# Dominated Convergence theorem

• steven187
In summary, the dominated convergence theorem allows for the interchange of the summation and integral signs around in a proof. This theorem is useful for proving relationships between different functions.
steven187
hello all

I have been researching about the dominated convergence theorem so that i could use it to prove the relationship between the gamma function and the Riemann zeta function, what i need is the simple version of the dominated convergence theorem to be able to swap the summation sign and the integral sign around in order to prove the relationship, after doing some research i have only found information relating to the measure theory but i don't understand how that relates to how I am going to use it, anybody have any ideas

Are you trying to prove something like $\sum_{n=1}^{\infty}\int_{0}^{\infty}x^{s-1}e^{-nx}dx=\int_{0}^{\infty}x^{s-1}/(e^x-1)dx?$

In this case your sequence of functions is bounded in absolute value by $$x^{\sigma-1}/(e^x-1)$$, where $$\sigma$$ is the real part of s and your sequence of functions is uniformly convergent on compact sets, so you should be able to apply even the strictest versions of dominated convergence (of course we're also assuming $$\sigma>1$$ here).

Or are you having trouble finding a statement for the Riemann integral? If a function is Riemann integrable, it is Lebesgue integrable and these integrals are equal, so you can translate a lebesgue theorem to a riemann one (provided you meet all hypotheses of course). In any case, the following is sufficient here (this is from Rudin but should be in most analysis texts in some form):

$$g,\ f_n$$ Riemann integrable on [t,T] for all $$0<t<T<\infty$$, $$|f_n|<g$$, $$f_n$$ converges to f uniformly on compact subsets and $\int_{0}^{\infty}g(x)dx$ is finite (and exists), then $\lim_{n\rightarrow\infty}\int_{0}^{\infty}f_n(x)dx=\int_{0}^{\infty}f(x)dx$

well yes I am trying to prove this

$\sum_{n=1}^{\infty}\int_{0}^{\infty}x^{s-1}e^{-nx}dx=\int_{0}^{\infty}x^{s-1}/(e^x-1)dx?$

what I am trying to do is look for a statement relating to the dominated convergence theorem which allows me to interchange the summation sign and the integral sign, like are there certain conditions that are to be met to do such a step, or is it just a statement which needs to be stated in the proof, i think an explanation in simple terms on what this theorem means and how it relates to proving this problem would help, in terms of
$$\sigma$$ i didnt understand it, especially if we only take the real part of s how do we know that it would be bounded in absolute value?

steven187 said:
what I am trying to do is look for a statement relating to the dominated convergence theorem which allows me to interchange the summation sign and the integral sign,

Then look at the statement I provided in my last post.

steven187 said:
like are there certain conditions that are to be met to do such a step, or is it just a statement which needs to be stated in the proof,

If you want to apply any theorem there are always conditions that have to be met first.

steven187 said:
i think an explanation in simple terms on what this theorem means and how it relates to proving this problem would help,

Well it gives sufficient conditions on when you can say the limit of the integrals of a sequence of functions is equal to the integral of the limit of those functions, i.e. the swapping of the limit with the integral sign. For this problem your infinite sum is hiding the limit, the $$f_n$$ in the theorem is just the nth partial sum, $f_n(x)=\sum_{k=0}^{n}x^{s-1}e^{-kx}$, f is the limit of these sums, $$x^{s-1}/(e^x-1)$$ and here we can take g to be $$x^{\sigma-1}/(e^x-1)$$ by the triangle inequality and what follows below. We have uniform convergence compact sets to the right of real part s equals 1. So the theorem applies and the swapping of the infinite sum and the integral is justified.

steven187 said:
in terms of $$\sigma$$ i didnt understand it, especially if we only take the real part of s how do we know that it would be bounded in absolute value?

You are familiar with complex exponents, right? If $$s=\sigma+it$$ where $$\sigma$$ and t are real then $$x^{s-1}=x^{\sigma}x^{it}=x^{\sigma-1}e^{it\log{x}}=x^{\sigma-1}(\cos{(t\log{x})}+i\sin{(t\log{x})})$$, so $$|x^{-1}s|=|x^{\sigma-1}|=x^{\sigma-1}$$. This shouldn't be new to you. (remember x>0 here)

## 1. What is the Dominated Convergence Theorem?

The Dominated Convergence Theorem is a fundamental theorem in real analysis that describes the behavior of a sequence of functions whose limit is dominated by another function. It is commonly used to evaluate the limit of integrals and series.

## 2. What is the significance of the Dominated Convergence Theorem?

The Dominated Convergence Theorem allows for the interchange of limits and integrals, which is a crucial tool in many areas of mathematics, including probability theory, harmonic analysis, and measure theory. It also provides a way to evaluate limits of functions that do not have explicit formulas.

## 3. What are the conditions for the Dominated Convergence Theorem to hold?

The Dominated Convergence Theorem requires that the sequence of functions is pointwise convergent, that is, the limit of the sequence at each point exists. Additionally, there must be a function that dominates the sequence of functions in absolute value, and this dominating function must be integrable.

## 4. Can the Dominated Convergence Theorem be used for infinite series?

Yes, the Dominated Convergence Theorem can be used for both infinite series and integrals. In the case of series, it states that if a sequence of functions is dominated by an integrable function, then the limit of the series is equal to the integral of the limit of the sequence of functions.

## 5. Are there any applications of the Dominated Convergence Theorem?

The Dominated Convergence Theorem has numerous applications in mathematics, including probability theory, functional analysis, and differential equations. It is also used in statistics, physics, and engineering to evaluate limits and integrals involving complicated functions.

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