Consider the following improper integral, whereby $a \neq 0 $ is real and the $C^\infty$ real function $f(x)$ is further such that the following integral vanishes: $$ \int_{-\infty} ^{\infty} \frac{f(x)}{a^2+x^2} \, dx \, = \, 0 $$ EDIT: I am now adding that an interesting general case would be for $f(x)$ characterized by an aperiodic oscillatory behaviour with a countable infinity of zero crossings. EDIT END
Further suppose that the above integral exists also when $f(x)$ is replaced by $x \, f(x)$. Would it be possible to state that therefore: $$ \int_{-\infty} ^{\infty} x \, \frac{f(x)}{a^2+x^2} \, dx \, \neq \, 0 \,\,\,\, ? $$ Let us start from the trivial case where $f(x)$ is an even function, though such that the integral vanishes. A vanishing first integral would then correspond to a second integral vanishing as well, as its integrand is thus an odd function (the denominator is even).
So, $f(x)$ even ==> the answer is NO.
Suppose now that $f(x)$ is an odd function, so that the first integral vanishes by definition. The second integral's integrand is then an even function. Whether its integral will vanish or not will depend on the sum of "positive areas" exactly matching the sum of "negative areas". In the general case $f(x)$ might cross the $x$ axis several times, and said zeros are unaffected by the multiplication $x \, f(x)$ (except possibly at $x=0$, were $f(x) \neq 0$ ), although the value of each "area" in between two consecutive zeros is indeed affected. But could their total sum ever vanish if it already does so for the first integral?
And what about the most general case whereby $f(x)$ is neither even nor odd ? (though still such that both integrals exist).
Were those above indefinite integrals, and defining the primitive of the first integral's integrand as:
$$ F(x) \, = \, \int \frac{f(x)}{a^2+x^2} \, dx $$
integration by parts would thus yield:
$$ \int x \, \frac{f(x)}{a^2+x^2} \, dx \, = \, x\int \frac{f(x)}{a^2+x^2} \, dx \, - \int \, F(x) \, dx \, $$ But switching then to improper integrals from $-\infty$ to $+\infty$ I was unable to make anything useful out of it ...
Thanks for any suggestion about how to best proceed with this investigation.
There are many more ways than even / odd to make the integral $0$, so it cannot work.
To get a proper example, I will just call $$ g(x) = \frac{1}{a^2+x^2}, $$ since its specific form does not matter (it just needs to not be identically $0$). Consider any positive function $\varphi$ which is positive that is $C^{\infty}$, strictly positive on $[0,1]$, and $0$ on $(-\infty, -1) \cup (2, + \infty)$ (like a plateau function). Then, for $b,c,d \in \mathbb{R}$, define $$ f_{c,d}(x) = b \: \varphi(x-1) + c \: \varphi(x-4) + d \: \varphi(x-7). $$ It has three bumps on $[0,3]$, $[3,6]$, $[6,9]$, with sizes given by $b,c,d$. Then we have $$ \int_{-\infty}^{+\infty} f_{b,c,d}(x) g(x) \: \mathrm{d}x = b \int_0^3 \varphi(x-1) g(x) \: \mathrm{d}x + c \int_3^6 \varphi(x-4) g(x) \: \mathrm{d}x + d \int_6^9\varphi(x-7) g(x) \: \mathrm{d}x $$ and \begin{align*} \int_{-\infty}^{+\infty} x f_{b,c,d}(x) g(x) \: \mathrm{d}x & = b \int_0^3 x \varphi(x-1) g(x) \: \mathrm{d}x \\ & \quad + c \int_3^6 x \varphi(x-4) g(x) \: \mathrm{d}x + d \int_6^9 x \varphi(x-7) g(x) \: \mathrm{d}x. \end{align*} You want both of these to be $0$. This gives you two linear equations, three unknown, so you can find $b,c,d$ not all $0$ that make this work.