$$ \int_{-\infty}^{\infty} \, \frac{\sin(k)}{k}e^{ikx} dk$$
I'm not too sure how to go about this. I think there is a pole at $z=0$ but then the residue is also$ =0 $ so i'm a bit lost :/
Any help would be appreciated!
$$ \int_{-\infty}^{\infty} \, \frac{\sin(k)}{k}e^{ikx} dk$$
I'm not too sure how to go about this. I think there is a pole at $z=0$ but then the residue is also$ =0 $ so i'm a bit lost :/
Any help would be appreciated!
The contribute given by the imaginary part of $e^{ikx}$ vanishes by symmetry and
$$ \int_{-\infty}^{+\infty}\frac{\sin(k)\cos(kx)}{k}\,dk = \frac{1}{2}\left[\int_{-\infty}^{+\infty}\frac{\sin((1+x)k)}{k}\,dk+\int_{-\infty}^{+\infty}\frac{\sin((1-x)k)}{k}\,dk\right]$$ equals $$ \frac{\pi}{2}\left[\text{sign}(1+x)+\text{sign}(1-x)\right]=\left\{\begin{array}{rcl}0 & \text{if} & |x|>1\\ \frac{\pi}{2} & \text{if} & |x|=1 \\ \pi & \text{if} & |x|<1\end{array}\right. $$ by the well-known lemma
$$ \forall x\in\mathbb{R},\qquad \int_{-\infty}^{+\infty}\frac{\sin(kx)}{k}\,dk = \pi\,\text{sign}(x).$$