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Given a function $f$ which is continuously differentiable over the real line, and given that $\displaystyle \lim_{|t|\to \infty}f(t)\exp(-t^2) = 0$ how does one prove that $\displaystyle \int_{-\infty}^{\infty}f'(t)\exp(-t^2)\text{d}t$ is finite?

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3 Answers 3

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I think you can't. Just looking at the positive side of your integral, consider $$f:t\mapsto \frac{e^{t^2}}{\sqrt{t+1}}$$ You do have $\lim_{t\to+\infty} f(t)e^{-t^2}=0$, but $$f'(t)e^{-t^2} = \frac{2t\sqrt{t+1}-\frac{1}{\sqrt{t+1}}}{t+1} \sim_{t\to+\infty} 2\sqrt t$$ which is not integrable on $[0,+\infty[$.

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The result is wrong as stated

By integration by parts for $a >0$ $$\int_{- a}^a f^\prime(t) e^{-t^2} \ dt = \underbrace{\left[f(t)e^{-t^2}\right]_{-a}^a}_{A}+2\underbrace{\int_{- a}^a tf(t) e^{-t^2} \ dt}_B $$

$A$ converges to zero as $a \to \infty$ by hypothesis.

Now take a function $f$ that vanishes for $x\le 1$ and such that $\int_1^\infty tf(t) e^{-t^2} \ dt$ diverges. For example $$f : x \mapsto t^\alpha e^{t^2}$$ with $-2 <\alpha <0$. This will provide a counterexample as $B$ diverges for $a \to \infty$ while $f$ satisfies the hypothesis $\displaystyle \lim_{|t|\to \infty}f(t)\exp(-t^2) = 0$.

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  • $\begingroup$ for $\alpha<0$ and real however this function is not continuously differentiable everywhere on the real axis. Not that it really matters, but just saying. $\endgroup$ Aug 8, 2018 at 17:29
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This cannot be true for all functions $f(t)$. Consider $f(t)=\frac{\sin t}{t}e^{t^2}$. This is continuously differentiable on the real line, and $f'(t)e^{-t^2}=2\sin t-\frac{\sin t}{t^2}+\frac{\cos t}{t}$. The $\sin t$ part makes the integral oscillate indefinitely and thus diverge.

More explicitly,

$\int^{\infty}_{-\infty}(2\sin t-\frac{\sin t}{t^2}+\frac{\cos t}{t})dt=2\int^{\infty}_{-\infty}dt\sin t+[\frac{\sin t}{t}]^{\infty}_{-\infty}=2\int^{\infty}_{-\infty}dt\sin t $ -(DNE)

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