Evaluate $\frac1{2\pi i}\int_{-\infty}^{\infty} \frac{e^{(iz+2)k}(1+e^{-\pi(iz+2)})}{(iz+2)^2+1}$

Question

Evaluate $$\frac1{2\pi i}\int_{-\infty}^{\infty} \frac{e^{(iz+2)k}(1+e^{-\pi(iz+2)})}{(iz+2)^2+1}\,dz, k\in \mathbb R_+$$ I think, we can use here residues and Jordan's lemma. Then we will take a semisircle $U_R=\{|z|=R, Imz\geq0\}$ centered at 0 in the upper half-plane, with radius $R\to \infty$ . And we got $\frac1{2\pi i}\int_{-\infty}^{\infty} \frac{e^{(iz+2)k}(1+e^{-\pi(iz+2)})}{(iz+2)^2+1}\,dz=\frac1{2\pi i}\lim_{R\to \infty}(\int_{[-R,R]} \frac{e^{(iz+2)k}(1+e^{-\pi(iz+2)})}{(iz+2)^2+1}\,dz+\int_{U_R} \frac{e^{(iz+2)k}(1+e^{-\pi(iz+2)})}{(iz+2)^2+1}\,dz)$. So the second summand $\to 0$ by Jordan's lemma. If it is right, we can compute the residues.In the upper half-plane we have 1 pole in the point $z=\frac{2-i}i=1+2i$.Then $Res=\lim_{z\to(1+2i)} \frac{e^{(iz+2)k}(1+e^{-\pi(iz+2)})(z-(1+2i)}{(iz+2)^2+1}$. And that's equal to $0$. Therefore< I think I have some mistakes here.

Answer 1

Your idea is correct, but there are indeed mistakes. First of all, rewrite the numerator as $e^{izk+2k}+e^{2k+2\pi+iz(k-\pi)}$. The first term is bounded in the upper half plane $\Im z>0$. For the second term, this depends on $k$. If $k\geq\pi$, then it is bounded on the upper half plane, whereas it is bounded on the lower half plane if $0\leq k\leq \pi$. Therefore the integral has to be split $$\int_{-\infty}^\infty\frac{e^{izk+2k}}{(iz+2)^2+1}dz+\int_{-\infty}^\infty\frac{e^{2k+2\pi+iz(k-\pi)}}{(iz+2)^2+1}dz.$$ For the first term we apply the method mentioned in the question. The denominator has two zeros satisfying $iz=-2\pm i$, that is $z=\pm 1+2i$ and obtain that the first integral equals $$2\pi i \mbox{Res}\left(\frac{e^{izk+2k}}{(iz+2)^2+1},z=1+2i\right)+2\pi i \mbox{Res}\left(\frac{e^{izk+2k}}{(iz+2)^2+1},z=-1+2i\right).$$ The first residue, for example, can be calculated as $$\lim_{z\to 1+2i}\frac{e^{izk+2k}}{(iz+2)^2+1}(z-1-2i),$$ but this limit is not 0, because the denominator also vanishes as $z\to1+2i$. Hence we obtain as limit $$\left.\frac{e^{izk+2k}}{\frac d{dz}\left((iz+2)^2+1\right)}\right|_{z=1+2i}=-\frac12e^{ik}.$$ The other residue is calculated similarly. I do not carry this out.
For the second integral $\int_{-\infty}^\infty\frac{e^{2k+2\pi+iz(k-\pi)}}{(iz+2)^2+1}dz$, we have to distinguish the two cases $k\geq\pi$ and $0\leq k\leq \pi$ (For $k=\pi$, we can choose). In the first case, since the exponential is bounded in the upper half plane, we proceed as for the first integral. In the second case, the exponential is bounded in the lower half plane and we use a semicircle $|z|=R$, $\Im z\geq 0$. As the denominator does not vanish in the lower half plane, there are no residues to calculate: the integral equals 0.
I hope, I can leave the remaining calculations to the readers...