# Rate of divergence of Lebesgue integrals.

• Feb 9th 2011, 03:04 AM
Jezuz
Rate of divergence of Lebesgue integrals.
Hi!
I have a couple of problems from the book: Mathematics for physics and physicists by W. Appel, which I can not solve.

The problems are:

1) Show that
$\displaystyle \int_1^x dt \frac{| \sin t|}{t} \underset{x\rightarrow \infty}{\sim} \frac{2}{\pi} \log(x)$

and

2) Show that
$\displaystyle \int_1^\infty dt \frac{e^{-xt}}{\sqrt{1+t^2}} \underset{x \rightarrow 0^+}{\sim} - \log(x)$.

Related to this is given first the defintion:
$\displaystyle f \underset{b}{\sim} g \Leftrightarrow \forall \epsilon>0, \exists \delta>0 \textrm{ s.t. } |x - b|<\delta \Rightarrow |f(x) - g(x)| < \epsilon |g(x)|$ ,

which in a slightly less general way, but perhaps more comprehensive way can be written as

$\displaystyle \frac{f(x)}{g(x)} \underset{x\rightarrow b}{\rightarrow} 1 .$

A theorem is also stated which says:
Let $\displaystyle g$ be a non-negative measurable function on $\displaystyle [ a, b [$, and let $\displaystyle f$ be a measurable function on $\displaystyle [ a, b [$. Assume that $\displaystyle g$ is not integrable and that $\displaystyle f$ and $\displaystyle g$ are both integrable on any interval $\displaystyle [ a, c ]$ where $\displaystyle c < b$. Then the asymptotic comparison between the functions extend to the integrals in the way

$\displaystyle f \underset{b}{\sim} g \Rightarrow \int_a^x f \underset{x \rightarrow b}{\sim} \int_a^x g$
and
$\displaystyle f \underset{b}{=} o(g) \Rightarrow \int_a^x f = \underset{x\rightarrow b}{o} \left( \int_a^x g \right).$

My attempt at a solution:
I have tried to solve 1) by noting that

$\displaystyle \frac{2}{\pi} \log (x) = \int_1^x dt \frac{2}{\pi t}$.

After writing it like this I should be able to utilize the first part of the theorem.
The next step would then be to show that $\displaystyle \frac{|\sin x|}{x} \underset{x \rightarrow \infty}{\sim} \frac{2}{\pi x}$. However, this does not seem to hold as $\displaystyle \frac{|\sin x|}{x} \left(\frac{2}{\pi x}\right)^{-1} = \frac{\pi |\sin x|}{2}$ and this does not tend to 1 as $\displaystyle x \rightarrow \infty$ so I cannot conclude that the integrands are asymptotically equivalent.

For problem 2) I have even more problems. In this case the x-dependence is not in the limits of the integral and the theorem cannot be applied directly. I have attempted to rewrite the integral by the use of different variable substitutions, but I cannot obtain the required form.

I have worked quite extensively on both of these problems, but I cannot seem to solve them. They are also the only two problems related to the theorem in the chapter I am studying. Either I do not fully comprehend the theorem, or there is something missing. I am a bit unsure about the definition for equivalence between two functions $\displaystyle f \underset{b}{\sim} g$ since, in the book, the exact meaning of this is only properly defined for sequences. However, the definition I give above is the direct generalization of this and should be correct.

I am grateful for any suggestions, or help I can get.
• Feb 10th 2011, 02:43 AM
Jezuz
I have made some further attempts to solve this.

It seems like the integrands in problem 1) are actually related as
$\displaystyle \frac{|\sin x|}{x} - \frac{2}{\pi x} \underset{x\rightarrow \infty}{=} \mathcal O \left( \frac{2}{\pi x} \right) \underset{x\rightarrow \infty}{=} \mathcal O \left( \frac{ |\sin x| }{x} \right).$
Could it be that the definition of what is meant by $\displaystyle f \underset{b}{\sim} g$ that I have used is wrong? What is the most common definition for this? In the book I am studying, the definition is only given for sequences. It is stated that for two sequences $\displaystyle { a_n }_{n \in \mathbb N}$ and $\displaystyle { b_n }_{n \in \mathbb N}$ we have the definition $\displaystyle a_n \sim b_n$ iff $\displaystyle a_n - b_n = o(b_n)$ or equivalently $\displaystyle a_n - b_n = o (a_n)$. Perhaps this is not generalized to functions in the way I thought.

Alternatively, there is some other trick that should be made in order to solve problems 1) and 2), or perhaps (but probably very unlikely), the problems are wrong.

