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I am interested in proving rigorously the use of the Leibniz rule for integration on the following integral:

$$\int_{0}^{\infty}\frac{\sin(x)}{x}dx$$ In this question the integral is written as

$$\int_{0}^{\infty}\frac{\sin(x)e^{-\alpha x}}{x}dx$$ and it is then differentiated with respect to alpha. The author of the answer mentions that "this is justified by uniform convergence of the "differentiated" integral for ≥>0". However, I am not really sure what this means.

Could you please provide a step by step proof of why we are allowed to use the Leibniz rule in this particular example? I would really appreciate any help.

Edit: I read that it is necessary to prove uniform convergence of the integral. But how would one go about doing that?

user2051058
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  • it’s not an indefinite integral, it is an improper integral. Next, it is not necessary to prove uniform convergence; it (and perhaps some other stuff) is a sufficient condition. Leibniz’s integral rule has varying levels of generality. If you know about Lebesgue integrals, and dominated convergence then this is pretty obvious. Otherwise, it requires some more finesse. – peek-a-boo Jan 11 '23 at 08:15
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    @peek-a-boo the improper Riemann integral $\int_0^\infty (\sin x)/x,dx$ converges but $(\sin x)/x$ on $[0,\infty)$ is not Lebesgue integrable (since $\int_0^\infty |(\sin x)/x|,dx = \infty$), so I am suspicious about using Lebesgue integrals and the dominated convergence theorem alone to justify the evaluation of $\int_0^\infty (\sin x)/x,dx$ by passage to the limit of $\int_0^\infty (\sin x)e^{-\alpha x}/x,dx$ as $\alpha \to 0^+$ without any finesse. – KCd Jan 11 '23 at 16:10
  • @KCd yes that’s certainly true; I was merely addressing the issue of justifying the differentiation under the integral sign. – peek-a-boo Jan 11 '23 at 16:27
  • Ah, okay. The next step, justifying that setting $\alpha = 0$ in the formula for $\int_0^\infty (\sin x)e^{-\alpha x}/x,dx$ from Leibniz when $\alpha > 0$ gives you the correct value of the integral at $\alpha = 0$ needs some more finesse as far as I can tell. – KCd Jan 11 '23 at 16:30

1 Answers1

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See https://kconrad.math.uconn.edu/blurbs/analysis/diffunderint.pdf: Section $3$, the second row of Table $2$ on page $16$ (this is related to the uniform convergence asked about), and the appendix.

KCd
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    While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review – Christian E. Ramirez Jan 11 '23 at 15:55
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    @C-RAM I have edited the answer to include the link explicitly; its details are too long for me to put here. I think it is a bit absurd for someone to be voting to delete this answer (2 so far) when it shows the OP precisely where to look to get an answer. – KCd Jan 11 '23 at 16:01
  • +1 I don't know what's happened to math.se over the last few years, but the site isn't what it used to be. – user7530 Jan 11 '23 at 16:15
  • Don't get me wrong, your answer is indeed a good resource for the asker, but as someone who has been silently using MSE for the past 6-7 years, the amount of old answer links that just don't work does get on my nerves, and that's precisely the problem with such link-only answers; sometimes they break over time. However, this seems to me like it would make an excellent comment, which perhaps someone may use to construct a full answer here. I don't claim to fully understand the reasoning behind the higher level users who voted to delete, but that was my logic. – Christian E. Ramirez Jan 11 '23 at 18:45
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    @C-RAM I don't have the time right now to write up a more detailed answer. I'll try to return to this later. – KCd Jan 11 '23 at 19:05