Can we prove that the function is integrable for s>1?

[steven187]

hello all

im having trouble trying to show that the function
fn(x)=[x^(s-1)]/[e^x-1] is integrable for s>1
anybody have any ideas

also is it true that if a function is integrable it is equivalent to saying that the same function is riemann integrable?

thanxs

steven

 

[whozum]

There are some integrals that are not reimann integrable. I would say induction, do you know what that is?

 

[steven187]

integrable functions

so would that mean that if something is integrable would that imply that it is riemann integrable?

yes i know what mathematical induction is but can't see how that would help to show that something is integrable,

now i want to show it over the interval [o,infinity) this problem is only a small part of a larger problem which I am working on which involves

Sn(x)=x^(s-1)*[1-e^-Nx]/[e^x-1]

where |Sn(x)|<=fn(x) (the n isn't suppose to be there but I am leaving it there for consistency) by doing this i will be able to satisfy the conditions of the dominated convergence theorem, what do i do from here?

 

[matt grime]

When you say "is integrable" you ought to have in mind in which sense already, ie Riemann integrable, or lebesgue or the other one (Stiltjes?).

Over what interval are you integrating? Or is it an indefinite interval? What is the n for? IT doesn't appear in the definition of f_n

 

[steven187]

hello Matt

well in accordance to the dominated convergence theorem it requires the function to be riemann integrable, so would that mean that to be integrable it wouldn't imply that a function is riemann integrable? I am integrating over the interval [0,infinity) yeah about the n that was a mistake sorry :(

 

[matt grime]

Again, what do you mean by "is integrable"?

 

[steven187]

what i mean by is integrable in the sense of riemann integrable

 

[matt grime]

Ok, well, there is another method for interagting things, called lebesgue integration. Anything that is Riemann integrable i lebesgue integrable but NOT vice versa. I was just confused as to which method you were supposed to use.

 

[steven187]

im not actually familiar with lebesgue but i never knew that there there was a 3rd method Stiltjes,

 

[shmoe]

Stieltjes is a rarer kind of integration to learn, but you're bound to run into it if you keep studying number theory.

Anyway, to show that

$$\int_0^\infty\frac{x^{s-1}}{e^x-1}dx$$

converges when s>1, you'll want to break it into 2 intervals, say from (0,1) and (1, infinity). On (0,1) show the integrand is bounded by a constant times $$x^{s-2}$$ and then show this integral converges. On (0, infinity), you should be able to show it's bounded by a constant times $$e^{-x}$$, which is very convergent.

By the way, you're assuming s is real here, but for the full result you're after you probably want s any complex number with real part greater than 1. Are you happy with why it's enough to consider real exponents?

 

[steven187]

well i only know the basics when it comes to analytic number theory but i will be looking forward to it,

so basically you are saying that if the integral of a function converges then it is integrable, and to find out if it converges we do this by recognising another function which is known to converge to bound the function we are working with, and by breaking up the interval it would make it simpler to find this other function is that right?

in terms of s yes i am assuming that it is real and yes I am happy to only consider real exponents

 

[matt grime]

so basically you are saying that if the integral of a function converges then it is integrable,

An integral of a function does not converge. It exists. Sequences converge.

and to find out if it converges we do this by recognising another function which is known to converge

now the function converges? Again, a sequence converges.

to bound the function we are working with, and by breaking up the interval it would make it simpler to find this other function is that right?

 

[steven187]

i can see that if the integral of a function exists then it is integrable, and to find out if the integral of a function exists we need to find another function which is integrable over the same interval to bound the function we are working with,

what i don't get is where these functions came from
$$x^{s-2}$$
$$e^{-x}$$
i tried to look for something else to bound it by but i didnt get anywhere taking that path,

but now if we are trying to bound
$$\frac{x^{s-1}}{e^x-1}$$ by
$$x^{s-2}$$ , $$e^{-x}$$ how do we know if these are also integrable on [0,1] and [1,infinity) respectively or is that if we are able to integrate these two functions over those intervals then it is integrable,

by the way how does convergence relate to functions being integrable?

 

[shmoe]

An integral of a function does not converge. It exists. Sequences converge.

"convergence" is pretty standard terminology for an improper integral, at least in my experience.


steven, remember that our integral actually means:

$$\int_0^\infty\frac{x^{s-1}}{e^x-1}dx=\lim_{\delta\rightarrow 0^{+}}\int_{\delta}^{1}\frac{x^{s-1}}{e^x-1}dx+\lim_{T\rightarrow \infty}\int_{1}^{T}\frac{x^{s-1}}{e^x-1}dx$$

because the integrand is not actually defined at 0 we needed this delta limit as well. Breaking it up at 1 was an arbitrary choice, you could have used any positive number there. Our integral over (0,infinty) converges (or exists, or our function is integrable here, all mean the same thing) if both of these limits converge (to something finite).

Our function is bounded by a constant times $$e^{-x}$$ on [1, infinity). I chose this function because the bound is easy enough to prove (can you do this?) and we can show that $\int_{1}^{\infty}e^{-x}dx$ converges, in fact we can find it's value exactly (can you do this?). Since our original integrand was positive, this will prove its integral on (1,infinty) exists, that is:
$$\lim_{T\rightarrow \infty}\int_{1}^{T}\frac{x^{s-1}}{e^x-1}dx$$
exists and is finite. The other function will handle the interval (0,1).

 

[steven187]

thanxs shmoe your response made things more clearer but I wanted to ask
will the intergrand say f(x) be integrable if there exists an integrable function that "behaves" like f(x)? or does it need to be bounded by another function?

also would it be true to say that if the function values of a function are convergent then it is integrable or are there more conditions that need to be satisfied?

