Suppose that $\chi(\xi)\in C_{c}^{\infty}(B(0,1))$, such that $\chi(\xi)=1$ for $\left|\xi\right|\le1/2$. Let $h=\mathcal{F}^{-1}\chi$, the inverse Fourier transform of $\chi$. If it is possible that $$ \int_{\mathbb{R}^{d}}h(x)\mathrm{d}x=1? $$ which means $$ \int_{\mathbb{R}}\int_{B(0,1)}\chi(\xi)\cos(\xi x)\mathrm{d} \xi\mathrm{d} x=1? $$ Does this $\chi$ exists?
I calculate in $1D$, in $(-l,l)$ for any $l>0$, we know that $$ \int_{\mathbb{R}}\int_{-l}^{l}\cos(\xi x)\mathrm{d}\xi\mathrm{d}x=2\pi $$ which is much bigger than $1$.