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Consider the following equality:


where $$f_n(x):=\frac{x^n}{1+x^n}\qquad x\in [0,1]$$

Since the sequence $(f_n(x))_{n=1}^{\infty}$ is not uniformly convergent, one cannot use the theorem about integration of uniformly convergent function sequences. So here is my question:

How to show that the equality is true?

I think it is equivalent to show that $$\lim_{n\to\infty}\int_{0}^1\frac{1}{1+x^n}dx=1$$ Then things boil down to calculating $$\int_{0}^1\frac{1}{1+x^n}dx$$ for every $n$, which is what I have no idea to do.

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@Qiaochu,@Eric:It seems that this quite "depends on knowing the dominated convergence theorem (which very few people prove for the Riemann integral, so usually has to wait until you are studying measure theory)". (the quote is from Mariano's comment to Terry Tao's answer to this question in MO. I happened to see it today.) – Jack Sep 10 '11 at 15:28
up vote 13 down vote accepted

$$ \int_0^1 {\frac{{x^n }}{{1 + x^n }}dx} \le \int_0^1 {x^n dx} = \frac{1}{{n + 1}}, $$ hence the left-hand side integral converges to $0$ as $n \to \infty$. On the other hand, $\frac{{x^n }}{{1 + x^n }} \to 0$ pointwise for $x \in [0,1)$, hence obviously $$ \int_0^1 {\bigg(\lim _{n \to \infty } \frac{{x^n }}{{1 + x^n }}\bigg)dx} = 0. $$ The key here is that $\frac{{x^n }}{{1 + x^n }} \leq x^n$, $x \in [0,1]$.

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+1: For the most simple and elegant solution – user17762 Jun 6 '11 at 23:59
@Sivaram: Thanks! – Shai Covo Jun 7 '11 at 0:00
I agree! This is nice, +1 – Eric Naslund Jun 7 '11 at 0:02
@Eric: Thanks... – Shai Covo Jun 7 '11 at 0:11

Any time you have a sequence of functions bounded in absolute value by the same constant on a finite interval, you can switch the limit and the integral. (Here they are all bounded above by $1$)

This is a special case of the Dominated Convergence Theorem.

You could also approach your problem more directly. The limit function is $0$ on $(0,1)$ so that the integral on the right hand side is zero. Now, we just have to show that $$\lim_{n\rightarrow \infty} \int_0^1 \frac{x^n}{1+x^n}dx=0.$$ Let $\epsilon>0$ be given. Then consider the decomposition of $[0,1]$ into the intervals $[0,1-\frac{\epsilon}{2}]$ and $[1-\epsilon/2,1]$. On the second interval, we get an error of at most $\epsilon/2$ since all of our functions are bounded by $1$. Since the sequence of functions is uniformly convergent on the first interval, we can make it so the integral over that interval is less then $\frac{\epsilon}{2}$ for sufficiently large $n$. Then all together, this will show that the integral is less then $\epsilon$ for sufficiently large $n$. This is equivalent to the limit equaling zero.

Hope that helps,

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Lebesgue Dominated Convergence Theorem

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Hint: On the right side, you have a pointwise limit. It looks like you can show the integral of this limit is zero. Then I would break the left hand integral at a point depending on $n$, bounding the integral on each side by a decreasing function of $n$ which goes to 0.

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