# Fourier transform of tempered distribution

I'm wondering whether anyone knows a reference or proof for finding the Fourier transform of $f(t):=(t+1)^{1/2}t_+^{1/2}$? (Here $t_+=\max (t,0)$.)

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$$\tfrac{1}{2}\int_{0}^{\infty}e^{i\omega t}dt=\tfrac{1}{2}\pi\delta(\omega)+\tfrac{1}{2}i\omega^{-1}$$
$$\int_{0}^{\infty}te^{i\omega t}dt=-i\frac{d}{d\omega}\int_{0}^{\infty}e^{i\omega t}dt=-i\pi\frac{d}{d\omega}\delta(\omega)-\omega^{-2}$$
$$\int_{0}^{\infty}[\sqrt{t(1+t)}-t-\tfrac{1}{2}]e^{i\omega t}dt=\omega^{-2}-\tfrac{1}{2}i\omega^{-1}+\tfrac{1}{2}i\omega^{-1}e^{-i\omega/2}K_{1}(-i\omega/2)$$ adding these three results gives the required Fourier transform $$\int_{0}^{\infty}\sqrt{t(1+t)}e^{i\omega t}dt=\tfrac{1}{2}\pi\delta(\omega)-i\pi\frac{d}{d\omega}\delta(\omega)+\tfrac{1}{2}i\omega^{-1}e^{-i\omega/2}K_{1}(-i\omega/2)$$ with $K_1$ a modified Bessel function of the second kind; the derivative of the Dirac delta function should be understood in the context of an integral, $$\int_{-\infty}^{\infty}f(\omega)\frac{d}{d\omega}\delta(\omega)d\omega=-\lim_{\omega\rightarrow 0}\frac{d}{d\omega}f(\omega)$$
The second integral is the output of Mathematica, with the condition ${\rm Im}\,\omega>0$. Since the integrand decays as $1/t$ for large $t$, this condition can be extended to ${\rm Im}\,\omega\geq 0$. – Carlo Beenakker Jul 3 '13 at 14:42
Alternatively, it can be justified by referring to one of the integral representations for $K_1$. Like this one: dlmf.nist.gov/10.32.E8 – Igor Khavkine Jul 3 '13 at 14:51
@Alex A: $\int_0^{\infty}t^{-1}\sin(\omega t)dt=\pi/2\times{\rm sign}(\omega)$ for $\omega$ real; similarly, the integral that decays as $t^{-1}\cos\omega t$ is finite. OK? – Carlo Beenakker Jul 12 '13 at 15:04
@Alex A: you are right, my mistake, I have corrected it; the final answer now contains an extra term $(\pi/2)\delta(\omega)$ – Carlo Beenakker Jul 24 '13 at 10:52