Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find $L=\lim\limits_{n \to \infty} \int_0^{n a} \exp\left(-\dfrac{t}{1+\frac{b t}{n}}\right) dt$, where $a>0$, $b>0$.

I can't see what is dominating function, but I feel that I have to use Dominated convergence theorem. Any kind of help is welcome.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Rewrite the integral as

$$\int_0^{n a} dt \: \exp\left(-\dfrac{t}{1+\frac{b t}{n}}\right) = n a \int_0^1 du \: \exp\left(-\dfrac{n a u}{1+a b u}\right)$$

As $n \rightarrow \infty$, the contributions from the integral come mainly from the neighborhood near $u=0$. In this limit, then

$$L = \lim_{n \rightarrow \infty} n a \int_0^{\infty} du \: e^{- n a u} = 1$$

share|improve this answer
    
i have a small q about your ans: i to got to the point where the integral of u is between 0 and 1/ in the next step you take L to be the same integral but u is now from o to infinity. why is that? is that a mistake? because my ans is 0... thanks! –  user61202 Feb 6 '13 at 11:03
    
We can set the upper limit to $\infty$ because the error in doing so is exponentially small as $n \rightarrow \infty$. In fcat, you can see that in evaluating the integral with the finite limit. –  Ron Gordon Feb 6 '13 at 11:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.