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Recently, when facing a baby Rudin's exercise, I proved that: $$ \int_{0}^{y}\frac{|\cos x\,|}{1+x}\,dx = \frac{2}{\pi}\log(1+y)+O(1) $$ holds by integration by parts.

Now I wonder if $$\color{red}{L}=\lim_{y\to +\infty}\left(-\frac{2}{\pi}\log(1+y)+\int_{0}^{y}\frac{|\cos x\,|}{1+x}\,dx\right)$$ has a nice closed form expression.

It is easy to argue that, since the Fourier series of $|\cos x\,|$ is given by: $$ |\cos x\,|=\frac{2}{\pi}+\frac{4}{\pi}\sum_{n\geq 1}\frac{(-1)^{n+1}}{4n^2-1}\cos(2nx)\tag{A} $$ our constant is given by: $$ \int_{0}^{+\infty} f(x)\cos(2x)\,dx,\tag{B} $$ where: $$ f(x) = \frac{4}{\pi}\sum_{n\geq 1}\frac{(-1)^{n+1}}{(4n^2-1)(x+n)},\tag{C} $$ hence $L$ is probably related with the Clausen function. Numerically, $$ L \approx 0.057813. \tag{D}$$ By using the identity stated in the comments and Fubini's theorem we also have: $$ L = \frac{2}{\pi}\int_{0}^{+\infty}\frac{y}{1+y^2}\left(-1+2\arctan(e^{-y})\cosh(y)\right)\,dy.\tag{E}$$

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  • $\begingroup$ Probably $$\sum_{n\geq 1}\frac{(-1)^{n+1}}{4n^2-1}e^{ny}=-\frac{1}{2}+\arctan(e^{y/2})\cosh(y/2)$$ is pretty useful. $\endgroup$ Apr 12 '15 at 21:27
  • $\begingroup$ Have you tried writing the integral as a sum $\int_0^{\pi /2}\frac{|\cos x|}{1+x}dx+ \sum_{n=0}^N \int_{(n+1/2)\pi}^{(n+3/2)\pi} \frac{|\cos x|}{1+x}dx$ and performing multiple integration by parts? $\endgroup$
    – Mark Viola
    Apr 12 '15 at 21:36
  • $\begingroup$ @Dr.MV: is that really different from just using the Fourier series of $|\cos x\,|$? $\endgroup$ Apr 12 '15 at 21:38
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    $\begingroup$ That is a good question. I have not "turned the crank" to determine if that is so. $\endgroup$
    – Mark Viola
    Apr 12 '15 at 21:41

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