# Integrability

• May 7th 2010, 11:33 AM
AidenMatthew
Integrability
Suppose $f$ is continuous and $F(x)=\int_a^x f(t)dt$ bounded on $[a,b)$. Given $g>0, g'\geq 0$ and $g'$ locally integrable on $[a,b)$ and $lim_{ x\rightarrow b^-} g(x) =$ infinity. prove
for p>1
$\displaystyle{lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} \int^x_a f(t)g(t)dt = 0}$

I think I should use integration by parts... but I am not sure. Any help would be greatly appreciated.
• May 7th 2010, 11:40 AM
AidenMatthew
In addition to this, I want to show that if $\int_a^b f(t)dt$ converges, then we have

$lim_{x\rightarrow b^-} \frac{1}{g(x)} \int^x_a f(t)g(t)dt = 0$
• May 7th 2010, 05:20 PM
BBrown
Quote:

Originally Posted by AidenMatthew
Suppose $f$ is continuous and $F(x)=\int_a^x f(t)dt$ bounded on $[a,b)$. Given $g>0, g'\geq 0$ and $g'$ locally integrable on $[a,b)$ and $lim_{ x\rightarrow b^-} g(x) =$ infinity. prove
for p>1
$\displaystyle{lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} \int^x_a f(t)g(t)dt = 0}$

I think I should use integration by parts... but I am not sure. Any help would be greatly appreciated.

If you know $lim_{ x\rightarrow b^-} g(x) =\infty$, don't you also know $lim_{ x\rightarrow b^-} \frac{1}{[g(x)]} = 0$ and therefore $lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p}=0$... I might be jumping the gun here
• May 8th 2010, 12:41 PM
Opalg
Quote:

Originally Posted by AidenMatthew
Suppose $f$ is continuous and $F(x)=\int_a^x f(t)dt$ bounded on $[a,b)$. Given $g>0, g'\geq 0$ and $g'$ locally integrable on $[a,b)$ and $lim_{ x\rightarrow b^-} g(x) =$ infinity. prove
for p>1
$\displaystyle{lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} \int^x_a f(t)g(t)dt = 0}$

I think I should use integration by parts.

I agree, integration by parts is definitely the right way to go about this. In fact, $\int_a^x\!\!\!f(t)g(t)\,dt = \Bigl[F(t)g(t)\Bigr]_a^x - \int_a^xF(t)g'(t)\,dt$. When you divide by $\bigl(g(x)\bigr)^p$, the term in square brackets will go to 0 as $x\nearrow b$. So you just need to estimate the size of the integral on the right.

You know that F(t) is bounded, say $|F(t)|\leqslant M$ for some constant M. Then $\Bigl|\int_a^xF(t)g'(t)\,dt\Bigr| \leqslant \int_a^xMg'(t)\,dt = M\bigl(g(x)-g(a)\bigr)$. Again, when you divide by $\bigl(g(x)\bigr)^p$, that expression will go to 0 as $x\nearrow b$.
• May 8th 2010, 01:45 PM
BBrown
Quote:

Originally Posted by AidenMatthew
In addition to this, I want to show that if $\int_a^b f(t)dt$ converges, then we have

$lim_{x\rightarrow b^-} \frac{1}{g(x)} \int^x_a f(t)g(t)dt = 0$

I agree with the above argument, do you see it AidenMattew? In my assumptions, I was not looking at the integral for boundedness.

How would you go about proving this second part now though (above), Opalg? Use instead that $F(x) = \int_x^b f(t)dt$ and do parts integration again?
• May 8th 2010, 01:47 PM
AidenMatthew
Thats what I was thinking, and then use the fact that $\int^b_b f(t)dt = 0$... I'm still a little uneasy about how to set it up with this difference though... but I think the arguments are similar... doesnt convergence imply boundedness so the same argument goes?
• May 8th 2010, 09:15 PM
casanova
using the same integration by parts, at the very end of moving things around with what you said above you have

$lim_{x\rightarrow b^-}\frac{1}{g(x)} \int^x_a F(t)g'(t)dt$

right? you know what $lim_{x\rightarrow b^-}\frac{1}{g(x)}$ is and you know $\int^x_a F(t)g'(t)dt$ is bounded, so you're done unless i'm missing something
• May 9th 2010, 02:04 AM
Opalg
Quote:

Originally Posted by AidenMatthew
In addition to this, I want to show that if $\int_a^b f(t)dt$ converges, then we have

$lim_{x\rightarrow b^-} \frac{1}{g(x)} \int^x_a f(t)g(t)dt = 0$

This is a bit more subtle than the case where p>1. But you have the additional information that $F(b) = \int_a^b\!\!\! f(t)dt$ exists as the limit $\textstyle\lim_{x\nearrow b}F(x)$. So F is continuous on the left at b, and given $\varepsilon>0$ there exists a point $x_0 such that $|F(x)-F(t)|<\varepsilon$ whenever x and t lie in the interval $[x_0,b)$.

