Estimate the principal term of the asymptotic integral $$\int\limits_{-1}^{1}\cos (kx)e^{i\lambda |x|}dx,\lambda \to \infty $$ At "moderate" values of $\lambda $, calculate the integral using the quadrature formula and compare with the asymptotic. On the parameter plane $\lambda$ , $N$ (number of nodes of the quadrature formula) indicate the area where the estimates diverge by less than $0.01$
I have tried a Taylor series of functions at $0$ $$\cos(kx)e^{i\lambda|x|} = \cos(kx)e^{i\lambda x} = \left(1 - \frac{(kx)^2}{2} + \frac{(kx)^4}{4!} - \ldots\right) \left(1 + i\lambda x - \frac{(\lambda x)^2}{2} + \ldots\right)$$
And then it is possible to make this assessment $$\cos(kx)e^{i\lambda|x|} \approx \left(1 - \frac{(kx)^2}{2}\right)\left(1 + i\lambda x - \frac{(\lambda x)^2}{2}\right)\Rightarrow \int_{-1}^{1} \cos(kx)e^{i\lambda|x|}dx \approx \int\limits_{-1}^{1} \left(1 - \frac{(kx)^2}{2}\right)\left(1 + i\lambda x - \frac{(\lambda x)^2}{2}\right)dx$$
But now how do I estimate the main term of the integral? How do I go to the integral for "moderate" values ? (I tried through the quadrature fom, but nothing worked)
I assume that $k$ is fixed. Note that your integral is $$ 2\int_0^1 {\cos (kx){\rm e}^{{\rm i}\lambda x} {\rm d}x}. $$ Integration by parts yields $$ 2\int_0^1 {\cos (kx){\rm e}^{{\rm i}\lambda x} {\rm d}x} = 2\frac{{\cos (k){\rm e}^{{\rm i}\lambda } - 1}}{{{\rm i}\lambda }} + \frac{{2k}}{{{\rm i}\lambda }}\int_0^1 {\sin (kx){\rm e}^{{\rm i}\lambda x} {\rm d}x} . $$ Show, using integration by parts, that $$ \int_0^1 {\sin (kx){\rm e}^{{\rm i}\lambda x} {\rm d}x} = \mathcal{O}\!\left( {\frac{1}{\lambda }} \right) $$ as $\lambda \to +\infty$, and hence conclude that $$ \int_{-1}^1 {\cos (kx){\rm e}^{{\rm i}\lambda |x|} {\rm d}x} = 2\frac{{\cos (k){\rm e}^{{\rm i}\lambda } - 1}}{{{\rm i}\lambda }} + \mathcal{O}\!\left( {\frac{1}{{\lambda ^2 }}} \right) $$ as $\lambda \to +\infty$.