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Can it be proven that for the integral $$\int_0^{\infty} e^{-x} f(x) dx $$ to equal zero, the function f (domain and codomain $\mathbb{R}$) has to be necessarily bounded?

Julian Rachman
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Sidd
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  • I'm sorry, but why $\mathbb{R}^n$ and not $\mathbb{R}$ if integration domain lies in $\mathbb{R}$? – Brightsun Jan 11 '15 at 19:46
  • @Brightsun: Sorry, that was a typo. Corrected. – Sidd Jan 11 '15 at 19:55
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    The integral is nonzero even when $f(x)=1$. If your asking if $f(x)$ has to be bounded when the integral is finite, this is also not true. Consider, $f(x)=x$. –  Jan 11 '15 at 20:04

2 Answers2

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No.

I started by trying $f(x)=1$. We can find $\int_0^\infty e^{-x}\,dx=1$, and then integration by parts gives $\int_0^\infty e^{-x}\,x\,dx = 1$.

So consider $f(x)=x-1$, and you have

$$\textstyle\int_0^\infty e^{-x}\,f(x)\,dx = \int_0^\infty e^{-x}\,x\,dx - \int_0^\infty e^{-x}\,dx = 0$$ but clearly $f$ is unbounded.

alexwlchan
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Take $$ f(x) =\frac{e^{-x}}{\sqrt{x}} \qquad as\ 0<x<1 $$ $$ f(x)=-\int_0^1\frac{e^{-t}}{\sqrt{t}}dt \qquad as\ 1<x<2 $$ and zero otherwhise. The integral appearing in the second branch converges since the leading term as $t\to0$ is $t^{-1/2}$ which is integrable. Then $$ \int_0^\infty f(x) dx = \int_0^1\frac{e^{-x}}{\sqrt{x}}dx -\int_0^1\frac{e^{-t}} {\sqrt{t}}dt \times \int_1^2dx + 0 = 0. $$ However $f$ is not bounded since $\lim_{x\to 0^+}f(x)=+\infty.$

Brightsun
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