I ended up solving with $x^n$ in this problem. Notice that the messy integration in part b. could have been avoided by using properties of the Fourier transform as is done later in the problem.
I take the definition of the Fourier transform $ \hat{f} $ for a function $ f $ to be: $ F\left(k\right) = \hat{f}\left(k\right) = \int_{-\infty}^{\infty}f\left(x\right) e^{-i k x}dx, $ with inverse Fourier transform $ f\left(x\right) = \check{F}\left(x\right) = \int_{-\infty}^{\infty}F\left(k\right) e^{i k x}\frac{dk}{2 \pi} $.
Problem
Let: $ \alpha > 0, \beta > 0, n\in\mathbb{N} $. Compute the Fourier transform for:
a. $e^{-\alpha |x|} $
Answer a.
$$ F\left(k\right) = \int_{-\infty}^{\infty}e^{-\alpha |x|} e^{-i k x}dx = \int_{0}^{\infty}e^{-\alpha \left(x\right)} e^{-i k x}dx + \int_{-\infty}^{0}e^{-\alpha \left(-x\right)} e^{-i k x}dx$$
$$= \int_{0}^{\infty}e^{-\left(\alpha + i k\right)x}dx + \int_{-\infty}^{0}e^{\left(\alpha - i k \right)x}dx = \frac{1}{-\left(\alpha + i k \right)}e^{-\left(\alpha + i k\right)x}\Bigr|_{x=0}^{\infty} + \frac{1}{\left(\alpha- i k\right)}e^{\left(\alpha - i k\right)x}\Bigr|_{x=-\infty}^{0} $$
$$ = \lim_{x \to +\infty} \frac{1}{-\left(\alpha + i k \right)}e^{-\left(\alpha + i k\right)x} - \frac{1}{-\left(\alpha + i k \right)}e^{-\left(\alpha + i k\right)\left(0\right)} + \frac{1}{\left(\alpha- i k\right)}e^{\left(\alpha - i k\right)\left(0\right)} - \lim_{x \to -\infty} \frac{1}{\left(\alpha- i k\right)}e^{\left(\alpha - i k\right)x}$$
$$= 0 + \frac{1}{\left(\alpha + i k \right)} + \frac{1}{\left(\alpha- i k\right)} - 0 = \frac{\left(\alpha- i k\right)+\left(\alpha + i k \right)}{\left(\alpha + i k \right)\left(\alpha- i k\right)} = \frac{2\alpha}{\alpha ^2 - k^2} $$
$$\implies \frac{2\alpha}{\alpha ^2 - k^2} = F\left(k\right) \text{ is our Fourier transform }\blacksquare$$
b. i) $ e^{-\alpha |x|}\cos\left(\beta x\right)$
Answer
$$ F\left(k\right) = \int_{-\infty}^{\infty}e^{-\alpha |x|} \cos\left(\beta x\right) e^{-i k x}dx = \int_{0}^{\infty}e^{-\alpha \left(x\right)} e^{-i k x} \cos\left(\beta x\right)dx + \int_{-\infty}^{0}e^{-\alpha \left(-x\right)} e^{-i k x} \cos\left(\beta x\right)dx$$
$$ \text{Now, } \forall\{a,b\}\in\mathbb{C}: \int e^{a x} \cos\left(b x\right) dx = \frac{e^{a x}}{a^2 + b^2}\left(b \sin\left(b x\right) + a \cos\left(b x\right)\right) \text{ so out integral is:}$$
$$ \int_{0}^{\infty}e^{-\left(\alpha + i k\right) x} \cos\left(\beta x\right)dx + \int_{-\infty}^{0}e^{\left(\alpha -i k\right) x} \cos\left(\beta x\right)dx $$
$$ = \frac{e^{\left(-\left(\alpha + i k\right)\right) x}}{\left(-\left(\alpha + i k\right)\right)^2 + \beta^2}\left(\beta \sin\left(\beta x\right) + \left(-\left(\alpha + i k\right)\right) \cos\left(\beta x\right)\right)\Bigr|_{x=0}^{\infty} + \frac{e^{\left(\alpha -i k\right) x}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta x\right) + \left(\alpha -i k\right) \cos\left(\beta x\right)\right)\Bigr|_{x=-\infty}^{0} $$
$$ = \lim_{x \to +\infty} \frac{e^{-\left(\alpha + i k\right) x}}{\left(\alpha + i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta x\right) - \left(\alpha + i k\right) \cos\left(\beta x\right)\right) - \frac{e^{-\left(\alpha + i k\right) \left(0\right)}}{\left(\alpha + i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta \left(0\right)\right) -\left(\alpha + i k\right) \cos\left(\beta \left(0\right)\right)\right)$$
