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I am attempting to prove Lobachevsky's integral formula, which states that for an integrable even function $f$ with a period of π, the following integral equality must hold,$$\int_{0}^{\infty} \frac{\sin(x)}{x}f(x)dx = \int_{0}^{\pi/2}f(x)dx$$ I am aware of a classic proof,but I want to try proving this integral formula from the perspective of Fourier series expansion,Due to the function being even, we can let $$f(x)=\frac{1}{2}a_0+\sum_{n\geq 1}a_n\cos(2nx)$$, and then switch the order of summation and integration, thus obtaining $$\begin{aligned}\int_{0}^{\infty} \frac{\sin(x)}{x}f(x)dx &= \sum_{n\geq 1}a_n\int_{0}^{\infty} \frac{\sin(x)}{x}\cos(2nx)dx+\frac{\pi}{4}a_0 \\ &= a_0\frac{\pi}{4} =\frac{\pi}{4}\frac{4}{\pi}\int_{0}^{\pi/2}f(x)dx \\ &= \int_{0}^{\pi/2}f(x)dx\end{aligned}$$.

  1. However, there is a problem here. The Fourier series of a continuous function may not necessarily converge pointwise to the function itself. But Carleson's theorem tells us that the Fourier series of continuous functions converge almost everywhere. Therefore, if we consider the Lebesgue version of the integral formula, then this proof should be feasible.
  2. My main question is, if we consider the Riemann integral, is the proof method using Fourier series expansion feasible?
Henry
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  • Similar question here – Jean Marie Mar 08 '24 at 16:51
  • @JeanMarie As I mentioned in the question, I know a classical proof, but I just want to try if it's feasible to use the Fourier series expansion directly and then switch the order of operations. – Henry Mar 08 '24 at 16:55
  • But the answer given there deals with Fourier expansion... – Jean Marie Mar 08 '24 at 17:14
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    @JeanMarie However, if you look closely, the answer that uses the Fourier expansion is to prove a series expansion of $1/\sin(\alpha)$, and that expansion can be proved by many other methods. What I'm talking about is the method of directly applying the Fourier expansion to f, which is different. – Henry Mar 08 '24 at 17:30

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