The Green's function of the beam deflection equation This is a problem in a textbook used in my class:

Suppose we have an infinite elastic beam, where the deflection $u(x)$
  satisfies the differential equation $$\frac{d^4 u}{dx^4}+k^4 u =
> f(x),$$ where $k^4$ is a positive constant regarded as known, and
  $f(x)$ is a load.
For part (a) of the question, we assume that the load is a unit
  concentrated load at $x = \xi$, so that it satisfies we equation
  $$\frac{d^4 u}{dx^4}+k^4 u = \delta(x - \xi),$$ where $\delta$ is the
  Dirac-delta equation.

So, for this problem I would like to find the deflection (which is equal to the free space Green's function). I've looked in several mechanics textbooks to find a detailed approach as to how to calculate this, but only found solutions for different problems.
 A: Finally found a way to do it using Fourier transform, but a friend of mine did it just using calculations (much more tedious).
Since the deflection $u(x)$ satisfies the above differential equation, the Green's function satisfies
\begin{equation}
 \frac{\partial^4 g}{\partial x^4}+\alpha^4 g = \delta(x - \xi),
 \end{equation}
and assuming that $g = \frac{\partial g}{\partial x} = \frac{\partial^2 g}{\partial x^2} = \frac{\partial g^3}{\partial x^3} = 0$ at $|x| = \infty$, we have the solution $$u(x) = \int_{-\infty}^{\infty}g(x,\xi)f(\xi)d\xi.$$
Next, define the Fourier transforms $$\hat{u}(k) = \int_{-\infty}^{\infty}u(\xi) e^{ik\xi}d\xi; \; u(\xi) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-ik\xi}\hat{u}(k)dk$$ and $$\hat{u'} = \int_{-\infty}^{\infty}u'(\xi)e^{ik\xi}d\xi = u(\xi)e^{ik\xi}|_{-\infty}^{\infty} - ik \int_{-\infty}^{\infty}u(\xi)e^{ik\xi}d\xi = (-ik)\hat{u}(k).$$
Applying the Fourier transform to our solution, we get $(k^4+\alpha^4)\hat{g} = e^{ik\xi},$ so 
    \begin{equation*}
 \hat{g} = \frac{e^{ik\xi}}{k^4 + \alpha^4}.
 \end{equation*}
Using the Fourier inversion formula, we obtain
    \begin{align}
  g(x,\xi)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{ikx}e^{ik\xi}}{k^4 + \alpha^4}dk\\
  & = \frac{1}{\pi}\int_0^\infty \frac{\cos k(x-\xi)}{k^4+\alpha^4}dk\\
  &=\frac{e^{-\alpha|x-\xi|/\sqrt{2}}}{2\alpha^3}\sin\left(\frac{\alpha|x-\xi|}{\sqrt{2}}+\frac{\pi}{4}\right).
 \end{align}
