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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.

Luca
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1 Answers1

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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.

Raoul
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  • I am probably missing something here, as I am wondering how adding up the areas of 3 positive bumps can ever results in $0$. – Luca Jun 02 '20 at 14:47
  • More in general, I would be particularly interested to the case $f(x)$ characterized by some kind of aperiodic oscillatory behaviour with an infinity of zero crossings. – Luca Jun 02 '20 at 14:53
  • This was a typo, I updated it, you naturally do not want $b,c,d > 0$. I am not quite sure I understand what you mean by characterizing $f$ in this way. – Raoul Jun 02 '20 at 15:16
  • You are right that I shall have been more accurate in explaining my aim. I just thought "I better start from the simplest approach". I am in fact interested in the study of that kind of oscillatory integrals, ultimately aiming at verifying whether there might be any hope to identify some general property, which when satisfied by $f(x)$ would then ensure that the first and second integral cannot simultaneously vanish. – Luca Jun 02 '20 at 16:07
  • I very much doubt that there would be something simple to say, except odd / even. There's typically no reason that one integral would be $0$ if the other is, but as I showed, you can very easily construct functions that make both of them $0$: add or remove a bit of weight where $x$ is small or large, and you can basically tweak any function to $f$ to make both integrals vanish. – Raoul Jun 02 '20 at 17:50