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How can we show that

  1. The integral is finite, i.e., $$\int_{0}^{\infty} e^{-\epsilon x} \frac{\sin x}{x} dx < \infty.$$

$$ \lim_{\epsilon \rightarrow 0} \int_{0}^{\infty} e^{-\epsilon x} \frac{\sin x}{x} dx = \int_{0}^{\infty} \frac{\sin x}{x} dx. $$

Actually, in [Stein-Weiss, Fourier analysis, p. 5], they give a more genernal statement as follows,

If $f$ is locally integrable and $\lim_{p\rightarrow \infty} \int_0^p f(x) dx=l$, then the Abel means $A_\epsilon = \int_0^{\infty} e^{-\epsilon x} f(x) dx$ converge to $l$.

I think integral by part will be helpful. Any hint will be helpful. Thanks in advance.

I will give proof of the above statement here, since it seems that I can't write an answer now.

Firstly, we have \begin{align} \int_0^\infty e^{-\epsilon x} f(x) dx & = \int_0^\infty e^{-\epsilon x} d\left[\int_0^x f(t) dt\right] \\ & = \left.e^{-\epsilon x} \int_0^x f(t) dt \right|_0^{\infty} + \epsilon \int_0^{\infty} \left[\int_0^x f(t) dt\right] e^{-\epsilon x} dx \\ & = \epsilon \int_0^{\infty} \left[\int_0^x f(t) dt\right] e^{-\epsilon x} dx =I_{\epsilon}. \end{align} Then letting $\epsilon x =y$, then we have $$ I_{\epsilon} = \int_0^{\infty} \left[ \int_0^{y/\epsilon} f(t)dt\right] e^{-y} dy. $$ By Lebesgue's dominated theorem, we have $$ \lim_{\epsilon \rightarrow 0_+} I_{\epsilon} = \int_{0}^{\infty} \int_0^{\infty} f(t) dt e^{-y} dt = l. $$

xxxg
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    @Gonçalo this question is not the same as what you link to because this question is concerned with the limiting behavior as $\varepsilon \to 0^+$ where the task is to show we can set $\varepsilon = 0$, and this is subtle because the integral at $\varepsilon = 0$ is not absolutely convergent. No answer in your linked question addresses the behavior at $\varepsilon = 0$, either directly or in the limit. Therefore I do not think this question should be closed. – KCd Dec 14 '24 at 04:01
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    It amazes me that this question has been listed as being a duplicate when I explained carefully in my previous comment why this question is not a duplicate. – KCd Dec 14 '24 at 05:00
  • @KCd, maybe you can vote for reopen. (I have done it.) And I add the proof of the statement. – xxxg Dec 14 '24 at 10:46

1 Answers1

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The function $(\sin x)/x$ is bounded (look at a graph!), so the first part when $\varepsilon > 0$ is a straightforward estimate. The finiteness at $\varepsilon = 0$ is more subtle, since in that case the integral is not absolutely convergent.

You wrote that you think integration by parts will be helpful. It is, since it lets us convert the integral at $\varepsilon = 0$ into an absolutely convergent integral. Set $u = 1/x$ and $dv = \sin x\,dx$. Then $du = -dx/x^2$, but don't use $v = -\cos x$: instead use $v = 1 - \cos x$. Then $\int_0^\infty v(x)\,du(x) = -\int_0^\infty (1-\cos x)/x^2 \,dx$ and $$ \int_0^\infty \frac{\sin x}{x}\,dx = \int_0^\infty \frac{1-\cos x}{x^2}\,dx, $$ with the integral on the right side being absolutely integrable. (You should work initially with truncated integrals over $[0,b]$ when doing integration by parts and let $b \to \infty$ at the end.)

Apply integration by parts to rewrite $\int_0^\infty e^{-\varepsilon x}(\sin x)/x \,dx$ when $\varepsilon > 0$ in a similar way. Further details are left to you.

KCd
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