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Any Idea how to integrate the Integral in the brackets and than applying limits:

$$\lim_{n\rightarrow \infty }\left ( \int_{0}^{2\pi}{\frac{\cos(nx)}{x^{2}+n^{2}}} dx\right ) $$

Integral $\int_{0}^{2\pi}{\frac{\cos(nx)}{x^{2}+n^{2}}}dx $ is the main issue here.

Mr.Gandalf Sauron
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Salz Engineering
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    Note that $\frac{cos(nx)}{x^2+n^2}\leq \frac{1}{x^2+n^2}$, and we know the integral of the right hand side. Use squeezing theorem to get the final limit. – Joshua Woo Nov 26 '21 at 12:22
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    Alternatively you can also use the dominated convergence theorem – Maximilian Janisch Nov 26 '21 at 12:38
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    $1/n^2$ is an even more useful upper bound on the integrand, since it implies the integral is between $\pm 2\pi/n^2$. – J.G. Nov 26 '21 at 15:14
  • @JoshuaWoo, Thanks for the tips. Interestingly Wolfram gives this result of Integral: $\frac{1}{2}cos(n)ln(x^{2}+n^{2})$, which when put to limits test of $n\rightarrow \infty$, the results becomes "Undefined". I will try methods to integrate this as told by many commenters here to get the exact integral just for curiosity. – Salz Engineering Nov 27 '21 at 11:05

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First try to show Uniform convergence of $\frac{\cos(nx)}{x^{2}+n^{2}}$ . We can do this by Weirestrass-M test.

Theorem:- If $\{f_{n}\}$ is a sequence of Riemann integrable functions on $[a,b]$ such that $f_{n}$ converges uniformly to $f$ , then the limiting function $f$ is Riemann integrable on $[a,b]$ and $\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dx=\int_{a}^{b}f(x)\,dx$ . That is we can change the order of integral and limit.

We have:-

$$\sup_{x\in[0,2\pi]}|\frac{\cos(nx)}{x^{2}+n^{2}}-0|\leq \sup_{x\in[0,2\pi]} \frac{1}{x^{2}+n^{2}}\leq \frac{1}{n^{2}}$$ .

As $\frac{1}{n^{2}}\to 0$ we have $\frac{\cos(nx)}{x^{2}+n^{2}}$ is uniformly convergent to $0$.

As for each $n\in\mathbb{N}$. $\frac{\cos(nx)}{x^{2}+n^{2}}$ is Riemann integrable in $[0,2\pi]$ as each of them is continuous on this compact interval. So applying the theorem We can interchange the order of limit and integral.

Thus we have $$ \lim_{n\to\infty}\int_{0}^{2\pi}{\frac{\cos(nx)}{x^{2}+n^{2}}} dx = \int_{0}^{2\pi}\lim_{n\to\infty}\frac{\cos(nx)}{x^{2}+n^{2}}dx=\int_{0}^{2\pi} 0\,dx = 0$$

Mr.Gandalf Sauron
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    Your proof is just perfect! (+1) – Laxmi Narayan Bhandari Nov 26 '21 at 15:46
  • Wonderful work @Mr.Gandalf, Can you point me to the book/reference for "Theorem of Riemann integrable function for interchanging order of limits and Integral" ? – Salz Engineering Nov 27 '21 at 11:07
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    Walter Rudin(McGraw hill 3rd edition). page 163 has it in terms of Riemann Stieltjes sense. Just take $\alpha=x$ to get this version of Riemann integral. Also T.M. Apostol(2nd edition) Page 225 has a proof. Basically any real analysis book will have a proof of this. Look under the chapter sequences of functions and under the section Uniform convergence and Riemann Integral. – Mr.Gandalf Sauron Nov 27 '21 at 12:08
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    If you just want a proof of this particular theorem then you can look at answers in this site. Here is one https://math.stackexchange.com/questions/595285/riemann-integrable-and-uniform-convergence – Mr.Gandalf Sauron Nov 27 '21 at 12:10
  • @Mr.GandalfSauron Got it. Really appreciate your kind effort. – Salz Engineering Nov 27 '21 at 12:16
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    You can mark as answer if it was able to satisfy your query – Mr.Gandalf Sauron Nov 27 '21 at 12:17
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Since $ 0\le x \le2\pi$, For large $n$ we have $x<<n$ so we can approximate for bounded $x$ values: $$ \int_{0}^{2\pi}{\frac{\cos(nx)}{x^{2}+n^{2}}} dx\approx \frac{1}{n^2}\int_{0}^{2\pi} \cos(nx)(1-\frac{x^2}{n^2} ) dx \\\to \frac{1}{n^2}\int_{0}^{2\pi} \cos(nx)dx=\frac1{n^2}\cdot0=0 $$

am301
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    You cannot just put arrows and justify the existence of limits. You either show the bounds and use Sandwich theorem or use Dominated Convergence Theorem or any other way. Putting $\to$ and saying that limit is $0$ such that the quadratic factor in the denominator mysteriously disappears is not a proof. – Mr.Gandalf Sauron Nov 26 '21 at 15:04