Fourier Transform of a function that is zero for $x<0$. I have this function
$$
f(x)=xH(x)\ln{(x)}e^{-x},
$$
of which I need the Fourier transform, where $H(x)$ is the Heaviside function. Maple tells me the Fourier transform is
$$
{\frac { \left({\omega}+i \right)^2 \left(\ln  {\left(\sqrt{
\omega^2+1}\right)} + i \arctan \left( \omega \right) + \gamma-1 \right) }{ \left( {\omega}^{2}+1
 \right) ^{2}}}
,
$$
where $\gamma$ is Euler's constant.
I have computed it several different ways through Maple and I believe it is the right formula. I tried to do it analytically but I never quite got there. If someone would know the steps to get there, I would really appreciate it.
I need this result because I need to check in which fractional Sobolev spaces $H^s(\mathbb{R})$ does $f$ belong to.
 A: The Fourier transform of a function $f$ is traditionally denoted ${\displaystyle {\hat {f}}}$, by adding a circumflex to the symbol of the function. There are several common conventions for defining the Fourier transform of an integrable function ${\displaystyle f:\mathbb {R} \to \mathbb {C} }$. One of them is:
$$\hat{f}\left(\xi\right):=\int_\mathbb{R}f(x)\exp\left(-2\pi x\xi i\right)\space\text{d}x\tag1$$
for any real number $\xi$ with $i^2=-1$.
Using your function, we can notice that:
$$\hat{f}\left(\xi\right)=\int_0^\infty x\ln\left(x\right)\exp\left(-x\right)\exp\left(-2\pi x\xi i\right)\space\text{d}x\tag2$$
Let's define a second integral:
$$\mathscr{S}_\text{n}\left(\text{s}\right):=\int_0^\infty x\ln\left(x\right)\exp\left(-\text{s}x\right)\exp\left(\text{n}x\right)\space\text{d}x\tag3$$
Using Laplace transform, we can write:
$$\mathscr{S}_\text{n}\left(\text{s}\right)=\mathscr{L}_x\left[x\ln\left(x\right)\exp\left(\text{n}x\right)\right]_{\left(\text{s}\right)}\tag4$$
Using the frequency-domain derivative and frequency shifting property of the Laplace transform, we can write:
$$\mathscr{S}_\text{n}\left(\text{s}\right)=-\frac{\partial}{\partial\text{s}}\left\{\mathscr{L}_x\left[\ln\left(x\right)\right]_{\left(\text{s}-\text{n}\right)}\right\}\tag5$$
Using table of selected Laplace transforms, we can write:
$$\mathscr{S}_\text{n}\left(\text{s}\right)=-\frac{\partial}{\partial\text{s}}\left\{\frac{\ln\left(\text{s}-\text{n}\right)+\gamma}{\text{n}-\text{s}}\right\}=\frac{1-\gamma-\ln\left(\text{s}-\text{n}\right)}{\left(\text{n}-\text{s}\right)^2}\tag6$$
Where $\gamma$ is the Euler–Mascheroni constant.
Going back to $(2)$ we can see that:
$$\hat{f}\left(\xi\right)=\int_0^\infty x\ln\left(x\right)\exp\left(-x\right)\exp\left(-2\pi x\xi i\right)\space\text{d}x=\frac{1-\gamma-\ln\left(1-\left(-2\pi\xi i\right)\right)}{\left(\left(-2\pi\xi i\right)-1\right)^2}=$$
$$\frac{1-\gamma-\ln\left(1+2\pi\xi i\right)}{\left(\left(-2\pi\xi i\right)-1\right)^2}=\frac{1-\gamma-\ln\left(1+2\pi\xi i\right)}{\left(1+2\pi\xi i\right)^2}\tag7$$
