# Thread: Integrability

1. ## 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.

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

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

3. 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

4. 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 $x\nearrow b$.

5. 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?

6. 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?

7. 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

8. 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 $\lim_{x\nearrow b}\frac{1}{g(x)} \int^x_a f(t)g(t)\,dt = 0$.

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.

9. 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

10. 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.

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.