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I wanted to prove that if I have to functions $f$ and $g$ such that $\int_{-\infty}^{\infty}f(x)dx <\infty$ and $\int_{-\infty}^{\infty}g(x)dx = 0$ then $\int_{-\infty}^{\infty}g(x)f(x)dx = 0$. I think it is intuitive obvious but I haven't found anything (maybe because is too obvious)

So first I integrated by parts

$\int_{-\infty}^{\infty}g(x)f(x)dx = f(x) \int g(x') dx' |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} f(x)dx \int_{-\infty}^{x}g(x')dx'$

The first term is zero by hypothesis and the second term reminds me the Abel's theorem for series convergence that states than if I have to series such that $\sum_n^{\infty} a_n = A$ and $\sum_n^{\infty} b_n = B$ and both with finite limit then

$\sum_n^{\infty} c_n = \sum_n^{\infty}\sum_r^{n} a_rb_{n-r+1} $

converges to $AB$. So generalizing this result to integral I could say that the second integral goes to zero and the integral of the product is zero.

Is this correct? If the reasoning is wrong, is there any way to prove this?

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    Take both $f$ and $g$ to be odd and equal, e.g. $f(x)=g(x)=xe^{-x^2}$. Then $\int f \, \mathrm{d}x= \int g \, \mathrm{d}x = 0$, but $\int fg \, \mathrm{d}x \neq 0$. – 2017-01-10
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    You would actually be hard pressed to find examples for which your statement is true (other than if, say both were odd functions, or g is zero). – 2017-01-10
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    So, what should I ask for to both functions to guarantee that $\int f g = 0$ ? – 2017-01-10

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