To estimate $\int^x_a\!\!\! f(t)g(t)\,dt$, split it up as $\int^{x_0}_a\!\!\! f(t)g(t)\,dt + \int_{x_0}^x\!\!\! f(t)g(t)\,dt$. The first of those two integrals is independent of x, so when we divide by g(x) and let $x\nearrow b$, it will go to 0. Thus we need only look at the second of the two integrals, $\int_{x_0}^x\!\!\! f(t)g(t)\,dt$.

Integrating by parts as previously, we get $\int_{x_0}^x\!\!\! f(t)g(t)\,dt = F(x)g(x) - F(x_0)g(x_0) - \int_{x_0}^x\!\!\! F(t)g'(t)\,dt$. Use the fact that $g(x) = g(x_0) + \int_{x_0}^x\!\!\! g'(t)\,dt$ to write this as $\int_{x_0}^x \!\!\! f(t)g(t)\,dt = F(x)g(x_0) - F(x_0)g(x_0) + \int_{x_0}^x \bigl(F(x)-F(t)\bigr)g'(t)\,dt$. The first two terms on the right side of that equation are bounded, so when we divide by g(x) and let $x\nearrow b$ they will go to 0. Thus we need only estimate the size of the integral term.

For that, using the inequality $|F(x)-F(t)|<\varepsilon$, we get $\Bigl|\int_{x_0}^x \bigl(F(x)-F(t)\bigr)g'(t)\,dt\Bigr|\leqslant \varepsilon|g(x)-g(x_0)|$. When we divide by g(x), that gives a result that is essentially less than $\varepsilon$.

Putting everything together, that should show that $\lim_{x\nearrow b}\frac{1}{g(x)} \int^x_a f(t)g(t)\,dt = 0$.

Quote:

Originally Posted by casanova
using the same integration by parts, at the very end of moving things around with what you said above you have

$lim_{x\rightarrow b^-}\frac{1}{g(x)} \int^x_a F(t)g'(t)dt$

right? you know what $lim_{x\rightarrow b^-}\frac{1}{g(x)}$ is and you know $\int^x_a F(t)g'(t)dt$ is bounded, so you're done unless i'm missing something

I don't think it's as simple as that, because you don't know that $\int^x_a\!\!\! F(t)g'(t)dt$ is bounded. In fact, $g'(x)\to \infty$ as $x\nearrow b$, so that integral might well be unbounded.
• May 9th 2010, 06:57 AM
AidenMatthew
Quote:

Originally Posted by Opalg
Integrating by parts as previously, we get $\int_{x_0}^x\!\!\! f(t)g(t)\,dt = F(x)g(x) - F(x_0)g(x_0) - \int_{x_0}^x\!\!\! F(t)g'(t)\,dt$. Use the fact that $g(x) = g(x_0) + \int_{x_0}^x\!\!\! g'(t)\,dt$ to write this as $\int_{x_0}^x \!\!\! f(t)g(t)\,dt = F(x)g(x_0) - F(x_0)g(x_0) + \int_{x_0}^x \bigl(F(x)-F(t)\bigr)g'(t)\,dt$. The first two terms on the right side of that equation are bounded, so when we divide by g(x) and let $x\nearrow b$ they will go to 0. Thus we need only estimate the size of the integral term.

Why can we bring F(x) into the integral equation? It's not a constant? And I know continuity implies integrability--is this an iff statement? I didnt know I could say for sure F was continuous...

Thanks so much! You really made the rest clear for me
• May 9th 2010, 07:22 AM
Opalg
Quote:

Originally Posted by AidenMatthew
Why can we bring F(x) into the integral equation? It's not a constant?

It's a constant as far as t is concerned, and the integral is with respect to t. After all, x is the upper limit of integration in the integral $\int_{x_0}\kern-5pt\mathop{\phantom{\big|}}^{\color{red}x}\!\!\!\l dots dt
$
, so x has to be regarded as a constant for the purposes of that integral.

Quote:

Originally Posted by AidenMatthew
And I know continuity implies integrability--is this an iff statement? I didnt know I could say for sure F was continuous...

The reason that F is continuous at b comes from the definition of an improper integral. To say that the integral $F(b) = \int_a^b\!\!\!f(t)\,dt$ converges means that $F(b) = \textstyle\lim_{x\nearrow b}F(x)$, which is the same as saying that F is continuous at b.