Calculation of $\int_{\mathbb R^n} e^{it|\xi + \frac{x}{2t}|^2} d \xi$

Question

Let $t > 0$ and $x \in \mathbb R^n$. I want to derive the following integral value $$ \int_{\mathbb R^n} \exp\left(it\left|\xi +\frac{x}{2t}\right|^2\right) d \xi $$ How can I calculate this?

Answer 1

Enforce the substitution $\xi_i+x_i/2t\to \xi_i$ to arrive at

$$\begin{align} \int_{\mathbb{R}^n}e^{it|\xi+x/2t|^2}\,d\xi&=\int_{\mathbb{R}^n}e^{it|\xi|^2}\,d\xi\\\\ &=\prod_{i=1}^n \int_{-\infty}^\infty e^{it\xi_i^2}\,d\xi_i\\\\ &=\left(\int_{-\infty}^\infty e^{it\xi_i^2}\,d\xi_i\right)^n \end{align}$$

Now, finish by recalling that the value of the Fresnel Integral is $\int_{-\infty}^\infty e^{it\xi_i^2}\,d\xi_i=\sqrt{\frac{\pi}{2t}}(1+i)$

Answer 2

A translation of the variable doesn't affect the value of the integral, to see this you can perform the substitution $z=\xi+\frac{x}{2t}$ and notice that nothing changes in the integral (the jacobian of the substitution is 1), thus

$$ \int_{\mathbb{R}^n}e^{it|\xi+\frac{x}{2t}|^2}\mathrm{d}\xi=\int_{\mathbb{R}^n}e^{it|z|^2}\mathrm{d}z=\int_{\mathbb{R}^n}e^{it<z,z>}\mathrm{d}z=\prod_{k=1}^{n}\int_{\mathbb{R}}e^{itz_k^2} \mathrm{d}z_k$$

substitute $itz_k^2$ with some $s^2$ to obtain the integral of a gaussian