The main problem is to evaluate $$\lim_{n\to\infty} \int_{-\infty}^{\infty} \frac{ne^{\cos x}}{1+n^2x^2}\ dx$$
The hint of this problem suggests to observe the graph of $y = \frac{n}{1+n^2x^2}$ as n increasing and evaluating its intrgral on $(-\infty, \infty)$, and then find out the above limit.
The following is my thoughts (may not be correct).
Denote $f_{n}(x) = \frac{n}{1+n^2x^2}$. Then
$$\lim_{n\to\infty} f_{n}(x) =
\begin{cases}
0& \text{if $x$} \neq 0 \\
\lim_{n\to\infty} n & \text{if $x$} = 0
\end{cases}$$
Define
$$f(x) =
\begin{cases}
0& \text{if $x$} \neq 0 \\
\lim_{n\to\infty} n & \text{if $x$} = 0
\end{cases}$$
Then $\lim_{n\to\infty} f_{n}(x) = f(x)$, for all real number $x$.
Similarly, denote $g_{n}(x) = \frac{ne^{\cos x}}{1+n^2x^2}$. Then
$$\lim_{n\to\infty} g_{n}(x) =
\begin{cases}
0& \text{if $x$} \neq 0 \\
\lim_{n\to\infty} ne = e \lim_{n\to\infty} n & \text{if $x$} = 0
\end{cases}$$
Define
$$g(x) =
\begin{cases}
0& \text{if $x$} \neq 0 \\
e\lim_{n\to\infty} n & \text{if $x$} = 0
\end{cases}$$
Then $\lim_{n\to\infty} g_{n}(x) = g(x)$, for all real number $x$. And $g(x) = ef(x)$.
A little bit calculations give
$$\int_{-\infty}^{\infty} f_{n}(x) \ dx = \int_{-\infty}^{\infty} \frac{n}{1+n^2x^2}\ dx = arctan(nx) \bigg|_{x=-\infty}^{x=\infty} = \frac{\pi}{2} – \frac{-\pi}{2} = \pi$$
Then,
$$\pi = \lim_{n\to\infty} \pi = \lim_{n\to\infty}\int_{-\infty}^{\infty} f_{n}(x) \ dx \stackrel{correct ?}{=} \int_{-\infty}^{\infty} \lim_{n\to\infty}f_{n}(x) \ dx= \int_{-\infty}^{\infty} f(x) \ dx$$
Similarly,
$$\lim_{n\to\infty} \int_{-\infty}^{\infty} \frac{ne^{\cos x}}{1+n^2x^2}\ dx = \lim_{n\to\infty}\int_{-\infty}^{\infty} g_{n}(x) \ dx \stackrel{correct ?}{=} \int_{-\infty}^{\infty} \lim_{n\to\infty}g_{n}(x) \ dx \\ = \int_{-\infty}^{\infty} g(x) \ dx = \int_{-\infty}^{\infty} ef(x) \ dx = e\int_{-\infty}^{\infty} f(x) \ dx = e\pi$$
In the above work, I have two questions. First, does commuting limit and improper integral here be correct? Is there any theory to support the idea? Second, we have the improper integral for the impulse function like $f(x)$ and $g(x)$ in the above, this is beyond my knowledge in Calculus. Can we define the improper integral of $f(x)$ and $g(x)$ over $(-\infty, \infty)$?
And of-course, I am hoping and seeking for a more simple method, like we don't need to exchange the limit and the improper integral sign or avoiding to deal with improper integral of an impulse function.
Best Answer
$$\int_{-\infty}^{+\infty}\frac{ne^{\cos x}}{1+n^2x^2}\,\mathrm dx=\int_{-\infty}^{+\infty}\frac{e^{\cos(y/n)}}{1+y^2}\,\mathrm dy.$$
Since $0\le\frac{e^{\cos(y/n)}}{1+y^2}\le h(y):=\frac e{1+y^2}$ and $h\in L^1(\Bbb R),$ the dominated convergence theorem gives: $$\lim_{n\to\infty}\int_{-\infty}^{+\infty}\frac{e^{\cos(y/n)}}{1+y^2}\,\mathrm dy= \int_{-\infty}^{+\infty}\lim_{n\to\infty}\frac{e^{\cos(y/n)}}{1+y^2}\,\mathrm dy=\int_{-\infty}^{+\infty}\frac e{1+y^2}\,\mathrm dy=e\pi. $$