Find the Fourier transform of the function $f(x) := x^k \cdot \chi_{[0,1]}(x)$, where $k = 1,2,3,\ldots$.
My attempt:
\begin{align*} \widehat{f}(\omega)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(x)e^{-i\omega x}dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{0}^{1}x^ke^{-i \omega x} dx\\ \end{align*}
According to Wolfram Alpha $\int_{0}^{1}x^ke^{-i \omega x} dx=(i\omega)^{-k-1}\Gamma(k+1)-\Gamma(k+1,i\omega)$, but I do not see how to obtain this. Could you please give me a hint?
Let $f(x)=x^k\xi_{[0,1]}(x)$. The Fourier transform $F(\omega)$ of $f(x)$ is given by
$$\begin{align} F(\omega)&=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty f(x) e^{-i\omega x}\,dx\\\\ &=\frac1{\sqrt{2\pi}}\int_0^1 x^k e^{-i\omega x}\,dx\\\\ &\overbrace{=}^{x=t/i\omega}\frac1{\sqrt{2\pi}} \int_0^{i\omega} \left(\frac{t}{i\omega}\right)^ke^{-t}\,\frac1{i\omega}\,dt\\\\ &=\frac1{\sqrt{2\pi}} (i\omega)^{-(k+1)}\int_0^{i\omega}t^k e^{-t}\,dt \\\\ &=\frac1{\sqrt{2\pi}} (i\omega)^{-(k+1)}\left(\int_0^\infty t^k e^{-t}\,dt-\int_{i\omega}^\infty t^k e^{-t}\,dt\right)\\\\ &=\frac1{\sqrt{2\pi}} (i\omega)^{-(k+1)}\left(\Gamma(k+1)-\Gamma(k+1,i\omega)\right) \end{align}$$
as was to be shown!