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Please, could somebody help me find a function $f(x)$ such that $| \frac{1}{n+n^2 \sin(xn^{-2})}| \le f(x)$ for each $n \in (0, \infty)$. $f(x)$ has to be $\ge 0$ for every $x \in (0, \infty)$ and integrable in $(0,\infty)$. I've been trying, but nothing came to my mind...

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How about when $n = 1$, then the function on the left is $1/(1 + \sin(x))$ which is itself unbounded and not integrable on $(0, \infty)$. So, it doesn't seem like its possible to find an $f(x)$ like you asked. – Eric Haengel Feb 21 '13 at 20:56
Ok, and can I find such function for almost every $n$? – Anne Feb 21 '13 at 21:32
It seems to me that as long as $n \geq 1$ the function on the left is unbounded, so there is no solution unless you assume $n \in (0, 1)$, and even then the function is not integrable because $|1/(n + n^2 \sin(xn^{-2})) |$ is periodic and integrates to a finite number on each period. Hence its integral over $(0, \infty)$ is infinite. – Eric Haengel Feb 21 '13 at 21:38
Ok, true. So maybe you have an idea how to find: $\lim_{n \rightarrow \infty} \int\limits_0^n \frac{1}{n+n^2\sin(xn^{-2})} dx$ ? I thought that it would go with Lebesgue theorem, but it doesn't... Could you help me solving this problem? – Anne Feb 21 '13 at 22:35
Oh, were you trying to use the dominated convergence theorem to switch the limit and integral? Is finding the value of that limit the original problem you started with? – Eric Haengel Feb 22 '13 at 18:29

It may be useful to know that if $a^2\lt b^2$ then $$\int{dx\over a+b\sin x}={-1\over\sqrt{b^2-a^2}}\log{b+a\sin x+\sqrt{b^2-a^2}\cos x\over a+b\sin x}$$

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Well, I don't actually get it. First of all, how do we obtain this result? And secondly, having $a=n$, $b=n^2$ and $\sin(x) = \sin(\frac{x}{n^2}) $ it is still very hard to find the limit... Please, help me. – Anne Feb 22 '13 at 11:48
The way I obtained this result was by looking it up in a table of integrals. There's another one for $a^2\gt b^2$, but I don't have the tables with me. You can make the substitution $u=(1/n)x^2$ to get your integral into this form. As to finding the limit, I don't know --- that's why I wrote it may be useful instead of it will be useful. If it's not useful, well, sorry --- just trying to help. – Gerry Myerson Feb 22 '13 at 22:19

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