Problem 2
Let: $ \alpha > 0, \beta > 0, n\in\mathbb{N} $. Compute the Fourier transform for:
a. $ \left(x^2 + \alpha^2\right)^{-1} $
Answer
Let $ f\left(z\right) $ be holomorphic on a domain $ D $. Cauchy's formula states that for any positively oriented piecewise smooth simple closed curve $ \gamma \in D $ whose inside $ \Omega $ lies in $ D $:
$$ \forall z \in \Omega: f\left(z\right) = \frac{1}{2 \pi i}\int_\gamma \frac{f\left(\zeta\right)}{\left(\zeta - z\right)} d\zeta $$
$$ \text{Claim: } F\left(k\right) = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{\left(x - i\alpha\right)\left(x + i\alpha\right)}dx = \frac{ \pi e^{-\alpha |k|} }{\alpha } $$
Proof: Let: $ \gamma_r^+ $ be the positively oriented semi-circle on the upper half-plane with radius $ r $, and $ \gamma_r^- $ be the positively oriented semi-circle on the lower half-plane with radius $ r $. Then both $ \{\gamma_r^+,\gamma_r^-\} $ are positively oriented piecewise smooth simple closed curves. Moreover, $ f^+\left(z\right)= \frac{e^{-i k z}}{\left(z + i\alpha\right)} $ and $ f^-\left(z\right) = \frac{e^{-i k z}}{\left(z - i\alpha\right)} $ are holomorphic on $ \gamma_r^+ $ and $ \gamma_r^- $ respectively, as $ e^{z} $ is entire and we avoid the poles of the rational functions on their respective domains. We take cases and apply Cauchy's formula:
$$ \text{Suppose: } k < 0. \text{ Then: } \int_{\gamma_r^+}\frac{\frac{e^{-i k x}}{\left(x + i\alpha\right)}}{\left(x - i\alpha\right)}dx = 2 \pi i f^+\left(i\alpha\right) = 2\pi i \frac{e^{-i k \left(i \alpha\right)}}{\left(\left(i \alpha\right) + i\alpha\right)} = \frac{ \pi e^{\alpha k} }{\alpha } $$
$$ \text{Suppose: } k > 0. \text{ Then: } \int_{\gamma_r^-}\frac{\frac{e^{-i k x}}{\left(x - i\alpha\right)}}{\left(x + i\alpha\right)}dx = 2 \pi i f^-\left(i\alpha\right) = 2\pi i \frac{e^{-i k \left(i \alpha\right)}}{\left(\left(-i \alpha\right) - i\alpha\right)} = -\frac{ \pi e^{- \alpha k} }{\alpha } $$
$$ \text{Now, } \int_{\gamma_r}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx \text{ is the sum of two integrals:} $$
$$ \int_{\gamma_r^+}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx = \int_{-r}^{r}\frac{e^{-i k x}}{x^2 + \alpha^2}dx + \int_{C_r^+}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx = \frac{ \pi e^{\alpha k} }{\alpha }$$
$$ \int_{\gamma_r^-}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx = \int_{r}^{-r}\frac{e^{-i k x}}{x^2 + \alpha^2}dx + \int_{C_r^-}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx = -\frac{ \pi e^{- \alpha k} }{\alpha } $$
Where $ C_r $ is the upper or lower half of the half circle. We show that the $C_r$ portion of the integral vanishes as $ r \rightarrow \infty $. Without loss of generality we set $ \alpha $ to $ 1 $ to simplify the inequalities:
$$ \mid \int_{C_r^+}\frac{e^{-i k x}}{x^2 + 1}dx \mid \le \int_{C_r^+} \mid \frac{e^{-i k x}}{x^2 + 1} \mid \mid dx \mid \le \int_{C_r^+}\frac{1}{r^2 - 1} \mid dx \mid = \frac{\pi r}{r^2 - 1} \rightarrow 0 \text{ as } r \rightarrow \infty $$
$$ \text{Then: } \lim_{r \to +\infty} \int_{\gamma_r^+}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx \rightarrow \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx + 0 = \frac{ \pi e^{\alpha k} }{\alpha } $$
$$ \implies \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \frac{ \pi e^{\alpha k} }{\alpha } \text{ when } k < 0 $$
And similarly for our $ \gamma_r^- $ integral, changing the bounds of integration and the sign. We have then: $ \frac{ \pi e^{\alpha k} }{\alpha } $ for $ k < 0 $ and $ \frac{ \pi e^{-\alpha k} }{\alpha } $ for $ k > 0 $. Thus:
$$ F\left(k\right) = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \frac{ \pi e^{-\alpha |k|} }{\alpha } \blacksquare $$
b. $ x\left(x^2 + \alpha^2\right)^{-1} $
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} $. Then:
$$ g\left(x\right) = x\left(x^2 + \alpha^2\right)^{-1} = x f\left(x\right) \implies G\left(k\right) = i\frac{dF}{dk} = i \partial_k \left(\frac{ \pi e^{-\alpha |k|} }{\alpha }\right) = - i \pi sgn\left(k\right) e^{-\alpha |k|} \blacksquare $$
Where $ f\left(x\right) = \left(x^2 + \alpha^2\right)^{-1} $ and $ F\left(k\right) = \frac{ \pi e^{-\alpha |k|} }{\alpha } $ as found in part a.
