How should I evaluate the integral $\lim_{n\rightarrow\infty}n\int\frac{\cos(nx)}{1+x^2}\,dx$? Q1: How should I evaluate the integral $$\lim_{n\rightarrow\infty}n\int_{\mathbb{R}}\frac{\cos(nx)}{1+x^2}\,dx$$ (Hint: dominated convergent theorem and integration by parts) ?
Possible Solution:
\begin{align}
n\int\frac{\cos(nx)}{1+x^2}\,dx&=n\left(\left[\frac{1}{1+x^2}\frac{\sin(nx)}{n}\right]_{-\infty}^\infty+\int\frac{\sin(nx)2x}{n(1+x^2)^2}\,dx\right)\\
&=2\int\frac{x\sin(nx)}{n(1+x^2)^2}\,dx=\frac{2}{n}\int \frac{-x}{(1+x^2)^2}\,d(\cos(nx))\\
&=\frac{2}{n}\left(\left[-\frac{x \cos(nx)}{(1+x^2)^2}\right]_{-\infty}^\infty+\int \cos(nx)\frac{(1+x^2)^2-2x(1+x^2)}{(1+x^2)^4}\,dx\right)\\
&=\frac{2}{n}\int \cos(nx)\frac{1+x^2-2x}{(1+x^2)^3}\,dx.
\end{align}
Hence
$$\lim\frac{2}{n}\int \cos(nx)\frac{1+x^2-2x}{(1+x^2)^3}\,dx=\int\lim\frac{2}{n}\cos(nx)\frac{1+x^2-2x}{(1+x^2)^3}\,dx=0$$
by Dominated Convergent Theorem.
I am still working on this hence I will update this post if there are more steps coming out.
 A: By $nx=t$ and using integration by parts $\int_{-\infty}^{\infty}n\frac{cos(nx)}{1+x^2}dx = \frac{sin(t)}{1+(t/n)^2}|_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\frac{sin(t)}{(1+(t/n)^2)}\frac{2t}{n^2}dt \\ $. Here the first term goes to zero and second term is bounded by $\int_{-\infty}^{\infty}\frac{1}{(1+(t/n)^2)}\frac{2t}{n^2}dt$. The last term is an odd function hence the integral is 0. Hence the answer should be zero.
A: This integral is a typical candidate for using the Residue Theorem, as others commented already. I don't know if the OP is allowed to use that, but anyways, I am posting this, it might be useful.
So your integral is the real part of this
$$
I=n\int_{-\infty}^{\infty}dx\frac{e^{inx}}{1+x^2}.
$$
Since you are interested in the $n\to\infty$ limit I will assume $n>0$ and close the contour in the upper half plane ($x$ now considered as a complex variable). The upper half plane is chosen because this way the contribution from the arc (with radius $R$) will vanish in the limit $R\to\infty$. Now, using Cauchy's theorem, the integral is equal to the contribution from the poles inside the closed contour, in this case from the pole $x=i$, and we get
$$
I=n\oint dx\frac{e^{inx}}{1+x^2}=n2\pi i\frac{e^{-n}}{2i}=\pi ne^{-n}.
$$
Finally, $\lim_{n\to\infty}I=0$.
