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I have an integral of the square of a function ($f(x)\cdot f(x)$ from $0$ to infinity) that's very hard to compute. I just need to know the sign of its value. Is it always non-negative? If not, what are the conditions for non-negativity?

Michael Hardy
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Hass
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2 Answers2

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The integrand will be non-negative, so the same goes for the integral (it may be infinite).

StrangerLoop
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suppose $$f(x)=\frac{i}{x+1},$$ $$f(x)\cdot f(x)=-\frac{1}{(x+1)^2}$$ $$\int_0^\infty \left(f(x)\right)^2 \,\mathrm{d}x = \int_0^\infty -\frac{1}{(x+1)^2} \,\mathrm{d}x$$ $$=-\int_0^\infty \frac{1}{(x+1)^2} \,\mathrm{d}x$$ $$=-\left[ -\frac{1}{(x+1)} \right]_0^\infty$$ $$=\left[ \frac{1}{(x+1)} \right]_0^\infty$$ $$=(0)-(1)$$ $$\therefore \quad \int_0^\infty \left(f(x)\right)^2 \,\mathrm{d}x=-1$$ from this, we can see that if $f(x)$ takes complex values, its integral is not always positive (though nor is it always negative)

Jam
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    The integral can easily be non-negative even if the integrand is sometimes not real. Consider a function $f$ that takes a small imaginary value for a small region and otherwise takes large real values. Overall the real parts will dominate the integral giving a non-negative result. – Dan Piponi Jul 07 '14 at 18:14
  • good point (I didn't mean to state all the conditions, only to show that there is a contradiction) – Jam Jul 07 '14 at 18:21