Solve Integral Limits w/ Monotone & Lebesgue Convergence

In summary, the conversation discusses how to solve for the limits of two integrals, using either the monotone convergence theorem or the Lebesgue dominated convergence theorem. The equations given approach 1 as n approaches infinity, but this does not mean the integrals equal infinity. Instead, it is possible to calculate the integrals exactly for n>0. One approach is to use a substitution and Euler functions, but this leads to a more complicated solution. Another approach is to use the fact that the limit of the integrals is equal to the limit of the function, and find a function g that is greater than or equal to the integrand. This allows us to use the Lebesgue dominated convergence theorem to solve the integral, giving a
  • #1
island-boy
99
0
hello, how do you solve for the limits of the following integrals:

lim as n -> infinity of the integral from 0 to n of [ (1 - x/n)^n times exp(x/2)]

and

lim as n -> infinity of the integral from 0 to n of [ (1 + x/n)^n times exp(-2x)]

Can you solve these using either the monotone convergence theorem or the lebesgue dominated convergence theorem?

and doesn't (1 - x/n)^n and (1 + x/n)^n approach 1 as n approaches infinty? so does that mean I just have to solve the integral of the exponential for both cases from 0 to infinity, which would mean it equals infinity?
 
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  • #2
One can calculate

[tex] \int_{0}^{n} \left(1-\frac{x}{n}\right)^{n} e^{\frac{x}{2}} \ dx [/tex]

exactly for n>0.

And the same goes for the second integral.

However, the limits seem to be 2 and 1, respectively.

Daniel.
 
  • #3
thanks for the quick reply.

am I correct in assuming this is done through integration by parts?
hold on, I'll try to solve it this way
 
  • #4
Well, a substitution [itex] y=1-\frac{x}{n} [/itex] is useful because it leads you to ans answer which uses gamma-Euler functions. However, i don't think the limit of the integral can be computed easily from there. I think those sequences of integrals do converge, though.

Daniel.
 
  • #5
forgive my ignorance, but what is a gamma-euler function?

also, if i use substitution and let y = 1 - x/n, which will give me the new integral: -ny^n exp[n(1-y)/2] dy...I can't seem to solve this integral also.

the topic we are studying btw is abstract integration and borel measures.
we are suppose to solve the limit of the integrals using convergence theorems (monotone or dominated).

since the first thing I should do is to let fn(x) = (1-x/n)^nexp(x/2)...however, I don't think this fn is positive for all x, so I can't use monotone convergence thm, I should therefore use dominated convergence and find a fcn g >= fn for every n and x. my problem is I can't find the function g.

If I were able to find g, then the problem becomes the simpler problem of finding the integral of f, where lim fn = f. For some reason I'm thinking the lim of fn is exp(x/2), which is leading me to the conclusion that its integral is infinite. Is this correct?

thanks again for helping.
 
  • #6
Something which might prove itself to be useful is

[tex] \lim_{n\rightarrow +\infty} \left(1-\frac{x}{n}\right)^{n} = e^{-x} [/tex]

Multiplying with [itex] e^{x/2} [/itex] gives [itex] e^{-x/2} [/itex] which, integrated on that domain (the real positive semiaxis) converges to 2...

The same argument could be applied to the second integral. However, there's still a problem to be dealed with

[tex] \forall n>0, \forall x \in\mathbb{R} \ , \left(1-\frac{x}{n}\right)^{n}< e^{-x} [/tex]

Daniel.
 
  • #7
dextercioby said:
Something which might prove itself to be useful is

[tex] \lim_{n\rightarrow +\infty} \left(1-\frac{x}{n}\right)^{n} = e^{-x} [/tex]

Multiplying with [itex] e^{x/2} [/itex] gives [itex] e^{-x/2} [/itex] which, integrated on that domain (the real positive semiaxis) converges to 2...

The same argument could be applied to the second integral. However, there's still a problem to be dealed with

[tex] \forall n>0, \forall x \in\mathbb{R} \ , \left(1-\frac{x}{n}\right)^{n}< e^{-x} [/tex]

Daniel.

hey, that IS useful...no wonder that expression looks familiar.

hmm, I think I can figure out the solution to my problem.

If I let fn = (1 - x/n)^n times exp(x/2)
then lim fn = exp(-x/2)
and since fn <= exp(-x/2) for every x element of X and n finite
I let g(x) = exp(-x/2)
g(x) is lebesgue integrable since its its integral is equal to 2.
hence I can use Dominated Convergence Thm and solve for the integral of lim fn, which is equal to the integral of exp(-x/2) which is 2.

Is this correct?
 
  • #8
Hi, i see this has been quiet for a while, still, i can't see how the problem is solved, since it isn't in fact true what daniel says for all n and all x. (take for example n=1, x=3 or -3).
 
  • #9
Actually, i think this should be done using the caracteristic function X_[0,n) letting f_n := (1-x/n)^n times X_[o,n). By doing this the f_n will still converge pointwise to e^(-x), and what daniel says is true for all x and for all n, since f_n will be cero whenever x>n. This will enable us to use dominated convergence theorem, but the problem can still be solved using monotne convergence theorem, since all f_n are geq than cero.
 

What is an integral limit?

An integral limit is the value that an integral approaches as the limits of integration approach a certain value or as the function being integrated approaches a particular behavior.

What does it mean for a function to be monotone?

A function is monotone if it either always increases or always decreases as the input variable increases. This means that the function has a consistent rate of change and does not have any local maxima or minima.

What is the Lebesgue convergence theorem?

The Lebesgue convergence theorem is a mathematical theorem that states that under certain conditions, the integral of a sequence of functions will converge to the integral of the limit of the sequence of functions. This theorem is often used in measure theory and analysis.

How do you solve integral limits using monotone functions?

To solve integral limits using monotone functions, you first need to determine the behavior of the function being integrated. If the function is monotone, then you can use the Monotone Convergence Theorem to show that the integral limit is equal to the integral of the limit of the sequence of functions.

What is the significance of Lebesgue convergence in mathematics?

The Lebesgue convergence theorem is significant in mathematics because it allows for the evaluation of integrals of functions that may not be continuous or well-behaved. It also extends the concept of convergence to more general settings, such as measure spaces, which allows for more powerful mathematical tools and applications.

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