - \int_b^z G''(\omega)\omega \mathrm{d}\omega$

3. May 17, 2012

### DonAntonio

Here: putting $\,\,u = G'\Longrightarrow u'=G''\,\,,\,\,v'=1\Longrightarrow v=w\,\,$ , integrating by parts gives $$wG'(w)\left.\right]_b^z-\int_b^zwG''(w)dw$$ I don't understand why you got $\,(z-w)\,$ in the right integral and also $\,G'(b)(z-b)\,$ instead of

$\,zG'(z)-bG'(b)$...?

DonAntonio

4. May 17, 2012

### Citan Uzuki

The estimate that $\left\vert \int_{a}^{b} f(x)\ \mathrm{d}x\right\vert \leq \int_{a}^{b} \vert f(x) \vert \ \mathrm{d}x$ only applies when a<b. You applied it to an integral where the lower limit is greater than the upper limit.

Last edited: May 18, 2012
5. May 17, 2012

### DonAntonio

He didn't do this: he applied absolute value to both sides...

DonAntonio

6. May 18, 2012

### Citan Uzuki

Sorry, that was a typo. Should be fixed now.

7. May 18, 2012

### DonAntonio

I see, but it still is correct, as (assuming the function $\,f\,$ is positive on [a,b] , $\,a<b\,$ ) : $$\left|\int_b^a fdx\right|=\left|-\int_a^bfdx\right|=\int_a^bfdx= \int_a^b|f|dx$$ and if the function's negative on that interval then $$\left|\int_b^a fdx\right|=\left|-\int_a^bfdx\right|=\left|\int_a^b(-f)dx\right|= \int_a^b(-f)dx=\int_a^b|f|dx$$ so for a general function with changes of sign in [a,b] we have the same inequality.

DonAntonio

8. May 18, 2012

### AxiomOfChoice

If one lets $v = w - z$, you still get $dv = dw$ as desired, but you also get what I wrote out. I'm allowed to do this, right?

9. May 18, 2012

### AxiomOfChoice

This makes a lot of sense to me. I'm still trying to hash out DonAntonio's objection to this statement, but it seems right.

10. May 18, 2012

### AxiomOfChoice

I don't think this is true. Consider $f(x) = 1$ on the interval $[0,1]$. We have

$$\left| \int_1^0 1 \cdot dx \right| = | 0 - 1| = 1.$$

Whereas

$$\int_1^0 |1| dx = 0 - 1 = -1 \not \geq 1.$$

So I think Citan is right.

11. May 18, 2012

### HallsofIvy

Staff Emeritus
His objection was that Citan Uzuki had originally written
$$\int_a^b f(x)dx\le \int_a^b |f(x)|dx$$
rather than
$$\left|\int_a^b f(x)dx\right| \le \int_a^b |f(x)|dx$$
to which he changed it after DonAntonio's post.

12. May 18, 2012

### theorem4.5.9

Choosing an additional constant doesn't change the integration by parts, as it will cancel out. I illustrate with the constant $z$ but it's true for any constant.

$$\int_b^z G'(w)dw = G'(w)(w-z)|_b^z - \int_b^z G''(w)(w-z) = 0 - \int_b^z G''(w)wdw + z\int_b^z G''(w) = zG'(w)_b^z - \int_b^z G''(w)wdw = zG'(z) - \int_b^z G''(w)wdw$$

13. Jun 1, 2012

### Boorglar

I think you're both right (theorem4.5.9 and AxiomOfChoice). Both results seem to be true, but the OP chose to write it with (w-z) in the integral (so he integrated by parts with z-b instead of just z). theorem's result doesn't contradict Axiom's answer, it simply doesn't lead to the same place as he wants. The mistake is what Citan mentioned: when he took absolute value of both sides, he didn't switch the bounds (from b--->z to z---->b), which of course gave a negative number.

(Just out of curiosity, what happens if we continue this integration by parts process as many times as we want? Won't it tell us that g(x) = 0 since each derivative vanishes as b and therefore the u(x)v(x) part will always cancel and we'll remain only with the integrals each time...? I know this is wrong but has anyone tried?)

14. Jun 1, 2012

### theorem4.5.9

I wasn't sure how axiom arrived at his integration by parts, though he cleared that up. My point was to show the equivalence of the two. The answer to his question had already been shared at that point.

You're right that the $u(x)v(x)$ is always zero, but the integral never goes away. Furthermore, there are not any precise bounds on the derivatives of $G$, so we cannot make the integral small.