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forFor large $a$ you can replace $\cos x$ by 1 (since only the interval $x\lesssim 1/\sqrt a$ contributes), and then the integral evaluates to $$\int_0^\infty J_0(rx) e^{-ax^2} x\,dx = \frac{e^{-r^2/4 a}}{2 a},$$
which is positive for all $r$.
for large $a$ you can replace $\cos x$ by 1 (since only the interval $x\lesssim 1/\sqrt a$ contributes), and then the integral evaluates to $$\int_0^\infty J_0(rx) e^{-ax^2} x\,dx = \frac{e^{-r^2/4 a}}{2 a},$$
which is positive for all $r$.
For large $a$ you can replace $\cos x$ by 1 (since only the interval $x\lesssim 1/\sqrt a$ contributes), and then the integral evaluates to $$\int_0^\infty J_0(rx) e^{-ax^2} x\,dx = \frac{e^{-r^2/4 a}}{2 a},$$
which is positive for all $r$.
for large $a$ you can replace $\cos x$ by 1 (since only the interval $x\lesssim 1/\sqrt a$ contributes), and then the integral evaluates to $$\int_0^\infty J_0(rx) e^{-ax^2} x\,dx = \frac{e^{-r^2/4 a}}{2 a},$$
which is positive for all $r$.
