To understand integral :$\lambda (x) = \int_{0}^{\infty} \frac{\sin^{2} \alpha x}{\alpha^{2}} d\mu(\alpha), (\mu(0)=0)$

Question

I wants to understand the integrals of the form

$$\lambda (x) = \int_{0}^{\infty} \frac{\sin^{2} \alpha x}{\alpha^{2}} d\mu(\alpha), (\mu(0)=0)$$ where $\mu(\alpha)$ is a non decreasing function such that the integral converges for all real $x.$

Trivial Example: If we take, $\mu(\alpha)= \alpha,$ then $\lambda (x)= |x|\frac{\pi}{2}.$

My Questions are:

(1) Can you give few more examples of non decreasing function $\mu(\alpha)$, so that we know the exact value of $\lambda (x).$ ?

(2) Suppose the above integral ($\lambda (x)$) converges for the given non decreasing function $\mu(\alpha).$ Can we expect(to evolute) to get the exact(precise) values of $\lambda(x), (x\in \mathbb R)$ ?

(3) Why $\lambda (x)$ in the literature it is known as "negative definite function" ; can you motivates me bit ? (For instance, Arne Beurling, in his paper, "on the spectral synthesis of bounded functions" called this $\lambda (x)$ as negative definite functions)

(4) Does there exists any well-known analogue of ``negative definite functions" defined in $\mathbb R^{2} (\mathbb R^{n})$ ? If not, what one can expect ?

Thanks,

Answer 1

(1) Denote by $\mu(\alpha)$ the Radon-Nikodym derivative $\frac{d\mu}{d\alpha}$. Notice that if we extend $\mu(\alpha)$ to the negative reals by $\mu(-\alpha)=\mu(\alpha)$, the integral becomes \begin{equation}-2\int_{-\infty}^{\infty}e^{-i\alpha x}\frac{\mu(\frac{\alpha}{2})}{\alpha^2}d\alpha+\frac{1}{2}\int_{0}^{\infty}\frac{\mu(\alpha)}{\alpha^2}d\alpha.\end{equation} Therefore, what needs to be computed is the Fourier transform of $\frac{\mu(\frac{\alpha}{2})}{\alpha^2}$. For example, if $\mu(\frac{\alpha}{2})=\alpha^{2-p}$ where $0<p<2$, the Fourier transform is \begin{equation} \frac{2(\sin \frac{\pi p}{2})\Gamma(1-p)}{|\alpha|^{1-p}}, \end{equation} (after that the original integral is easy to compute), and you can find more examples here. For example, the case $\mu(\alpha)=\alpha^2\log(1+|\alpha|)$ can be computed explicitly (although the result will be a distribution).

(2) Obviously, there is no general method for these integrals, but if you know the function $\mu$ well enough, and if it is analytic, you can compute the integral by the residue theorem. So for exaple if $\mu(\alpha)$ is an even rational function such that the integral converges, it can be computed.

(3) By Bochner's theorem, the Fourier transform of a nonnegative function is positive definite, so its negative is negative definite.

(4) Negative definite functions can even be defined on any group, see this.