Asymptotic behavior of an integral $ \int_{-\infty}^{\infty} \frac{e^{i \lambda \cos x}}{1 + x^2} dx $ using (preferably) the stationary phase method As I assume, since only the first term of the asymptotic behavior needs to be found, we should find the behavior at each stationary point $ m = \pi k, \; k \in \mathbb{Z} $
$$  \int_{m - \varepsilon}^{m + \varepsilon} \frac{e^{i \lambda \cos x}}{1 + x^2} dx \; \sim \; ?, \quad \lambda \to \infty$$
The main trouble is that the exponential function has an infinite number of stationary points. Even if we find an approximation at each of them I don't know how to sum it up clearly.
 A: Let's denote $\displaystyle I(\lambda)=\int_{-\infty}^\infty\frac{e^{i\lambda\cos x}}{1+x^2}dx$.
Splitting the interval of integration and using the periodicity of $\cos x$
$$I(\lambda)=2\int_0^{\pi/2}\frac{e^{i\lambda\cos x}}{1+x^2}dx+2\sum_{k=1}^\infty\int_{\pi k-\pi/2}^{\pi k+\pi/2}\frac{e^{i\lambda\cos x}}{1+x^2}dx$$
$$=2\int_0^{\pi/2}\frac{e^{i\lambda\cos x}}{1+x^2}dx+2\sum_{k=1}^\infty \int_{-\pi/2}^{\pi/2}\frac{e^{i\lambda(-1)^k\cos t}}{1+(t+\pi k)^2}dt=2I_0+2\sum_{k=1}^\infty I_k\tag{1}$$
To find first asymptotic terms, we use the stationary phase approach:
$$I_k\sim\int_{-\pi/2}^{\pi/2}\frac{e^{i\lambda(-1)^k(1-t^2/2)}}{1+(\pi k)^2}dt\sim\frac{e^{(-1)^ki\lambda}e^{(-1)^{(k+1)}\frac{\pi i}{4}}}{\sqrt\lambda}\int_{-\infty}^\infty\frac{e^{-t^2/2}}{1+(\pi k)^2}dt$$
$$I_k\sim\sqrt\frac{2\pi}{\lambda}e^{-\frac{\pi i}{4}}\,\frac{e^{(-1)^ki(\lambda+\frac{\pi}{4})}}{1+(\pi k)^2}\tag{2}$$
$$I_0\sim\frac{1}{2}\sqrt\frac{2\pi}{\lambda}e^{(\lambda-\frac{\pi i}{4})i}\tag{3}$$
Putting (2) and (3) into (1), we can present the asymptotics in the form
$$I(\lambda)\sim 2I_0+2\sqrt\frac{2\pi}{\lambda}e^{-\frac{\pi i}{4}}\,\left(\cos\left(\lambda+\frac\pi4\right)\sum_{k=1}^\infty\frac{1}{1+(\pi k)^2}+i\sin\left(\lambda+\frac\pi4\right)\sum_{k=1}^\infty\frac{(-1)^k}{1+(\pi k)^2}\right)\tag{4}$$
Using complex integration, we can get
$$\sum_{k=1}^\infty\frac{1}{a^2+k^2}=-\frac{1}{2a^2}+\frac{\pi}{2a}\,\coth\pi a$$
$$\sum_{k=1}^\infty\frac{(-1)^k}{a^2+k^2}=-\frac{1}{2a^2}+\frac{\pi}{2a}\,\frac{1}{\sinh\pi a} $$
Taking $a=\frac{1}{\pi}$ and putting into (4), we get the main asymptotic term
$$\boxed{\,\,I(\lambda)\sim \sqrt\frac{2\pi}{\lambda}e^{-\frac{\pi i}{4}}\,\left(e^{i\lambda}+\frac{2\cos\left(\lambda+\frac\pi4\right)}{e^2-1}+i\frac{(1+2e-e^2)\sin\left(\lambda+\frac\pi4\right)}{e^2-1}\right)\,\,}$$
After straightforward transformations, $I(\lambda)$ can also be presented in the form
$$I(\lambda)\sim \sqrt\frac{2\pi}{\lambda}\left(e^{i\left(\lambda-\frac\pi4\right)}+\frac{2}{e^2-1}e^{i\lambda}-\frac{(e-1)}{e+1}e^{\frac{\pi i}{4}}\sin\left(\lambda+\frac\pi4\right)\right)$$
