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Prove that:

$$ -\dfrac{\pi}{2}\;\log(t)=\dfrac{\pi}{2}(1-\log(\pi))+\sum_{n=1}^{\infty}\dfrac{Si(n\pi)}{n}\cos(nt) $$ where $Si(x)$ denote the sine integral function, which is defined by: $$Si(x)=\int_0^x \dfrac{\sin(u)}{u}du$$

My work: the sum term looks like a Fourier series of an even function, since: $$ \dfrac{Si(n\pi)}{\pi n}=\int_0^1\dfrac{\sin(\pi n x)}{\pi nx}dx. $$ Also, the Fourier series should be related with the logarithm function, but compute it Fourier coefficients looks quite difficult.

Sergio Ferrer
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  • I believe $sen$ should be $\sin$ in the second formula, and the third formula should be $\frac{\text{Si}(\pi n)}{\pi n}=\int_0^1 \frac{\sin (\pi n x)}{\pi n x} , dx$. I believe the Fourier series is valid for $0<t<\pi$. – Steven Clark Apr 23 '23 at 15:37
  • Sorry, all of you are right. I just correct it. – Sergio Ferrer Apr 23 '23 at 15:48
  • Your third formula still has $\text{Si}$ instead of $\sin$ in the integrand on the right side. – Steven Clark Apr 23 '23 at 15:54
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    consider the function $\log |t|, |t| < \pi, t \ne 0$ which is integrable and even so write its cosine Fourier series $n$th term and integrate by parts for $n \ge 1$ (which works as $\sin nt/t$ is integrable at $0$ while it's straight computation for $n=0$ and you get the (correct) result; or you can start with the Fourier stieltjes series of $1/t$ (which is not integrable at $0$ but is odd and has a sine Fourier stieltjes series because $\sin nt /t$ is integrable as above)) in terms of $\sin nt$ and integrate it term by term – Conrad Apr 23 '23 at 15:54
  • Have a look to this answer. – Jean Marie Apr 23 '23 at 15:55

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