Problem: Let $f:\mathbb R\to\mathbb R$ be given by $f(x)=e^{-\vert x\vert}$.
a) Compute the Fourier transform of $f$, where we define the Fourier transform by
$$\widehat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-ix\xi}\,dx.$$
My Attempt: We have that
\begin{align*}
\widehat{f}(\xi)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\vert x\vert}e^{-ix\xi}\,dx\\
&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^0 e^xe^{-ix\xi}\,dx+\frac{1}{\sqrt{2\pi}}\int_0^\infty e^{-x}e^{-ix\xi}\,dx\\
&=\frac{1}{\sqrt{2\pi}}\left[\frac{e^{x(1-i\xi)}}{1-i\xi}\Bigg\vert_{-\infty}^0+\frac{-e^{-x(1+i\xi)}}{1+i\xi}\Bigg\vert_0^\infty\right]\\
&=\frac{1}{\sqrt{2\pi}}\frac{1}{1+\xi^2}.
\end{align*}
b) Use the above to solve the differential equation $-u''(x)+u(x)=g(x)$ where $g\in L^2(\mathbb R).$
My Attempt: Applying the Fourier transform to the equation we see that
$$\frac{\xi^2}{\sqrt{2\pi}}\widehat{u}(\xi)+\widehat{u}(\xi)=\widehat{g}(\xi),$$
and doing a bit of algebra yields
$$\widehat{u}(\xi)=\widehat{g}(\xi)\cdot\frac{1}{1+\xi^2/\sqrt{2\pi}}.$$
Using the Fourier inversion theorem, the convolution theorem, and the first part of the problem we have
$$u(x)=g\ast\left[\frac{1}{\sqrt{2\pi}}e^{\vert x\vert/\sqrt{2\pi}}\right](x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty g(x-y)e^{\vert y\vert/\sqrt{2\pi}}\,dy.$$
My question is: Do you agree with my calculation above?
Thank you for your time and feedback.