$$ + \frac{e^{\left(\alpha -i k\right) \left(0\right)}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta \left(0\right)\right) + \left(\alpha -i k\right) \cos\left(\beta \left(0\right)\right)\right) - \lim_{x \to -\infty} \frac{e^{\left(\alpha -i k\right) x}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta x\right) + \left(\alpha -i k\right) \cos\left(\beta x\right)\right)$$
$$ = 0 - \frac{-\left(\alpha + i k\right)}{\left(\alpha + i k\right)^2 + \beta^2} + \frac{\left(\alpha -i k\right)}{\left(\alpha -i k\right)^2 + \beta^2} - 0 $$
$$ \implies \frac{\left(\alpha + i k\right)}{\left(\alpha + i k\right)^2 + \beta^2} + \frac{\left(\alpha -i k\right)}{\left(\alpha -i k\right)^2 + \beta^2} = F\left(k\right) \text{ is our Fourier transform } \blacksquare$$
b. ii) $e^{-\alpha |x|}\sin\left(\beta x\right)$
Answer
$$ F\left(k\right) = \int_{-\infty}^{\infty}e^{-\alpha |x|} \sin\left(\beta x\right) e^{-i k x}dx = \int_{0}^{\infty}e^{-\alpha \left(x\right)} e^{-i k x} \sin\left(\beta x\right)dx + \int_{-\infty}^{0}e^{-\alpha \left(-x\right)} e^{-i k x} \sin\left(\beta x\right)dx$$
$$ \text{Now, } \forall\{a,b\}\in\mathbb{C}: \int e^{a x} \sin\left(b x\right) dx = \frac{e^{a x}}{a^2 + b^2}\left(a \sin\left(b x\right) - b \cos\left(b x\right)\right) \text{ so out integral is:}$$
$$ \int_{0}^{\infty}e^{-\left(\alpha + i k\right) x} \sin\left(\beta x\right)dx + \int_{-\infty}^{0}e^{\left(\alpha -i k\right) x} \sin\left(\beta x\right)dx $$
$$ = \frac{e^{\left(-\left(\alpha + i k\right)\right) x}}{\left(-\left(\alpha + i k\right)\right)^2 + \beta^2}\left(\left(-\left(\alpha + i k\right)\right) \sin\left(\beta x\right) - \beta \cos\left(\beta x\right)\right)\Bigr|_{x=0}^{\infty} $$
$$ + \frac{e^{\left(\alpha -i k\right) x}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\left(\alpha -i k\right) \sin\left(\beta x\right) - \beta \cos\left(\beta x\right)\right)\Bigr|_{x=-\infty}^{0} $$
$$ = \lim_{x \to +\infty} \frac{e^{-\left(\alpha + i k\right) x}}{\left(\alpha + i k\right)^2 + \beta^2}\left(-\left(\alpha + i k\right) \sin\left(\beta x\right) - \beta \cos\left(\beta x\right)\right) - \frac{e^{-\left(\alpha + i k\right) \left(0\right)}}{\left(\alpha + i k\right)^2 + \beta^2}\left(-\left(\alpha + i k\right) \sin\left(\beta \left(0\right)\right) - \beta \cos\left(\beta \left(0\right)\right)\right)$$
$$ + \frac{e^{\left(\alpha -i k\right) \left(0\right)}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\left(\alpha -i k\right) \sin\left(\beta \left(0\right)\right) - \beta \cos\left(\beta \left(0\right)\right)\right) - \lim_{x \to -\infty} \frac{e^{\left(\alpha -i k\right) x}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\left(\alpha -i k\right) \sin\left(\beta x\right) - \beta \cos\left(\beta x\right)\right)$$
$$ = 0 - \frac{-\beta}{\left(\alpha + i k\right)^2 + \beta^2} + \frac{- \beta}{\left(\alpha -i k\right)^2 + \beta^2} - 0 $$
$$ \implies \frac{\beta}{\left(\alpha + i k\right)^2 + \beta^2} - \frac{\beta}{\left(\alpha -i k\right)^2 + \beta^2} = F\left(k\right) \text{ is our Fourier transform } \blacksquare$$
c. $ x^n e^{-\alpha |x|}$
Answer
We have for for any function $ f\left(x\right) $ with Fourier transform $ F\left(k\right), $ the transform of $ g\left(x\right) = x f\left(x\right) $ is given by: $ G\left(k\right) = i\frac{dF}{dk} $. Moreover, it's clear that for: $ g\left(x\right) = x^n f\left(x\right) $, $ G\left(k\right) = i^n\frac{d^n F}{d k^n} $.
