Is this a valid asymptotic approximation to $\int_0^\pi \frac{\cos(kn)}{1+k^2}dk$?

Question

I want to obtain an asymptotic expansion for $\int_0^\pi \frac{\cos(kn)}{1+k^2}dk$ in which $n\to \infty$.

Per Gary's suggestion, examine https://dlmf.nist.gov/2.11#i

I do not understand where the trigonmetric terms go when $m\to \infty$ yielding their $(2.11.2)$ series. How is that series obtained?

Answer 1

Integrating by parts $4$ times: $$ \begin{align} &\int_0^\pi\frac{\cos(nx)}{1+x^2}\,\mathrm{d}x\tag{1a}\\ &=\frac1n\int_0^\pi\frac1{1+x^2}\,\mathrm{d}\sin(nx)\tag{1b}\\ &=\frac1n\frac{\sin(n\pi)}{1+\pi^2}+\frac2n\int_0^\pi\frac{x\sin(nx)}{\left(1+x^2\right)^2}\,\mathrm{d}x\tag{1c}\\ &=\frac1n\frac{\sin(n\pi)}{1+\pi^2}-\frac2{n^2}\int_0^\pi\frac{x}{\left(1+x^2\right)^2}\,\mathrm{d}\cos(nx)\tag{1d}\\ &=\frac1n\frac{\sin(n\pi)}{1+\pi^2}-\frac2{n^2}\frac{\pi\cos(n\pi)}{\left(1+\pi^2\right)^2}+\frac2{n^2}\int_0^\pi\frac{1-3x^2}{\left(1+x^2\right)^3}\cos(nx)\,\mathrm{d}x\tag{1e}\\ &=\frac1n\frac{\sin(n\pi)}{1+\pi^2}-\frac2{n^2}\frac{\pi\cos(n\pi)}{\left(1+\pi^2\right)^2}+\frac2{n^3}\int_0^\pi\frac{1-3x^2}{\left(1+x^2\right)^3}\,\mathrm{d}\sin(nx)\tag{1f}\\ &=\scriptsize\frac1n\frac{\sin(n\pi)}{1+\pi^2}-\frac2{n^2}\frac{\pi\cos(n\pi)}{\left(1+\pi^2\right)^2}+\frac2{n^3}\frac{1-3\pi^2}{\left(1+\pi^2\right)^3}\sin(n\pi)+\frac{24}{n^3}\int_0^\pi\frac{x\left(1-x^2\right)}{\left(1+x^2\right)^4}\sin(nx)\,\mathrm{d}x\tag{1g}\\ &=\scriptsize\frac1n\frac{\sin(n\pi)}{1+\pi^2}-\frac2{n^2}\frac{\pi\cos(n\pi)}{\left(1+\pi^2\right)^2}+\frac2{n^3}\frac{1-3\pi^2}{\left(1+\pi^2\right)^3}\sin(n\pi)-\frac{24}{n^4}\int_0^\pi\frac{x\left(1-x^2\right)}{\left(1+x^2\right)^4}\,\mathrm{d}\cos(nx)\tag{1h}\\ &=\scriptsize\frac1n\frac{\sin(n\pi)}{1+\pi^2}-\frac2{n^2}\frac{\pi\cos(n\pi)}{\left(1+\pi^2\right)^2}+\frac2{n^3}\frac{1-3\pi^2}{\left(1+\pi^2\right)^3}\sin(n\pi)-\frac{24}{n^4}\frac{\pi\left(1-\pi^2\right)}{\left(1+\pi^2\right)^4}\cos(n\pi)\\ &+\frac{24}{n^4}\int_0^\pi\frac{1-10 x^2+5x^4}{\left(1+x^2\right)^5}\cos(nx)\,\mathrm{d}x\tag{1i}\\ \end{align} $$ The total variation of $\frac{1-10 x^2+5x^4}{\left(1+x^2\right)^5}$ on $[0,\pi]$ is bounded by $\frac{59}{32}$; $f(0)=1,f\!\left(\frac1{\sqrt3}\right)=-\frac{27}{64},f(\pi)\approx0$. Therefore, $$ \begin{align} &\left|\,\frac{24}{n^4}\int_0^\pi\frac{1-10 x^2+5x^4}{\left(1+x^2\right)^5}\cos(nx)\,\mathrm{d}x\,\right|\tag{2a}\\ &=\left|\,\frac{24}{n^5}\int_0^\pi\frac{1-10 x^2+5x^4}{\left(1+x^2\right)^5}\,\mathrm{d}\sin(nx)\,\right|\tag{2b}\\ &=\left|\,\frac{24}{n^5}\frac{1-10\pi^2+5\pi^4}{\left(1+\pi^2\right)^5}-\frac{24}{n^5}\int_0^\pi\sin(nx)\,\mathrm{d}\frac{1-10 x^2+5x^4}{\left(1+x^2\right)^5}\,\right|\tag{2c}\\ &\le\frac{24}{n^5}\left(\frac{1-10\pi^2+5\pi^4}{\left(1+\pi^2\right)^5}+\frac{59}{32}\right)\tag{2d}\\ &\lt\frac{45}{n^5}\tag{2e} \end{align} $$ Thus, we get the asymptotic expansion of $\int_0^\pi\frac{\cos(nx)}{1+x^2}\,\mathrm{d}x$ to be $$ \frac1n\frac{\sin(n\pi)}{1+\pi^2}-\frac2{n^2}\frac{\pi\cos(n\pi)}{\left(1+\pi^2\right)^2}+\frac2{n^3}\frac{1-3\pi^2}{\left(1+\pi^2\right)^3}\sin(n\pi)-\frac{24}{n^4}\frac{\pi\left(1-\pi^2\right)}{\left(1+\pi^2\right)^4}\cos(n\pi)\tag3 $$ with an error smaller than $\frac{45}{n^5}$.