c. i) $ \left(x^2 + \alpha^2\right)^{-1} \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) $
$$ \text{Now, } \cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right) \text{ so we write:} $$
$$ \left(x^2 + \alpha^2\right)^{-1}\cos\left(\beta x\right) = \frac{1}{2}\left(e^{i \beta x} \left(x^2 + \alpha^2\right)^{-1} + e^{- i \beta x} \left(x^2 + \alpha^2\right)^{-1}\right) $$
$$ \text{Then our transform is given by: } G\left(k\right) = \frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right) \text{ where: } $$
$$ F\left(k\right) = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \frac{ \pi e^{-\alpha |k|} }{\alpha } \text{ is our transform from part a.} $$
$$ \implies G\left(k\right) = \frac{1}{2} \left( \frac{ \pi e^{-\alpha |k - \beta|} }{\alpha } + \frac{ \pi e^{-\alpha |k + \beta|} }{\alpha }\right) \blacksquare $$
ii) $ \left(x^2 + \alpha^2\right)^{-1} \sin\left(\beta x\right)$
Answer
$$ \text{Proceeding as in part c. i): } \sin{\beta x} = \frac{1}{2i}\left(e^{i \beta x} - e^{- i \beta x}\right) \text{ so we write:} $$
$$ \left(x^2 + \alpha^2\right)^{-1}\sin\left(\beta x\right) = \frac{1}{2i}\left(e^{i \beta x} \left(x^2 + \alpha^2\right)^{-1} - e^{- i \beta x} \left(x^2 + \alpha^2\right)^{-1}\right) $$
$$ \text{Then our transform is given by: } G\left(k\right) = \frac{1}{2i}\left(F\left(k-\beta\right) - F\left(k+\beta\right)\right) \text{ where: } $$
$$ F\left(k\right) = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \frac{ \pi e^{-\alpha |k|} }{\alpha } \text{ is our transform from part a.} $$
$$ \implies G\left(k\right) = \frac{1}{2i} \left( \frac{ \pi e^{-\alpha |k - \beta|} }{\alpha } - \frac{ \pi e^{-\alpha |k + \beta|} }{\alpha }\right) \blacksquare $$
d. i) $ x \left(x^2 + \alpha^2\right)^{-1} \cos\left(\beta x\right)$
Answer
$$ \text{Proceeding as in part c. i): } \cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right) \text{ so we write:} $$
$$ x \left(x^2 + \alpha^2\right)^{-1}\cos\left(\beta x\right) = \frac{1}{2}\left(e^{i \beta x} x \left(x^2 + \alpha^2\right)^{-1} + e^{- i \beta x} x \left(x^2 + \alpha^2\right)^{-1}\right) $$
$$ \text{Then our transform is given by: } G\left(k\right) = \frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right) \text{ where: } $$
$$ F\left(k\right) = \int_{-\infty}^{\infty}\frac{x e^{-i k x}}{x^2 + \alpha^2}dx = - i \pi sgn\left(k\right) e^{-\alpha |k|} \text{ is our transform from part b.} $$
$$ \implies G\left(k\right) = \frac{1}{2} \left( - i \pi sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} - i \pi sgn\left(k + \beta\right) e^{-\alpha |k + \beta|}\right) $$
$$ = - \frac{i \pi}{2} \left(sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} + sgn\left(k + \beta\right) e^{-\alpha |k + \beta|}\right) \blacksquare $$
ii) $ x \left(x^2 + \alpha^2\right)^{-1} \sin\left(\beta x\right)$
Answer
$$ \text{Proceeding as in part c. ii): } \sin{\beta x} = \frac{1}{2i}\left(e^{i \beta x} - e^{- i \beta x}\right) \text{ so we write:} $$
$$ x \left(x^2 + \alpha^2\right)^{-1}\sin\left(\beta x\right) = \frac{1}{2i}\left(e^{i \beta x} x \left(x^2 + \alpha^2\right)^{-1} - e^{- i \beta x} x \left(x^2 + \alpha^2\right)^{-1}\right) $$
$$ \text{Then our transform is given by: } G\left(k\right) = \frac{1}{2i}\left(F\left(k-\beta\right) - F\left(k+\beta\right)\right) \text{ where: } $$
$$ F\left(k\right) = \int_{-\infty}^{\infty}\frac{x e^{-i k x}}{x^2 + \alpha^2}dx = - i \pi sgn\left(k\right) e^{-\alpha |k|} \text{ is our transform from part b.} $$
$$ \implies G\left(k\right) = \frac{1}{2i} \left( - i \pi sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} {\alpha } + i \pi sgn\left(k + \beta\right) e^{-\alpha |k + \beta|} {\alpha }\right) $$
$$ = - \frac{\pi}{2} \left(sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} - sgn\left(k + \beta\right) e^{-\alpha |k + \beta|}\right) \blacksquare $$