 

[shmoe]

Oh boy, I notice now that I had made a hasty typo in my first post that's propogated itself. The integrand is not bounded by a constant times $$e^{-x}$$ on (1, infinity). I was aiming for $$e^{-x/2}$$, though any $$e^{-ax}$$ where 0<a<1 would do nicely. Silly mistake that I should have noticed, sorry if you've been trying to prove my original claim!

thanxs shmoe your response made things more clearer but I wanted to ask
will the intergrand say f(x) be integrable if there exists an integrable function that "behaves" like f(x)? or does it need to be bounded by another function?

Depends what you mean by "behaves". If you have two positive continuous functions that are asymptotic as x->infinity then their integrals will both either converge or both diverge. Proving this would really amount to just showing the functions are bounded by constant multiples of one another though, which may be the case for most reasonable definitions of "behaves".

also would it be true to say that if the function values of a function are convergent then it is integrable or are there more conditions that need to be satisfied?

I'm not sure what you mean here, specifically the part in bold. If your sequence of functions converges to some function f, you can make no claims in general about whether f is integrable or not. You need some stronger conditions like dominated convergence.

 

[steven187]

thats ok I ended up bounding it by $$x^{s-2}e^{-x/2}$$ and it worked out nicely,what I mean by "function values of a function are convergent" for example
$$f(x)= \frac{1}{x}$$ where the values of f(x) approach 0 , so would it be true to say if f(x) is any function that converges to 0 (as x goes to infinity) then it is integrable? also if a continuous function f(x) has a vertical asymptote at b>0 then can we call f(x) integrable on the interval [0,b]

 

[Benny]

Isn't the integral from 1 to infinity of (1/x) not convergent? I think lim(x->inf)f(x) = 0 is a necessary but not sufficient condition.

 

[steven187]

well I can see how $$f(x)= \frac{1}{x}$$ is a function that converges to 0 but isn't integrable since the integral is not finite, I could come up with a few functions that look like $$f(x)= \frac{1}{x}$$ which are integrable for example $$e^{-ax}$$ would this have anything to do with the harmonic series being a divergent series? also if a continuous function f(x) has a vertical asymptote at b>0 then can we call f(x) integrable on the interval [0,b]? i tried it for $$f(x)= \frac{1}{x-1}$$ but the integral for this function is not finite, could such a function be ever integrable?

 

[shmoe]

$\int_1^\infty\frac{1}{x} dx$ is very much related to the harmonic series, you have:

$$\sum_{i=2}^{M}\frac{1}{i}\leq\int_1^M\frac{1}{x} dx\leq\sum_{i=1}^{M-1}\frac{1}{i}$$

so the harmonic series and the integral will either both converge or diverge. Of course it's the latter. Look up the "Integral Comparison Test" for infinite series.

$$e^{-x}$$ and $$x^{-1}$$ have the same overall decreasing shape heading to 0 as x->infinity but $$e^{-x}$$ is decreasing *much* faster. Consider that

$$\lim_{x\rightarrow\infty}\frac{x^M}{e^x}=0$$

for any value of M you like, no matter how large, and you can get some idea on what I mean by *much* faster. Exponential growth (or decay) is just mind bogglingly faster than any polynomial you can dream up. You have to examine the functions on a finer scale than just heading to 1 as x goes to infinity to determine convergence of their integral, the 'speed' at which they go is important.

The situation for a vertical asymptote is the same, it depends on how it's heading up to infinity. Find the integral

$$\int_\delta^1\frac{1}{x^r}dx$$

for various values of r and see what happens as you let delta->0.

 

[steven187]

well this $$\lim_{x\rightarrow\infty}\frac{x^M}{e^x}=0$$ is an interesting limite how would one prove such a limite? I tried to do it with L' hospital's rule but i didnt get anywhere that path, if the 'speed' at which they go is important i tried to think of a function that would converge faster than the $$e^{-x}$$ such as $$f(x)= \frac{1}{x^{x^x}}$$ (by the way how would you differentiate and integrate $$f(x)=x^x$$), so how would we prove if such a function is integrable? also in terms of $$\int_\delta^1\frac{1}{x^r}dx$$ I tried it for r=1,2 and 3 but they all are not finite so i guess i would conclude that it is not finite for all values of r

 

[shmoe]

For $$x^M/e^x$$, if M<=0 the result is obvious. We can assume M is an integer (if not take any integer larger than it), so apply l'hopital M times.

Alternatively knowing $$x<e^x$$ you can write:

$$\frac{x^M}{e^x}=(M+1)^M\frac{(x/(M+1))^M}{(e^{x/(M+1)})^{M+1}}\leq(M+1)^M\frac{1}{e^{x/(M+1)}}$$

and the limit follows.

Something like $$x^x$$ grows *much* faster than $$e^x$$ as well. $$x^x=e^{x\log x}$$ is how you could differentiate it (it doesn't have a 'nice' integral, there's a recent thread in the general discussion about this) and gives you some idea of how much larger it is. It will grow faster than any $$e^{Mx}$$ for any constant choice of M, but loses out to any higher power in the exponent like $$e^{x^r}$$ where r>1. This just amounts to comparing log(x) with powers of x, which is very much related to comparing powers of x to e^x. Since $$1/x^{x^x}<1/x^x<1/e^x$$ for large enough x, the integral of the first two over (1, infinty) will also converge.

Try smaller values of r for that other integral. What if r<1? Also you might want to compare the graphs of $$1/x, 1/x^2, 1/x^{1/2}$$ etc. near 0.