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. $e(x) = e^{2 \pi i x}$.



Is $\int_{\mathbb{R}} \int_{\mathbb{R}^n} \alpha w(t) e(\alpha (a_1t_1 + \dotsb + a_n t_n)) dt\,d \alpha = 0$?
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.
