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While playing with this sum posted on Quora, I came across this integral: $$\int\limits^{2\pi}_{0} \frac{\operatorname{Cl}_{2}(z)}{z}dz$$ which I have tried to evaluate, but to no avail. I have tried using the fact that $\frac{\operatorname{Cl}_{2}(z)}{z}=\Re\left(\ln(1-e^{iz})\ln z\right)$ but that did not work either.

Related integrals on MSE include one appearing in this question featuring an identical integrand, and this class of integrals. Unfortunately, playing around with them did not give me any insight.

Kisaragi Ayami
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1 Answers1

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Using $$\operatorname{Cl}_{2}(z)=\sum_{k=1}^\infty \frac{\sin (k z)}{k^2}$$

$$\int\limits^{2\pi}_{0} \frac{\operatorname{Cl}_{2}(z)}{z}\,dz=\int\limits^{2\pi}_{0}\sum_{k=1}^\infty \frac{\sin (k z)}{k^2z}\,dz$$

We can swap the summation and the integration $$\int\frac{\sin (k z)}{k^2z}\,dz=\frac 1{k^2}\int\frac{\sin (t)}{t}\,dt=\frac 1{k^2}\text{Si}(t)=\frac 1{k^2}\text{Si}(kz)$$ $$\int_0^{2\pi}\frac{\sin (k z)}{k^2z}\,dz=\frac 1{k^2}\text{Si}(2k\pi)$$ $$\int\limits^{2\pi}_{0} \frac{\operatorname{Cl}_{2}(z)}{z}\,dz=\sum_{k=1}^\infty \frac{\text{Si}(2 k \pi )}{k^2}$$

The summation will be extremely slow since, if $a_k$ is the summand, for large $k$, $$\frac {a_{k+1}}{a_k}=1-\frac{2}{k}+\frac{1+3 \pi ^2}{\pi ^2 k^2}+O\left(\frac{1}{k^3}\right)$$

Computing the partial sums $$\left( \begin{array}{cc} p & \sum_{k=1}^p a_k \\ 1000 & 2.397763 \\ 2000 & 2.398548 \\ 3000 & 2.398810 \\ 4000 & 2.398941 \\ 5000 & 2.399019 \\ 6000 & 2.399072 \\ 7000 & 2.399109 \\ 8000 & 2.399137 \\ 9000 & 2.399159 \\ 10000 & 2.399176 \\ \end{array} \right)$$

The number can be approximated using the simplest form of Euler-Maclaurin summation since we know that $$\int\frac{\text{Si}(2 k \pi )}{k^2}\,dk=2 \pi \text{Ci}(2 k \pi )-\frac 1 k(\text{Si}(2 k \pi )+\sin (2 \pi k))$$ It would give,as an approximation, $$\frac{341}{210} \text{Si}(2 \pi )-2 \pi \text{Ci}(2 \pi )-\frac{\pi \left(927-148 \pi ^2+10 \pi ^4\right)}{18900}=2.37136$$ which corresponds to a relative error of $1.2$%.

Edit

We could also write $$\sum_{k=1}^\infty \frac{\text{Si}(2 k \pi )}{k^2}=\sum_{k=1}^n \frac{\text{Si}(2 k \pi )}{k^2}+\sum_{k=n+1}^\infty \frac{\text{Si}(2 k \pi )}{k^2}$$ and use the asymptotic $$\frac{\text{Si}(2 k \pi )}{k^2}=\frac{\pi }{2 k^2}-\frac{1}{2 \pi k^3}+\frac{1}{4 \pi ^3 k^5}-\frac{3}{4 \pi ^5 k^7}+\frac{45}{8 \pi ^7 k^9}+O\left(\frac{1}{k^{11}}\right) \tag 1$$ and $$\sum_{k=n+1}^\infty \frac 1{k^m}=\zeta (m,n+1)$$ where the rhs is the Hurwitz zeta function.

Using $n=1000$, this would give $$2.39776349908777+0.00156993169250=2.39933343078027$$

while is the value obtained numerically for $$\int_0^{2\pi} \Re\left(\log \left(1-e^{i z}\right) \log (z)\right)\,dz=2.39933343078039$$

We could still do better using more terms in $(1)$.

Claude Leibovici
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