Something weird with the fundamental theorem of calculus

[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?

 

[theorem4.5.9]

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)

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

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

 

[DonAntonio]

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)

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

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

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

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?

DonAntonio

 

[Citan Uzuki]

What on Earth have I done wrong here?

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.

 

[DonAntonio]

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

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

DonAntonio

 

[Citan Uzuki]

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

 

[DonAntonio]

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

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

 

[AxiomOfChoice]

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

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

 

[AxiomOfChoice]

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.

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.

 

[AxiomOfChoice]

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

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.

 

[HallsofIvy]

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.

 

[theorem4.5.9]

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

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.

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

 

[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?)

 

[theorem4.5.9]

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.

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.

(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?)

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.