Fourier transform with the derivative of a function

Question

I have to identify the Fourier transform, defined as

$\widehat f(x)=\displaystyle \int_{\mathbb R} e^{-ixy}f(y) dy$

As a task, I have to calculate the the fourier transform of

$g(x)= \frac{32}{1875}(3x^2e^{-\frac{4}{5}x}-\frac{4}{5}x^3e^{-\frac{4}{5}x})\mathbb{1}_{[0,\infty)}(x)$

As a hint, the task says I could use the fact, that:

$\widehat {f^{(k)}}(x)=i^kx^k\widehat{f}(x)$

So I used $k=1$ and defined ${f^{'}}(x)=g(x)$

Then we get

$f(x)=\frac{32}{1875}\Big[-\frac{15}{32}e^{-\frac{4}{5}x}(8x^2+20x+25) +\frac{1}{32}e^{-\frac{4}{5}x}(32x^3+120x^2+300x+375)\Big]$

which is a good sign, because $32$ reduces the fraction...

So, we get:

$f(x)=-\frac{1}{125}e^{-\frac{4}{5}x}(8x^2+20x+25)+\frac{1}{1875}e^{-\frac{4}{5}x}(32x^3+120x^2+300x+375)$

Now I could expand the brackets and apply in the definition of fourier transform but only with an integral from $0$ to $\infty$

$\widehat f(x)=-\frac{1}{125}(\widehat {e^{-\frac{4}{5}x}8x^2}+\widehat {e^{-\frac{4}{5}x}20x}+\widehat {e^{-\frac{4}{5}x}25})+\frac{1}{1875}(\widehat {e^{-\frac{4}{5}x}32x^3}+\widehat {e^{-\frac{4}{5}x}120x^2}+\widehat {e^{-\frac{4}{5}x}300x}+\widehat {e^{-\frac{4}{5}x}375})$

But it doesn't get easier. Is there anything, I do not see, which could make this more elegant?

Answer 1

Hint, too long for a comment.

1. It's really bad practice to use the same variable in both the spatial and Fourier domain, and will lead to massive confusion. Switching to more usual notation, note that the result is $\frac{d^k}{d \omega^k} \hat{f}(\omega) = \mathcal{F}[{i^k x^k f(x)}](\omega) $, where $\hat{f}(\omega) = \mathcal{F}[f(x)](\omega)$.

2. You know that $\mathcal{F}[e^{-ax} u(x)] = \frac{1}{j\omega + a}$. This is easy to differentiate. Try applying the above result to find $\mathcal{F}[ x^3 e^{-4x/5}]$ and $\mathcal{F}[ x^2 e^{-4x/5}]$

Answer 2

First write $$ \int_{0}^{\infty}e^{-\frac{4}{5}x}e^{-isx}dx =\frac{1}{\frac{4}{5}+is}. $$ Then $$ \begin{align} \int_{0}^{\infty}x^{2}e^{-\frac{4}{5}x}e^{-isx}dx & =-\frac{d^{2}}{ds^{2}}\int_{0}^{\infty}e^{-\frac{4}{5}x}e^{-isx}dx \\ & = -\frac{d^2}{ds^2}\frac{1}{\frac{4}{5}+is} \\ & = -(-1)(-2)(i)^{2}\frac{1}{(\frac{4}{5}+is)^{3}} \\ & = \frac{2}{(\frac{4}{5}+is)^{3}} \end{align} $$ Continuing, \begin{align} \int_{0}^{\infty}x^{3}e^{-\frac{4}{5}x}e^{-isx}dx & = i\frac{d}{ds}\frac{2}{(\frac{4}{5}+is)^{3}} \\ & = \frac{6}{(\frac{4}{5}+is)^{4}} \end{align}