I would be very thankful for any help, suggestions or comments.
• Feb 10th 2011, 05:45 AM
roninpro
I only had time to look at part one, and I think that what you are doing in correct. (And actually, you could use L'Hopital's rule to take the limit you describe, and get the same result as your theorem.) Asymptotic and Big-O notation seem pretty flexible, so I'm wondering if it is permissible to just put an upper bound on the expression: $\displaystyle \displaystyle \frac{\pi}{2}|\sin x|\leq \frac{\pi}{2}$ and then conclude.
• Feb 10th 2011, 07:28 AM
Opalg
Quote:

Originally Posted by Jezuz
1) Show that
$\displaystyle \int_1^x dt \frac{| \sin t|}{t} \underset{x\rightarrow \infty}{\sim} \frac{2}{\pi} \log(x)$

2) Show that
$\displaystyle \int_1^\infty dt \frac{e^{-xt}}{\sqrt{1+t^2}} \underset{x \rightarrow 0^+}{\sim} - \log(x)$.

For $\displaystyle t$ in the interval $\displaystyle [k\pi,(k+1)\pi]$, $\displaystyle \frac{\left|\sin t\right|}{(k+1)\pi} \leqslant\frac{\left|\sin t\right|}{t} \leqslant\frac{\left|\sin t\right|}{k\pi}$. But the integral of $\displaystyle \left|\sin t\right|$ over that interval is 2. Therefore $\displaystyle \displaystyle \frac2{(k+1)\pi} \leqslant \int_{k\pi}^{(k+1)\pi}\frac{\left|\sin t\right|}t\,dt \leqslant \frac2{k\pi}$. Sum that from $\displaystyle k=1$ to $\displaystyle n$, and use the fact that $\displaystyle \sum_{k=1}^n\frac1k \sim \log n$, to see that $\displaystyle \displaystyle \int_1^{n\pi}\frac{\left|\sin t\right|}t\,dt \sim \frac2\pi\log (n\pi)$ (I'm using $\displaystyle \sim$ in an informal way here, to mean "approximately equal", but you ought to be able to reconcile that with the formal definition). Once you have the result for $\displaystyle x=n\pi$, it should be easy to deduce it for general $\displaystyle x$.

The key point in problem 1) is that the mean value of $\displaystyle \left|\sin t\right|$ over any interval of length $\displaystyle \pi$ is $\displaystyle \frac2\pi$.

The result in 2) looks harder. The only route into it that I can see is to think in terms of Laplace transforms. If we write the Laplace transform of the function $\displaystyle f(t)$ as $\displaystyle F(x) = \mathcal{L}\bigl(f(t)\bigr) = \displaystyle\int_0^\infty\!\!\! f(t)e^{-xt}\,dt$, then it is known that $\displaystyle \mathcal{L}\bigl(\frac{f(t)}t\bigr) = -\int F(x)\,dx$ (with a suitable constant of integration).

The Laplace transform of the constant function 1 (on the interval $\displaystyle [0,\infty)$) is $\displaystyle 1/x$. So the Laplace transform of $\displaystyle 1/t$ should be $\displaystyle -\int \frac1x\,dx = -\log x$. But $\displaystyle \frac1{\sqrt{1+t^2}} \sim \frac1t$ for large values of $\displaystyle t$, and hence $\displaystyle \displaystyle \mathcal{L}\Bigl(\tfrac1{\sqrt{1+t^2}}\Bigr) = \int_0^\infty \frac{e^{-xt}}{\sqrt{1+t^2}}\,dt \sim -\log x$.

Unfortunately, that argument is highly suspect because $\displaystyle 1/t$ does not have a Laplace transform (the improper integral $\displaystyle \int_0^\infty\frac{e^{-xt}}t\,dt$ does not converge). I do not know whether the argument can be patched up to look convincing. But at least it gives the right answer!
• Feb 10th 2011, 01:14 PM
Jezuz
I try to follow the steps you have outlined. I am more and more suspecting that the definition $\displaystyle f \underset{b}{\sim} g$ really means that $\displaystyle f = \mathcal O (g)$ in the limit $\displaystyle x \rightarrow b$.

Using L'Hopital's rule it seems to be the case, since

$\displaystyle \lim_{x\rightarrow b} \frac{ \int_a^x f }{ \int_a^x g} = \lim_{x \rightarrow b} \frac{f(x)}{g(x)}$

and on the other hand if
$\displaystyle f \underset{ b}{=} \mathcal O (g) \Rightarrow \lim_{x \rightarrow x} \frac{ f(x)}{g(x)} \leq K$

for some constant $\displaystyle K > 0$ and we see that

$\displaystyle f \underset{b}{=} \mathcal O (g) \Rightarrow \int_a^x f \underset{x\rightarrow b}{=} \mathcal O \left( \int_a^x g \right)$ .

Thank you very much for the help!

Problem 2) seems to be a bit harder, but I will try fiddling around with Laplace transforms and see where I get.