Proof:} We proceed by induction. Our base case is obvious: for $ n = 1 $, $ g\left(x\right) = x^n f\left(x\right) = x f\left(x\right) \implies G\left(k\right) = i\frac{dF}{dk} = i^n\frac{d^n F}{d k^n} $. Suppose we have for some $ n \in \mathbb{N} $, $ h\left(x\right) = x^n f\left(x\right) \implies H\left(k\right) = i^n\frac{d^n F}{d k^n}.$ Then, $ g\left(x\right) = x^{n+1} f\left(x\right) = x h\left(x\right) \implies G\left(k\right) = i \cdot i^n\partial_k H\left(k\right) = i^{n+1} \partial_k \frac{d^n F}{d k^n} = i^{n+1}\frac{d^{n+1} F}{d k^{n+1}} $ and $ \forall n \in \mathbb{N}: g\left(x\right) = x^n f\left(x\right) \implies i^n\frac{d^n F}{d k^n} $, as needed $ \square $
Now, we found in part a. that the Fourier transform for $ f\left(x\right) = e^{-\alpha |x|} $ is given by: $ F\left(k\right) = \frac{2\alpha}{\alpha ^2 - k^2}$. Examining the derivatives of $ F\left(k\right)$:
$$ F\left(k\right) = \frac {2\alpha} {\alpha^2-k^2}, F'\left(k\right) = \frac{4 k \alpha }{\left(\alpha ^2-k^2\right)^2}, F''\left(k\right) = 2 \left(\frac{-1}{\left(\alpha-k \right)^3}+\frac{1}{\left(\alpha+k \right)^3}\right) $$
$$ F'''\left(k\right) = 6 \left( \frac{1}{\left(\alpha-k\right)^4}-\frac{1}{\left(\alpha+k \right)^4}\right), F^{\left(4\right)}\left(k\right) = 24 \left(\frac{-1}{\left(\alpha-k \right)^5}+\frac{1}{ \left(\alpha+k \right)^5}\right)$$
$$ \text{So for: } n \in \mathbb{N}: F^n \left(k\right) = n! \left(\frac{\left(-1\right)^{n+1}}{\left(\alpha-k\right)^{n+1}}+\frac{\left(-1\right)^n}{\left(\alpha+k\right)^{n+1}} \right) \text{, and:} $$
$$ g\left(x\right) = x^n f\left(x\right) = x^n e^{-\alpha |x|} \implies G\left(k\right) = i^n\frac{d^n F}{d k^n} = i^n n! \left(\frac{\left(-1\right)^{n}}{\left(\alpha-k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+k\right)^{n+1}} \right) \text{ is our transform } \blacksquare $$
d. i) $ x^n e^{-\alpha |x|}\cos\left(\beta x\right)$
Answer
Two properties of the Fourier transform are that for: $ g\left(x\right) = e^{i a x} f\left(x\right) $ the Fourier transform of $ g\left(x\right) $ is given by $ G\left(k\right) = F\left(k-a\right) $, and that the transform is linear: $ h\left(x\right) = a f\left(x\right) + b g\left(x\right) \implies H\left(k\right) = a F\left(k\right) + b G\left(k\right) $
$$ \text{Now, } \cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right) \text{ so we write:} $$
$$ x^n e^{-\alpha |x|}\cos\left(\beta x\right) = \frac{1}{2}\left(e^{i \beta x} x^n e^{-\alpha |x|} + e^{- i \beta x} x^n e^{-\alpha |x|}\right) $$
$$ \text{Then our transform is given by: } G\left(k\right) = \frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right) \text{ where: } $$
$$ F\left(k\right) = i^n n! \left(\frac{\left(-1\right)^{n}}{\left(\alpha-k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+k\right)^{n+1}} \right) \text{ is our transform from part c. for } x^n e^{-\alpha |x|}$$
$$ \implies G\left(k\right) = \frac{1}{2} i^n n! \left( \frac{\left(-1\right)^{n}}{\left(\alpha+ \beta - k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha -\beta + k\right)^{n+1}} + \frac{\left(-1\right)^{n}}{\left(\alpha -\beta -k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+\beta +k\right)^{n+1}} \right) \blacksquare $$
ii) $x^n e^{-\alpha |x|}\sin\left(\beta x\right)$
Answer
Proceeding as in part i) we use that: $ \sin{\beta x} = \frac{1}{2 i}\left(e^{i \beta x} - e^{- i \beta x}\right) $ to write:
$$ g\left(x\right) = x^n e^{-\alpha |x|}\sin\left(\beta x\right) = \frac{1}{2i}\left(e^{i \beta x} x^n e^{-\alpha |x|} - e^{- i \beta x} x^n e^{-\alpha |x|}\right) $$
$$ \text{Which yields the transform: } G\left(k\right) = \frac{1}{2i}\left(F\left(k-\beta\right) - F\left(k+\beta\right)\right) \text{ with: } F\left(k\right) = i^n n! \left(\frac{\left(-1\right)^{n}}{\left(\alpha-k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+k\right)^{n+1}} \right) \text{ and so:} $$
$$ G\left(k\right) = \frac{1}{2 i} i^n n! \left( \frac{\left(-1\right)^{n}}{\left(\alpha+ \beta - k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha -\beta + k\right)^{n+1}} - \frac{\left(-1\right)^{n}}{\left(\alpha -\beta -k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+\beta +k\right)^{n+1}} \right) \blacksquare $$
