Is $\int_{\mathbb{R}} \int_{\mathbb{R}^n} \alpha w(t) e(\alpha (a_1t_1 + \dotsb + a_n t_n)) dt\,d \alpha = 0$? Let $a_i$ be a nonzero real number for each $1 \leq i \leq n$. $w$ a smooth nonnegative with compact support. I would like to understand the following integral.
$$
I = \int_{\mathbb{R}} \int_{\mathbb{R}^n} \alpha w(t) e(\alpha (a_1t_1 + \dotsb + a_n t_n)) dt\,d \alpha.
$$
If the factor $\alpha$ were not present in the integrand, then the contribution comes from where $\alpha = 0$, but with this additional factor, I was thinking that perhaps $I$ is $0$? or something with very small absolute value. Any comment or suggestions to estimate this integral would be appreciated!
PS. $e(x) = e^{2 \pi i x}$.
 A: The OP asks whether the integral is 0, the answer is no in general. For example, let me take $n=2$, $a_1,a_2>0$, $w(t_1,t_2)=\theta(t_1)\theta(1-t_1)\theta(1-|t_2|)$, with $\theta(x)$ the unit step function. Then
$$I=\begin{cases}
\frac{i}{2\pi a_1a_2}&\text{if}\; a_1>a_2\\
0&\text{if}\; a_2>a_1.
\end{cases}
$$


Since there was some discussion in the comments, let me check the convergence of the integral:
$$I = \int_{\mathbb{R}} \int_{\mathbb{R}^n} \alpha w(t) \exp\bigl(2\pi i\alpha (a_1t_1 + \cdots + a_n t_n)\bigr) \,d \alpha\prod_{k=1}^n dt_k $$
$$\qquad=-\frac{1}{2\pi ia_1}\int_{\mathbb{R}} \int_{\mathbb{R}^n} \frac{dw}{dt_1} \exp\bigl(2\pi i\alpha (a_1t_1 + \cdots + a_n t_n)\bigr) \,d \alpha\prod_{k=1}^n dt_k$$
$$\qquad=-\frac{1}{2\pi ia_1}\int_{\mathbb{R}^n} \frac{dw}{dt_1} \delta\bigl(a_1t_1 + \cdots + a_n t_n\bigr) \,\prod_{k=1}^n dt_k, $$
which converges when $w$ has a compact support.

A: Example for divergence.  $n=1$, $a_1=1$, $w(t) = \mathbf1_{[0,1]}(t)$.  Note $w$ is not "smooth"; it is not even continuous.
Then
$$
\int_{\mathbb{R}^n} \alpha w(t) e(\alpha (a_1t_1 + \dotsb + a_n t_n)) dt
=\int_0^1(\alpha\exp(2\pi i(\alpha t))\;dt
=\frac{i}{2\pi}\big(1-e^{i 2 \pi \alpha}\big)
\\
\int_{-K}^K \frac{i}{2\pi}\big(1-e^{i 2 \pi \alpha}\big)\;d\alpha
=\frac{i}{2\pi^2}\big(2\pi K -\sin(2\pi K)\big)
\\
\lim_{K\to+\infty} \frac{i}{2\pi^2}\big(2\pi K -\sin(2\pi K)\big) = +i\infty
$$

According to Carlo's argment, we do have convergence when $w$ is $C^1$ with compact support.