If $n$ is an integer, then $\sin(n\pi)=0$ and $\cos(n\pi)=(-1)^n$, which accounts for any missing $\sin$’s and $\cos$’s.

Answer 2

Start writing $$\frac{\cos (k n)}{k^2+1}=\frac{\cos (k n)}{(k+i)(k-i)}=\frac i 2\left(\frac{\cos (k n)}{k+i}-\frac{\cos (k n)}{k-i} \right)$$

Let $k+i=x$

$$\int\frac{\cos (k n)}{k+i}\,dk=\int \frac{\cos (n (x-i))}{x}\,dx$$

Expand the cosine to face the sine and cosine integrals.

For the deinite integral, you should end with $$2\int_0^\pi \frac{\cos (k n)}{k^2+1}\,dk= (\text{Si}(i n-n \pi )-\text{Si}(n (i+\pi ))) \sinh (n)-i (\text{Ci}(-n (-i+\pi ))-\text{Ci}(n (i+\pi ))) \cosh (n)$$

Assuming that $n$ is an integer $$\int_0^\pi\frac{\cos (k n)}{k^2+1}\,dk=\frac \pi 2 \left(e^{-n}-4(-1)^n A\right)$$where $$A=\frac{1}{\left(1+\pi ^2\right)^2 n^2}-\frac{12 \left(\pi ^2-1\right)}{\left(1+\pi ^2\right)^4 n^4}+O\left(\frac{1}{n^6}\right)$$

Edit

If $n$ is not an integer $$\int_0^\pi\frac{\cos (k n)}{k^2+1}\,dk=\frac{\pi e^{-n}}{2}+\frac{A \cos(n\pi)+B \sin(n\pi)}{\left(1+\pi ^2\right)^4 n^4}$$ where $$A=-2 \pi \left(\left(1+\pi ^2\right)^2 n^2-12 \left(\pi ^2-1\right)\right)$$ $$B=\left(1+\pi ^2\right) n \left(\left(1+\pi ^2\right)^2 n^2-6 \pi ^2+2\right)$$

Answer 3

I will assume that $n$ is a positive integer. From Section $6.2$ on page $77$ of F. W. J. Olver's book Asymptotics and Special Functions, we obtain $$\int_0^\pi \frac{\cos(nt)}{1+t^2}\mathrm{d}t= \frac{\pi }{2}{\rm e}^{ - n} + ( - 1)^n \sum\limits_{k = 0}^{N - 1} {( - 1)^k \frac{{q^{(2k + 1)} (\pi )}}{{n^{2k + 2} }}} + R_N (n)\tag{1}$$ for any $N\ge 1$, with $$ \left| {R_N (n)} \right| \le \frac{1}{{n^{2N + 1} }}\int_\pi ^{ + \infty } {\left| {q^{(2N + 1)} (t)} \right|{\rm d}t}. $$ Here $q(t)=(1+t^2)^{-1}$. It is not difficult to show that $$ q^{(2k + 1)} (t) = -\frac{{(2k + 1)!}}{{(1 + t^2 )^{k + 1} }}\sin \left( {(2k + 2)\arcsin \left( {\frac{1}{{\sqrt {1 + t^2 } }}} \right)} \right), $$ whence \begin{align*} \left| {R_N (n)} \right| & \le \frac{{(2N + 1)!}}{{n^{2N + 1} }}\int_\pi ^{ + \infty } {\frac{{\rm d}t}{{(1 + t^2 )^{N+ 1} }}} \\ & = \frac{1}{2}\frac{{(2N + 1)!}}{{n^{2N + 1} }}\int_{\log (1 + \pi ^2 )}^{ + \infty } {{\rm e}^{ - Ns} \frac{{{\rm d}s}}{{\sqrt {{\rm e}^s - 1} }}} \\ & \le \frac{1}{{2\pi }}\frac{{(2N + 1)!}}{{n^{2N + 1} }}\int_{\log (1 + \pi ^2 )}^{ + \infty } {{\rm e}^{ - Ns} {\rm d}s} \\ & = \frac{1}{{2\pi N}}\frac{{(2N + 1)!}}{{(1 + \pi ^2 )^N n^{2N + 1} }}. \tag{2} \end{align*} Using the Chebyshev polynomials of the second kind, \begin{align*} q^{(2k + 1)} (t) & = -\frac{{(2k + 1)!}}{{(1 + t^2 )^{k + 1} }}\frac{1}{{\sqrt {1 + t^2 } }}U_{2k + 1} \!\left( {\frac{t}{{\sqrt {1 + t^2 } }}} \right) \\ & = ( - 1)^{k+1} \frac{{(2k + 1)!}}{{(1 + t^2 )^{2k + 2} }}\sum\limits_{j = 0}^k {( - 1)^j \binom{2k + 2}{2j + 1}t^{2j + 1} } . \end{align*} This formula provides an explicit expression for the coefficients of the expansion $(1)$.

By Stirling's formula and the bound $(2)$, it may be shown that with the choice $N = \left\lfloor {n/2} \right\rfloor$, $R_N (n)= \mathcal{O}(n^{ - 1/2} (\sqrt {1 + \pi ^2 } {\rm e})^{ - n})$.