How do I solve this:
Newest image: After i corrected myself
Compute the Fourier transform of $$f(x)=xe^{-|x|}$$
My answer is not correct:
\begin{align*} \sqrt{2\pi}\hat f(w)&=\int_0^\infty xe^{-x}e^{-iwx}\mathop{dx}+\int_{-\infty}^0 xe^xe^{-iwx}\mathop{dx}\\ &=\int_0^\infty xe^{-(1+iw)x}\mathop{dx}+\int_{-\infty}^0 xe^{-(1+iw)x}\mathop{dx} \end{align*}
\begin{align*} x&&&e^{(1-iw)x}\\ -1&&&-\frac{e^{(1-iw)x}}{1-iw}\\ 0&&&\frac{e^{(1-iw)x}}{(1-iw)^2} \end{align*}
\begin{align*} \left[-\frac{xe^{-(1+iw)x}}{1+iw}-\frac{e^{-(1+iw)x}}{(1+iw)^2}\right]_0^\infty+\left[\frac{xe^{(1+iw)x}}{1+iw}-\frac{e^{(1+iw)x}}{(1+iw)^2}\right]_0^\infty&=\frac{1}{(1+iw)^2}-\frac{1}{(1-iw)^2}\\ &=\frac{(1-iw)^2-(1+iw)^2}{(1+iw)^2(1-iw)^2}\\ &=\frac{-4iw}{(1+iw)^2(1-iw)^2}\\ \implies \hat f(w)=\frac{-4iw}{\sqrt{2\pi}(1+iw)^2(1-iw)^2} \end{align*}
I think that easiest is to use the fact that $$ \int_{-\infty}^{\infty} x f(x) e^{-iwx} \, dx = i \frac{d}{dw} \int_{-\infty}^{\infty} f(x) e^{-iwx} \, dx. $$
So first we calculate $$ \int_{-\infty}^{\infty} e^{-|x|} e^{-iwx} \, dx = \int_{-\infty}^{0} e^{x} e^{-iwx} \, dx + \int_{0}^{\infty} e^{-x} e^{-iwx} \, dx = \int_{-\infty}^{0} e^{(1-iw)x} \, dx + \int_{0}^{\infty} e^{(-1-iw)x} \, dx \\ = \left[ \frac{1}{1-iw} e^{(1-iw)x} \right]_{-\infty}^{0} + \left[ \frac{1}{-1-iw} e^{(-1-iw)x} \right]_{0}^{\infty} = \frac{1}{1-iw} - \frac{1}{-1-iw} = \frac{2}{1+w^2}. $$
Then we take the derivative: $$ \int_{-\infty}^{\infty} x e^{-|x|} e^{-iwx} \, dx = i \frac{d}{dw} \frac{2}{1+w^2} = i \frac{4w}{(1+w^2)^2}. $$
Thus, $$ \hat{f}(w) = \frac{i}{\sqrt{2\pi}} \frac{4w}{(1+w^2)^2}. $$