When can the limit be brought inside the integral?

1. May 13, 2005

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

?

2. May 13, 2005

dextercioby

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.

3. May 13, 2005

Galileo

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:

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

4. May 13, 2005

saltydog

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.

5. May 13, 2005

dextercioby

Did u mean

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

Daniel.

6. May 13, 2005

saltydog

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

Is:

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

7. May 13, 2005

Hoplite

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,

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

8. May 13, 2005

quetzalcoatl9

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: May 13, 2005
9. May 13, 2005

shmoe

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.

10. May 14, 2005

Hoplite

Thanks.

11. May 15, 2005

HallsofIvy

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.

12. May 15, 2005

mathwonk

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 fucntion 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: May 15, 2005