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Find the integral $I$.....it looks like a good problem which I was not able to solve ....please help...

$$I=\lim_{n \to \infty} \int_0^1 {{1 + nx^2}\over{(1 + x^2)^n}} \log(2 + \cos(x/n))\,dx.$$

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closed as off-topic by user21820, user26857, drhab, mathreadler, user223391 Apr 5 '17 at 18:28

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, user26857, drhab, mathreadler, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ – tired Jul 17 '16 at 19:24
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    $\begingroup$ the leading order as $n\rightarrow\infty$ is given by $I_n\sim\frac{3\log(3)\sqrt{\pi}}{4}\frac{1}{\sqrt{n}}$ $\endgroup$ – tired Jul 17 '16 at 19:37
  • $\begingroup$ No I am not familiar with dominated convergence sequence....... $\endgroup$ – COOLGUY Jul 17 '16 at 19:43
  • $\begingroup$ Yes I am familiar with uniform convergence..... $\endgroup$ – COOLGUY Jul 17 '16 at 19:54
  • $\begingroup$ @tired Perhaps you should post a solution. It would be instructive to present the development of the leading term in the asymptotic series. -Mark $\endgroup$ – Mark Viola Jul 17 '16 at 20:00
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By Bernoullis inequality, we have $\frac{1+nx^2}{(1+x^2)^n}\leq 1$.

Also, $\cos{t}\leq 1$.

Therefore, $$|f_n(x)|=\left|{{1 + nx^2}\over{(1 + x^2)^n}} \log(2 + \cos(x/n))\right|\leq\log{3}=g(x)$$

Now, by DCT, we have $I=0$.

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  • $\begingroup$ (+1) for the nice answer and to remove the silly downvote $\endgroup$ – tired Jul 17 '16 at 22:41
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The main contributions to the integral will be from the interval $[0,1/\sqrt{n}]$. In this region $ \cos(x/n)\approx 1$ and therefore

$$ I_n\sim\log(3)\int_0^{1/\sqrt{n}}dx e^{-n\log(1+x^2)}(1+nx^2)\sim\log(3)\int_0^{1/\sqrt{n}}dx e^{-n x^2}(1+nx^2) $$

because we are only introducing an exponentially small error by pushing the limits of integration up to infinity (Laplace method) we get

$$ I_n\sim\log(3)\int_0^{\infty}dx e^{-n x^2}(1+nx^2)=\log(3)\frac{3\sqrt{\pi}}{4\sqrt{n}} $$

and the limit is $0$.

Note that it is important to keep $1$ as well as $nx^2$ in the integrand because their contributions turn out to be of the same order!

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    $\begingroup$ The remarks made there apply here, mutatis mutandis. $\endgroup$ – Did Jan 15 '17 at 11:56
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Note that after enforcing the substitution $x\to x/\sqrt{n}$, we have

$$\begin{align} \left|\int_0^1 \frac{1+nx^2}{(1+x^2)^n}\log\left(2+\cos\left(\frac{x}{n}\right)\right)\,dx\right|&=\frac{1}{\sqrt{n}}\int_0^\sqrt{n} \frac{1+x^2}{(1+x^2/n)^n}\log\left(2+\cos\left(\frac{x}{n^{3/2}}\right)\right)\,dx\\\\ &\le \frac{\log(3)}{\sqrt{n}} \int_0^\infty \frac{1+x^2}{\left(1+\frac{x^2}{n}\right)^n}\,dx \tag 1\\\\ &\le \frac{\log(3)}{\sqrt{n}} \int_0^\infty \frac{1+x^2}{\left(1+\frac{x^2}{2}\right)^2}\,dx \tag 2\\\\ &=\frac{\log(3)}{\sqrt{n}}\,\left(\frac{3\sqrt{2}\pi}{4}\right)\tag 3 \end{align}$$

where we used the fact that $\left(1+\frac {x^2}n\right)^n\ge \left(1+\frac{x^2}{2}\right)^2$ for $n\ge2$ in going from $(1)$ to $(2)$.

The right-hand side of $(3)$ clearly approaches zero as $n\to \infty$. Therefore, the squeeze theorem guarantees that the integral of interest approaches zero also. And we are done!

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  • $\begingroup$ @tired Not nearly as good as yours, which actually develops the leading term of the asymptotic expansion. But thank you! $\endgroup$ – Mark Viola Jul 17 '16 at 23:44
  • $\begingroup$ great solution.....atleast for me (a high school student.........) $\endgroup$ – COOLGUY Jul 18 '16 at 6:17
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    $\begingroup$ @Viku Thank you! Much appreciated. $\endgroup$ – Mark Viola Jul 18 '16 at 6:33

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