Integral without residues

Question

How do I do this integral without using complex variable theorems? (i.e. residues)

$$\lim_{n\to \infty} \int_0^{\infty} \frac{\cos(nx)}{1+x^2} \, dx$$

Answer 1

Use integration by parts

$$\int_0^\infty \frac{\cos(nx)}{1+x^2} dx = \frac{1}{n} \int_0^\infty \frac{2\sin(nx)x}{(1+x^2)^2} dx$$

Now use the triangle inequality

$$\left|\int_0^\infty \frac{2\sin(nx)x}{(1+x^2)^2} dx\right|\le \int_0^\infty \frac{2x}{(1+x^2)^2} dx<\infty$$

So the limit is $0$.

Answer 2

Consider the function $f(t)=e^{-a|t|}$, then the Fourier transform of $f(t)$ is given by $$ \begin{align} F(x)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-ix t}\,dt\\ &=\int_{-\infty}^{\infty}e^{-a|t|}e^{-ix t}\,dt\\ &=\int_{-\infty}^{0}e^{at}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-at}e^{-ix t}\,dt\\ &=\lim_{u\to-\infty}\left. \frac{e^{(a-ix)t}}{a-ix} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(a+ix)t}}{a+ix} \right|_{0}^{t=v}\\ &=\frac{1}{a-ix}+\frac{1}{a+ix}\\ &=\frac{2a}{x^2+a^2}. \end{align} $$ Next, the inverse Fourier transform of $F(x)$ is $$ \begin{align} f(t)=\mathcal{F}^{-1}[F(x)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(x)e^{ix t}\,dx\\ e^{-a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2a}{x^2+a^2}e^{ix t}\,dx\\ \frac{\pi e^{-a|t|}}{a}&=\int_{-\infty}^{\infty}\frac{e^{ix t}}{x^2+a^2}\,dx\\ \frac{\pi e^{-a|t|}}{2a}&=\int_{0}^{\infty}\frac{e^{ix t}}{x^2+a^2}\,dx. \end{align} $$ Thus, taking the real part, putting $a=1$ and $t=n$, then $$ \Re\left[\int_{0}^{\infty}\frac{e^{inx}}{x^2+1}\,dx\right]=\int_{0}^{\infty}\frac{\cos nx}{x^2+1}\,dx=\large\color{blue}{\frac{\pi e^{-|n|}}{2}}. $$ Other method using double integral technique can be seen here. Consequently $$ \lim_{n\to\infty}\int_{0}^{\infty}\frac{\cos nx}{x^2+1}\,dx=\lim_{n\to\infty}\frac{\pi e^{-|n|}}{2}=\large\color{blue}0. $$