Oscillating integral I want to calculate 
$$
\int _0^\infty e^{-iyx}\sqrt{x(x+2)}\, dx
$$
in the sense of distributions, at least for $y\ne 0$. Now, I happen to know the following integral representation for the modified Bessel function $K_1$:
$$
K_1(z) = ze^{-z}\int _0^\infty e^{-zx}\sqrt{x(x+2)}\, dx
$$
which, unfortunately, is only valid for $|\arg z|<\pi /2$. Otherwise I would of course have liked to put $z=iy$ with $y$ real. 
I'm wondering, if, in the case $y\ne 0$, I am somehow permitted to do so anyway? 
Perhaps I could do something with the convergent integral
$$
\int _0 ^\infty e^{-iyx}(\sqrt{x(x+2)} -x-1)\, dx \quad ?
$$
 A: Firstly, I will prove that the integral $\int_0^\infty  \exp(-i y x) \sqrt{x(x+2)} d x$ defines the distribution. That means the functional
$T:~\mathcal{D}(\mathbb{R}) \ni \phi \mapsto \int_0^\infty \mathcal{F}(\phi)(x) \sqrt{x(x+2)} d x \in \mathbb{C}$ 
is bounded ($\mathcal{F}(\phi)(x)=\int_\mathbb{R} \exp(-i y x) \phi(y) d y $). 
We have 
$|(T - \mathcal{F}(\Theta) - \mathcal{F}(x \Theta)) \phi|^2 \leq \left|\int_0^\infty \mathcal{F}(\phi)(x) (\sqrt{x(x+2)} - x - 1) d x\right|^2 \leq \\ \leq\left(\int_{-\infty}^\infty |\mathcal{F}(\phi)(x)|^2 d x \right) \left(\int_0^\infty | (\sqrt{x(x+2)} - x - 1)|^2 d x \right) \leq \\\leq \textrm{const} \left(\int_{-\infty}^\infty |\phi(y)|^2 d y \right) \leq \textrm{const} |\textrm{supp}(\phi)|^2 \|\phi \|_\infty^2$.
(I used Schwartz inequality, Plancherel theorem and the fact that support of $\phi$ is compact, $\Theta$ is the Heaviside theta function, $|\textrm{supp}(\phi)|$ is Lebesgue measure of the support of $\phi$) 
So $T -\mathcal{F}(\Theta) - \mathcal{F}(x\Theta)$ is distribution of order 0. Since $ \mathcal{F}(\Theta)$, $\mathcal{F}(x\Theta)$ are well-defined Schwartz distributions, $T$ is also distribution.
Now let's assume that $\phi\in \mathcal{D}(]0,\infty[)$. Because $\phi$ has compact support its Fourier transform is entire function. Moreover because $\textrm{supp}(\phi) \subset ]0,\infty[$, 
$\mathcal{F}(\phi)(z) = \int_\mathbb{R} \exp(-i y z) \phi(y) d y$ 
decays exponentially for $\textrm{Im} z < 0$. Thus using the Cauchy theorem we obtain
$T\phi = \int_0^\infty \mathcal{F}(\phi)(x) \sqrt{x(x+2)} d x = \int_0^{-i \infty} \mathcal{F}(\phi)(z) \sqrt{z(z+2)} d z = \\ =
(-i) \int_0^\infty \mathcal{F}(\phi)(-i x) \sqrt{(-i x)((-i x)+2)}  d x 
~~~~~~~(1)$ 
(I change the contour of integration. Because of mentioned exponential decay of integrand the integral over the part of the contour in infinity vanishes.)
Now, for $y>0$ we have
$K_1(i y) = y \exp(-i y) \int_0^\infty \exp(- x y) \sqrt{(-i x)((-i x)+2)}  d x =~~~~~~~(2)\\=
y \exp(-i y) \int_0^{e^{+i\pi/4}\infty} \exp(- x y) \sqrt{(-i x)((-i x)+2)}  d x =\\= 
e^{+i\pi/4} y \exp(-i y) \int_0^\infty \exp(- e^{+i \pi/4} x y) \sqrt{(e^{-i\pi/4} x)((e^{-i\pi/4} x)+2)}  d x$.
To see that the above formula really defines the analytic extension of the function $K_1$, let's perform the following computation
$K_1(y) =  y \exp(-y) \int_0^\infty \exp(- x y) \sqrt{x(x+2)}  d x =\\=
 y \exp(-y) \int_0^{e^{-i \pi/4}\infty} \exp(-  x y) \sqrt{x(x+2)}  d x =\\=
e^{-i \pi/4} y \exp(-y) \int_0^{\infty} \exp(- e^{-i \pi/4} x y) \sqrt{(e^{-i \pi/4}x)((xe^{-i \pi/4})+2)}  d x $.
(the representation of $K_1$ in the last line above allows to extend this function analytically on upper half of the imaginary axis).
Using equality (2) and analicity of $\mathcal{F}(\phi)$ we can write
$ \int \frac{K_1(i y)}{y \exp(i y)} \phi(y) d y =  \int_0^\infty \mathcal{F}(\phi)(-i x) \sqrt{(-i x)((-i x)+2)}  d x$
(note that we have assumed that $\textrm{supp}\phi\subset]0,\infty[$ so the first integral above is well defined).
After comparing above expression with (1) we conclude that
$T \phi = (-i)\int \frac{K_1(i y)}{y \exp(i y)} \phi(y) d y $,
so on $\mathcal{D}(]0,\infty[)$ we have $T(y) = \frac{-i K_1(i y)}{y \exp(-i y)}$. Using similar technique one can show that this equality holds also on $\mathcal{D}(]-\infty,0[)$. 
It's easy to guess that actually $T(y)=\lim_{\epsilon\searrow 0} \frac{-i y K_1(i y)}{(y-i \epsilon)^2 \exp(-i y)}$. Let's assume that it's not the case. Then $T(y)= \lim_{\epsilon\searrow 0} \frac{-i y K_1(i y)}{(y-i \epsilon)^2 \exp(-i y)} + S(y)$, where $S(y)$ is some nonzero distribution. We have
$\mathcal{F}(\Theta)(y) - \mathcal{F}(x \Theta))(y) = \lim_{\epsilon\searrow 0} \frac{-i y-1}{(y-i \epsilon)^2}$ and we know that $|(T - \mathcal{F}(\Theta) - \mathcal{F}(x \Theta)) \phi|\leq\textrm{const}|\textrm{supp}(\phi)|\|\phi\|_\infty$. The above condition implies that $|S\phi|\leq\textrm{const}|\textrm{supp}(\phi)|\|\phi\|_\infty$. On the other hand we know that $|S\phi|=0$ for $\phi\in\mathcal{D}(\mathbb{R}\setminus\{0\})$. That implies $S=0$ which contradicts the assumption.
