I'm having trouble computing the Fourier transform of the following function:
$$y(x)=De^{-\lambda \lvert x\rvert}$$
It mainly has to do with the integration, I think, but I'll try to attempt it and illustrate where it all goes wrong.
$$\bar f(k) = \frac{D}{2\pi}\int_{-\infty}^{\infty}e^{ikx}e^{-\lambda \lvert x \rvert} \ dx$$
$$\bar f(k) = \frac{D}{2\pi}\int_{-\infty}^{\infty}e^{ikx-\lambda \lvert x \rvert} \ dx$$
From here I don't know how to integrate this. I tried a substitution, and quickly realized I couldn't due to the modulus. I was thinking it may be possible to convert the complex exponential into sines and cosines to try and integrate it by parts for each trigonometric component, but I don't think that will be doable since differentiating $e^{-\lambda \lvert x \rvert}$ will not allow for the typical way to integrate an exponential mutipled by a sine or cosine.
\begin{equation} \begin{split} \bar f(k) &\stackrel{(a)}{=} \frac{D}{2\pi}\int_{-\infty}^{\infty}e^{ikx}e^{-\lambda \lvert x \rvert} \ dx\\ &\stackrel{(b)}{=} \frac{D}{2\pi}\int_{-\infty}^{0}e^{ikx}e^{\lambda x } \ dx + \frac{D}{2\pi}\int_{0}^{\infty}e^{ikx}e^{-\lambda x} \ dx \\ &\stackrel{(c)}{=}\frac{D}{2\pi}\big( \frac{1}{\lambda + ik}e^{(\lambda+ik)x}\big]_{-\infty}^0 - \frac{1}{-\lambda + ik}e^{(-\lambda+ik)x}\big]^{\infty}_{0} \big)\\ &\stackrel{(d)}{=}\frac{D}{2\pi}\big( \frac{1}{\lambda + ik} - \frac{1}{-\lambda + ik} \big) \end{split} \end{equation} In $(a)$ we used the definition of Fourier transform.
In $(b)$ we have split the integral and made use of $\vert x \vert = -x$ if $x < 0$ and $\vert x \vert = x$ if $x \geq 0$.
In $(c)$, we just integrated.
In $(d)$, we replaced the limits.