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How would I evaluate these limits? $$ \lim_{n \to \infty} \int_0^\infty \frac{n}{1+(nx)^2} \ dx$$ and $$ \lim_{n \to \infty} \int_0^\infty \frac{(1+(nx)^2)}{(1+nx^2)^n} \ dx$$

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@jessica: Where are they from? What theorems do you know? What do you know about the pointwise limit of the integrands? –  Jonas Teuwen Feb 5 '11 at 23:13
    
@PEV: Thanks for editing. Even i was looking to edit it. –  anonymous Feb 5 '11 at 23:16
    
@jessica: What convergence theorems are you aware of? For the first one at least, you could evaluate it directly. –  user17762 Feb 5 '11 at 23:22
    
right now im learning fatou's lemma, mct, and dct. the integrand of the first question goes to 0. i was thinking of finding a dominating function, ie 1/nx^2. im not sure how to proceed from here. –  user6683 Feb 5 '11 at 23:26
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The first integral is trivial; Doing a change of variable $u=nx$ you get the integral of $\frac{1}{1+u^2}$, so $$\int_0^{\infty}\frac{n}{1+(nx)^2}\,dx = \lim_{a\to\infty}\int_0^{na}\frac{du}{1+u^2} = \lim_{a\to\infty}\arctan(na) = \frac{\pi}{2}$$ so the sequence is constant. –  Arturo Magidin Feb 5 '11 at 23:34

4 Answers 4

As already noted, the integrals considered in the first case do not depend on $n$ hence their limit is not very mysterious.

Let $I_n$ denote the integral considered in the second case. By the change of variable $u=nx$, $$ nI_n=\displaystyle\int_0^\infty(1+u^2)f_n(u)\mathrm{d}u,\qquad\mbox{with}\ f_n(u)=(1+u^2/n)^{-n}. $$ The sequence of positive functions $(f_n)_n$ is nonincreasing (*) and $f_n\to f$ pointwise when $n\to\infty$, with $f(u)=\mathrm{e}^{-u^2}$. Since $f_2\ge f_n$ for very $n\ge2$ and $I_2$ is finite, by Lebesgue dominated convergence theorem, $$ nI_n\to J=\displaystyle\int_0^\infty(1+u^2)f(u)\mathrm{d}u=1+(3\pi/2). $$ In particular, $I_n\to0$ when $n\to\infty$.

(*) This is implied by the fact that, for every $c\ge0$, the function $z\mapsto(1+c/z)^z$ is nondecreasing on $z\ge0$. To show this last fact, differentiate twice the logarithm of this function.

Added later on Following Shai's suggestion, I mention that the proof above shows that the sequence $(nI_n)_n$ is nonincreasing. For instance $nI_n\le2I_2$ for every $n\ge2$ (note that $I_1=+\infty$ but that $I_2$ is finite).

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From $(1+a)^n\geq1+na\;\;\;\forall a\geq0$ with $a=nx^2$, you have: $\frac{1+n^2x^2}{(1+nx^2)^n}\leq1$.

Furthermore, you have:$\frac{1+n^2x^2}{(1+nx^2)^n}<\frac{1+n^2x^2}{(nx^2)^n}=\frac{1}{n^nx^{2n}}+\frac{1}{n^{n-2}x^{2(n-1)}}<\frac{2}{x^2}$ if $n\geq2$ and $x>0$.

Hence, by the Lebesgue dominated convergence theorem, you need only consider the integral of the limit function to evaluate the limit (let $g(x)=\begin{cases} 1 & \text{ if } 0\leq x\leq1 \\ \frac{2}{x^2}& \text{ if } x>1 \end{cases}$ ).

Now $0\leq\frac{1+n^2x^2}{(1+nx^2)^n}\leq\frac{n^2+n^2x^2}{(1+x^2)^n}=\frac{n^2}{(1+x^2)^{n-1}}$.

Hence $lim_{n\rightarrow\infty}\frac{1+n^2x^2}{(1+nx^2)^n}=0$ by the squeeze theorem (we can neglect $x=0$ since it constitutes a set of measure zero).

Hence the desired integral is equal to $0$.

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the hypothes of dominated convergence states that the dominating function must be integrable. however, 1 is not integrable over 0 infinity –  user6689 Feb 6 '11 at 2:40
    
@jessica: Dear Jessica, I converted your answer to a comment (since I think it was directed to the current answer). –  Akhil Mathew Feb 6 '11 at 2:41
    
@jessica: You were correct. I have edited my answer accordingly. –  Benji Feb 6 '11 at 3:42

Actually, the second integral, like the first, could have been a "calc 1 problem" (but it is useful to practice the dominated convergence theorem). Indeed, the inequality $\frac{{1 + n^2 x^2 }}{{(1 + nx^2 )^n }} \le 1$ (also used by bobobinks) gives $$ \int_0^\varepsilon {\frac{{1 + n^2 x^2 }}{{(1 + nx^2 )^n }}\,{\rm d}x} \le \varepsilon, $$ for any $\varepsilon > 0$ (arbitrarily small, but fixed). On the other hand, for any $n \geq 4$, $$ \int_\varepsilon ^\infty {\frac{{1 + n^2 x^2 }}{{(1 + nx^2 )^n }}\,{\rm d}x} \le \int_\varepsilon ^\infty {\frac{{\varepsilon ^{ - 2} x^2 + n^2 x^2 }}{{(2\sqrt n x)^n }}\,{\rm d}x} \le \frac{{\varepsilon ^{ - 2} + n^2 }}{{(2\sqrt n )^n }}\int_\varepsilon ^\infty {x^{2 - n} \,{\rm d}x} = \frac{{\varepsilon ^{ - 2} + n^2 }}{{(2\sqrt n )^n }}\frac{{\varepsilon ^{3 - n} }}{{n - 3}}. $$ Since, for $\varepsilon > 0$ fixed, the expression on the right-hand side tends to $0$ as $n \to \infty$, it follows that $\int_0^\infty {\frac{{1 + n^2 x^2 }}{{(1 + nx^2 )^n }}\,{\rm d}x} \to 0$ too.

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Since $I_n\le(2/n)I_2$ for every $n\ge2$ (see my answer), one can skip these $\varepsilon$ estimates. –  Did Feb 6 '11 at 12:47
    
@Didier: Nice observation (maybe add it to your answer?). Anyway, both approaches are equally elementary. –  Shai Covo Feb 6 '11 at 13:52
    
Sure. To find a non elementary proof could be a challenge... :-) –  Did Feb 6 '11 at 14:36
    
@Didier: Probably, most students consider Dominated Convergence Theorem not elementary. –  Shai Covo Feb 6 '11 at 14:44
    
See jessicas's comment where she describes what she knows. –  Did Feb 6 '11 at 15:06

Limit for Intergral from 0 to infinity of [1-(cosx to power n)] over 1+(x to the power 2) of dx

$$ \lim_{n\to\infty}\int_0^\infty\frac{1-\cos^n(x)}{1+\cos^n(x)}\,\mathrm{d}x $$

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