We have a function to transform (get $F[f(x)]$):
$$f(x)=xe^{-\alpha|x|}$$
Using this formula (v.p. meaning):
$$F[f(y)]=v.p.\frac1{\sqrt{2\pi}}\int_{-\infty}^{+\infty}{f(t)e^{-ity}dt}$$
In my school, I went to the board and solved this. But now I tried to solve it again, and got other answer (other sign).
My way:
$$F[f(y)]= \frac1{\sqrt{2\pi}}\left(\int_{-\infty}^{0}{(f(t)e^{-ity}dt)}+\int_{0}^{+\infty}{(f(t)e^{-ity}dt)}\right)= \frac1{\sqrt{2\pi}}(\int_{-\infty}^{0}{(te^{\alpha t}e^{-ity}dt)}+\int_{0}^{+\infty}{(te^{-\alpha t}e^{-ity}dt)})= \frac1{\sqrt{2\pi}}(\int_{-\infty}^{0}{(te^{\alpha t-ity}dt)}+\int_{0}^{+\infty}{(te^{-\alpha t-ity}dt)}) $$
Replace with: ($\alpha t-ity=t\beta$) and ($-\alpha t-ity=t\gamma$) and continue:
$$F[f(y)]= \frac1{\sqrt{2\pi}}(\int_{-\infty}^{0}{(te^{t\beta}dt)}+\int_{0}^{+\infty}{(te^{t\gamma}dt)}) $$
And I just computed this two integrals using Wolframalpha:
$$F[f(y)]= \frac1{\sqrt{2\pi}}(-\frac1{\beta^2}+\frac1{\gamma^2}) $$
Now we should just recover replaced, simplify, and result will be like:
$$F[f(y)]=-\frac{4\alpha i y}{\sqrt{2\pi}(\alpha^2+y^2)^2}$$
Is it correct?
Result is correct (and there is no need for the principal value as the integrand is continuous and integrable). Wolfram can check 1D Fourier transforms (pic). $\renewcommand{\xi}y$
Another way to get the result by hand. Perhaps you know the result ($\mathcal Ff(\xi):=\int_{-\infty}^\infty f(x)e^{-ix\xi}dx) $ $$ F_1(\xi):= \mathcal F(\exp(-|x|))(\xi)=\frac2{\xi^2+1}$$ Scaling: $$ F_\alpha(\xi):=\mathcal F(\exp(-\alpha|x|)(\xi)=\frac1\alpha F_1\left(\frac\xi \alpha\right) = \frac{2\alpha}{\xi^2+\alpha^2}$$ Since $$\mathcal F (-ixf) = \int_{-\infty}^\infty f(x) (\partial_\xi e^{-ix\xi}) dx = \partial_\xi \mathcal F(f)(\xi) $$ we have $$ \mathcal F(x\exp(-\alpha|x|))(\xi)= i\partial_\xi F_\alpha(\xi) = i \cdot 2k \cdot \frac1{(\xi^2+\alpha^2)^2}\cdot (-1) \cdot 2\xi = \frac{-4i\alpha\xi}{(\xi^2+\alpha^2)^2}$$
which is your answer (up to convention for defining $\mathcal F$)