# Limit of an integral

1. Mar 6, 2006

### island-boy

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?

Last edited: Mar 6, 2006
2. Mar 6, 2006

### dextercioby

One can calculate

$$\int_{0}^{n} \left(1-\frac{x}{n}\right)^{n} e^{\frac{x}{2}} \ dx$$

exactly for n>0.

And the same goes for the second integral.

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

Daniel.

3. Mar 6, 2006

### island-boy

am I correct in assuming this is done through integration by parts?
hold on, I'll try to solve it this way

4. Mar 6, 2006

### dextercioby

Well, a substitution $y=1-\frac{x}{n}$ 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. Mar 6, 2006

### island-boy

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. Mar 6, 2006

### dextercioby

Something which might prove itself to be useful is

$$\lim_{n\rightarrow +\infty} \left(1-\frac{x}{n}\right)^{n} = e^{-x}$$

Multiplying with $e^{x/2}$ gives $e^{-x/2}$ 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

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

Daniel.

7. Mar 6, 2006

### island-boy

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. Nov 2, 2011

### sert

Hi, i see this has been quiet for a while, still, i cant 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. Nov 2, 2011

### sert

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.