How to integrate $\int_{\mathbb R^n}\left(1-e^{i(x,y)}+i(x,y)1_{|y|\leq1}\right)|y|^{-\alpha-n}dy$?

Question

I came across this integration when reading some material about levy process $$\int_{\mathbb R^n}\left(1-e^{i(x,y)}+i(x,y)1_{|y|\leq1}\right)|y|^{-\alpha-n}dy$$ where $\alpha\in(0,2)$ and $(x,y)$ is the inner product.

I don't know how to integrate it. Any help, please!

Answer 1

Before you start with the computations, I recommend you to check that the integral is well-defined, i.e. that

$$\int |1-e^{i(x,y)}+i(x,y) 1_{|y| \leq 1}| \cdot |y|^{-\alpha-n} \, dy < \infty.$$

This follows for example by applying Taylor's formula. Define

$$\psi(x) := \int (1-e^{i(x,y)}+i(x,y) 1_{|y| \leq 1}) \cdot |y|^{-\alpha-n} \, dy.$$

Hints: (Calculation of $\psi$ up to a multiplicative constant)

1. Using symmetry, show that $\text{Im} , \psi(x)=0$ for all $x \in \mathbb{R}^n$. Hence, $$\psi(x) = \text{Re} \, \psi(x) = \int (1-\cos((x,y)) |y|^{-\alpha-n} \, dy.$$
2. Let $R \in \mathbb{R}^{n \times n}$ be an orthogonal matrix. Using the identity $$\psi(R \cdot x) = \int (1-\cos((x,R^T y))) |y|^{-\alpha-n} \, dy$$ show that $\psi(R \cdot x) = \psi(x)$, i.e. that $\psi$ is a rotational invariant function.
3. It follows from step 2 that $$\psi(x) = \psi(|x| \cdot e_1)$$ where $e_1 := (1,0,\ldots,0) \in \mathbb{R}^n$ denotes a unit vector in $\mathbb{R}^n$. By the very definition of $\psi$, we have $$\psi(|x| \cdot e_1) = \int (1-\cos(|x| y_1)) |y|^{-\alpha-n} \, dy.$$ Now perform a change of variables to conclude that $$\psi(|x| \cdot e_1) = c |x|^{\alpha}$$ for some constant $c>0$.

Hints: (Calculation of $\psi$)

1. It follows from a well-known formula for the Fourier transform of the normal distribution that $$\frac{1}{(4\pi r)^{n/2}} \int \exp \left(- \frac{|y|^2}{4r} \right) \,d y =1. \tag{1} $$
2. Use the identity $$1-e^{-ur} = \int_0^r u e^{-us} \, ds$$ and Tonelli's theorem to show that $$\int_0^{\infty} (1-e^{-ur}) \, \frac{dr}{r^{1+\beta}} = \frac{\Gamma(1-\beta)}{\beta} u^{\beta} \tag{2}$$ for any $\beta \in (0,1)$ and $u \geq 0$.
3. By $(2)$, we have $$|x|^{\alpha} = (|x|^2)^{\alpha/2} = \frac{\frac{\alpha}{2}}{\Gamma(1-\frac{\alpha}{2})} \int_0^{\infty} (1-e^{-r |x|^2}) \, \frac{dr}{r^{1+\alpha/2}}.$$ Hence, by $(1)$, $$|x|^{\alpha} = \frac{\frac{\alpha}{2}}{\Gamma(1-\frac{\alpha}{2})}\int_0^{\infty} \int (1-e^{ix y}) e^{-|y|^2/4r} \frac{1}{(4\pi r)^{n/2}} \, dy \frac{dr}{r^{1+\alpha/2}}.$$ Since the left-hand side is a real-number, also the right-hand side is a real number, i.e. the integral equals the real part of the integral. Thus, $$|x|^{\alpha} = \frac{\frac{\alpha}{2}}{\Gamma(1-\frac{\alpha}{2})}\int_0^{\infty} \int (1-\cos(xy)) e^{-|y|^2/4r} \frac{1}{(4\pi r)^{n/2}} \, dy \frac{dr}{r^{1+\alpha/2}}.$$
4. Apply Fubini's theorem and perform a change of variables ($s:= |y|^2/(4r)$ for fixed $y$) to conclude that $$|x|^{\alpha} = \frac{\frac{\alpha}{2} 4^{\alpha/2}}{\Gamma(1-\alpha/2) \pi^{n/2}} \int (1-\cos(xy)) \frac{dy}{|y|^{n+\alpha}} \underbrace{\int_0^{\infty} e^{-s} s^{n/2+\alpha/2} \frac{ds}{s}}_{=\Gamma((n+\alpha)/2)}.$$ Finally, we conclude $$|x|^{\alpha} = \frac{\alpha 2^{\alpha-1} \Gamma((n+\alpha)/2)}{\Gamma(1-\alpha/2) \pi^{n/2}} \int (1-\cos((x,y)) \frac{dy}{|y|^{n+\alpha}}.$$