Solve Integral Limits w/ Monotone & Lebesgue Convergence

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Homework Help Overview

The discussion revolves around evaluating the limits of two integrals involving exponential functions and sequences as n approaches infinity. The integrals are framed within the context of abstract integration and Borel measures, specifically utilizing convergence theorems such as the monotone and dominated convergence theorems.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various methods for solving the integrals, including substitution and integration by parts. There is discussion on the convergence of sequences of integrals and the applicability of different convergence theorems.

Discussion Status

Some participants have suggested potential limits for the integrals, while others are questioning the validity of these limits and the assumptions made. There is an ongoing exploration of the conditions under which the dominated convergence theorem can be applied, with some participants expressing uncertainty about the positivity of the functions involved.

Contextual Notes

Participants note that the topic of study is abstract integration and Borel measures, with specific homework constraints requiring the use of convergence theorems. There is a recognition of the need to find a dominating function for the application of the dominated convergence theorem.

island-boy
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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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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.
 
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
 
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.
 
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.
 
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.
 
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?
 
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).
 
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.
 

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