# When can the limit be brought inside the integral?

• Hoplite
In summary, the criteria for when the limit and integral can be interchanged depend on the type of integral being used (Riemann or Lebesgue). For uniformly convergent sequences of Riemann integrable functions, the integral of the limit is equal to the limit of the integrals. For Lebesgue integrable functions, there are various criteria such as the dominated convergence theorem that can determine when the limit and integral can be interchanged. However, for functions like the zeta function that are not continuous or bounded, these criteria may not apply.
Hoplite
When is it alright to bring the limit within the integral?

In other words, when is it true to say...

lim∫f(x)dx = ∫limf(x)dx

?

Hmm,well,a Riemann integration involves itself.Sometimes,for improper integrals another limit appears explicitely,so the situation is under debate,there's no general case.

Be more specific.

Daniel.

You should be more specific, the limit $\lim_{x\to a}f(x)$ if it exists is just a number and so is a definite integral.

In the case where you have a sequence of continuous functions $\{f_n\}_{n=0}^\infty$ which converge uniformly to f on a bounded interval [a,b], then:

$$\lim_{n\to \infty}\int_a^b f_n(x)dx=\int_a^bf(x)dx$$

Is this what you mean:

Under what conditions is the following equality true? (can't find a question mark over equal sign symbol)

$$\mathop\lim\limits_{c\to h}\int_a^b f(x,c)dx=\int_a^b\mathop\lim\limits_{c\to h}f(x,c)dx$$

I dont'n know.

Did u mean

$$\substack{\displaystyle{!}\\ \displaystyle{=}}$$...? (click on the code)

Daniel.

dextercioby said:
Did u mean

$$\substack{\displaystyle{!}\\ \displaystyle{=}}$$...? (click on the code)

Daniel.

Thanks Daniel. Always nice to learn new LaTex code. For the record then:

Is:

$$\mathop\lim\limits_{c\to h}\int_a^b f(x,c)dx\substack{\displaystyle{ ? }\\ \displaystyle{ = }}\int_a^b\mathop\lim\limits_{c\to h}f(x,c)dx$$

What I mean is:

$$\lim_{k\rightarrow\infty} \int f_{k}(z)dz = \int\lim_{k\rightarrow\infty}f_{k}(z)dz$$

In this case, $$f_{k}(z)$$ is the zeta function,

$$f_{k}(z) = \sum^k_{n=1}\frac{1}{n^z}$$

My guess would be that this is not allowed:

$$\mathop\lim\limits_{c\to h}\int_a^b f(x,c)dx = \mathop\lim\limits_{c\to h} \mathop\lim\limits_{n\to \infty} \sum^n_{i=1}f(x_i,c)\Delta{x}_i \left\{\begin{array}{cc}x_i\in[a,b]\\\Delta{x}_i = x_{i+1} - x_i \end{array}\right\}$$

and the limits don't "commute" in general, for example:

$$\mathop\lim\limits_{x\to 0}\mathop\lim\limits_{y\to 0} 1 + \frac{y}{x} = 1$$

but,

$$\mathop\lim\limits_{y\to 0}\mathop\lim\limits_{x\to 0} 1 + \frac{y}{x} = \infty$$

Last edited:
Hoplite said:
What I mean is:

$$\lim_{k\rightarrow\infty} \int f_{k}(z)dz = \int\lim_{k\rightarrow\infty}f_{k}(z)dz$$

In this case, $$f_{k}(z)$$ is the zeta function,

$$f_{k}(z) = \sum^k_{n=1}\frac{1}{n^z}$$

This Dirichlet series is uniformly convergent in half planes real part of z >=d, where d>1, so you're good as long as your path of integration lies in one of these.

Thanks.

In general you can interchange "limit" and "integral" as long as both are uniformly convergent. I won't say (because I don't know) if that is necessary as well as sufficient.

This is not my expertise, but the following may be useful to someone.

there are two common definitions of an integral riemanns and lebesgues. riemann's integral is defined for bounded functions which are continuous almost everywhere, and lebesgue's is defined for essentially all (positive) functions (but it can be infinite).

for uniformly convergent sequences of riemann integrable functions, the integrals do converge to the integral of the limit.

for lebesgue integrable functions, there are various sufficient criteria. the most useful is probably the dominated convergence theorem: if there is a lebesgue integrable function with finite integral which dominates the absolute values of all function in the sequence, then pointwise convergence a.e. is sufficient for the integrals to converge to the integral of the limit.

these matters can be read up in any book on measure theory or integration.

of course the zeta function is of primary interest near points like z=1 where it is not continuous or even bounded.

Last edited:

## 1. What is the general rule for bringing the limit inside the integral?

The general rule is that if the integrand is continuous and the limits of integration are finite, then the limit can be brought inside the integral.

## 2. Can the limit be brought inside the integral if the integrand is discontinuous?

No, the limit cannot be brought inside the integral if the integrand is discontinuous. In such cases, the integral must be evaluated using methods such as the Cauchy principal value.

## 3. Are there any exceptions to the rule for bringing the limit inside the integral?

Yes, there are some exceptions to the rule. For example, if the integral is improper, the limit may need to be evaluated separately before bringing it inside the integral. Also, if the integrand has an essential singularity at one of the limits of integration, the limit cannot be brought inside the integral.

## 4. How does bringing the limit inside the integral affect the value of the integral?

Bringing the limit inside the integral does not change the value of the integral as long as the general rule is followed. This is because the limit is evaluated after the integration has been performed.

## 5. Why is it sometimes necessary to bring the limit inside the integral?

Bringing the limit inside the integral can be useful when dealing with certain types of integrals, such as those involving infinite limits or integrands with many terms. It can also simplify the integration process by allowing us to use known integration rules and techniques.

• Calculus
Replies
31
Views
891
• Calculus
Replies
2
Views
261
• Calculus
Replies
3
Views
957
• Calculus
Replies
8
Views
286
• Calculus
Replies
4
Views
299
• Calculus
Replies
3
Views
1K
• Calculus
Replies
11
Views
2K
• Calculus
Replies
16
Views
2K
• Calculus
Replies
20
Views
2K
• Calculus
Replies
3
Views
2K