Define $$\operatorname{Ci}(x)=-\int_{x}^{
\infty} \frac{\cos(y)}{y}\text{d}y.$$
It is easy to show
$$
\mathcal{L}(1)=\int_{0}^{\infty}\operatorname{Ci}(x)\text{d}x=0
$$
and
$$\mathcal{L}(2)=\int_{0}^{\infty}\operatorname{Ci}(x)^2\text{d}x
=\frac{\pi}{2}.$$
$\mathcal{L}(3),\mathcal{L}(4)$ is a little bit non-trivial. We have two claims(take a look here to find more details):
$$\begin{aligned}
&\mathcal{L}(3)=-\frac{3\pi}{2}\ln2 \\
&\mathcal{L}(4)=3\pi\operatorname{Li}_2
\left ( \frac{2}{3} \right )+\frac{3\pi}{2}\ln^23
\end{aligned}$$
Where $\operatorname{Li}$ are polylogarithms, they are defined by $\displaystyle{\operatorname{Li}_n(z)
=\sum_{k=1}^{\infty} \frac{z^k}{k^n}}$ for $|z|<1$.
$\mathcal{L}(5)$ is much more non-trivial. We have
$$
\mathcal{L}(5)=\int_{0}^{\infty}\operatorname{Ci}(x)^5\text{d}x
=-\frac{15\pi^3}{8}\ln(2)-\frac{15\pi}{2}\ln(2)^3
-\frac{45\pi}{4}\operatorname{Li}_2\left ( \frac{1}{4} \right )\ln(2)
-\frac{45\pi}{4}\operatorname{Li}_3\left ( \frac{1}{4} \right )
-\frac{15\pi}{16}\zeta(3).
$$
Where $\zeta(n)=\operatorname{Li}_n(1)$ for $\Re(n)>1$.
My question:
How can we find alternate generalizations? I believe that $\mathcal{L}(6)$ can be expressed by using ordinary polylogarithms($\mathcal{L}(7)$ seems impossible). We can also find the closed-forms of following integrals:
$$\int_{0}^{\infty}\operatorname{Ci}(x)^4\cos(x)\text{d}x,\int_{0}^{\infty}\operatorname{Ci}(x)^2\frac{\operatorname{Si}(2x)}{x} \cos(x)\text{d}x$$
where $\displaystyle{\operatorname{Si}(x)=\int_{0}^{x} \frac{\sin(t)}{t}\text{d}t}.$
Update 1: Define $\operatorname{si}(x)+\operatorname{Si}(x)=\frac{\pi}{2}$. Here are some results: $$\begin{aligned} &\int_{0}^{\infty}\operatorname{si}(x)\text{d}x=1\\ &\int_{0}^{\infty}\operatorname{si}(x)^2\text{d}x=\frac{\pi}{2}\\ &\int_{0}^{\infty}\operatorname{si}(x)^3\text{d}x=\frac{\pi^2}{4} -\frac{3}{2}\ln^22-\frac{3}{4} \operatorname{Li}_2\left ( \frac{1}{4} \right )\\ &\int_{0}^{\infty}\operatorname{si}(x)^4\text{d}x= \frac{\pi^3}{4} -3\pi\ln^22-\frac{3\pi}{2} \operatorname{Li}_2\left ( \frac{1}{4} \right ) \end{aligned}$$
Update 2: A useful fourier transform $$\int_{0}^{\infty}\operatorname{Ci}(x)^3\cos(a x)\text{d}x =\begin{cases} \color{Red}{\frac{\pi \text{Li}_2\left(\frac{1-a}{3}\right)}{4 a}}+\frac{\pi \text{Li}_2\left(\frac{a-1}{a-2}\right)}{2 a}+\frac{\pi \text{Li}_2\left(\frac{a+1}{3 (a-1)}\right)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{a-1}{a+1}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a+1}{a+2}\right)}{2 a}-\frac{\pi \text{Li}_2\left(\frac{a+1}{3}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a+1}{a-1}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a-1}{3 (a+1)}\right)}{4 a}+\frac{\pi \log ^2(2-a)}{4 a}-\frac{\pi \log ^2(a+2)}{4 a}+\frac{\pi \log (3) \log (a-1)}{4 a}+\frac{\pi \log (3) \log \left(\frac{a+2}{a+1}\right)}{4 a}-\frac{\pi \log (3) \log (a-2)}{4 a}-\frac{\pi \log (3) \tanh ^{-1}\left(\frac{a}{2}\right)}{2 a} & (0\le a\le1),\\ \color{Red}{\frac{\pi \text{Li}_2\left(\frac{a^2}{a^2-1}\right)}{4 a}}+\frac{\pi \log \left(-\frac{a}{a+1}\right) \log \left(\frac{1}{1-a^2}\right)}{4 a}+\frac{\pi \text{Li}_2\left(-\frac{a}{2}\right)}{2 a}+\frac{\pi \text{Li}_2(1-a)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{a+2}{2 (1-a)}\right)}{4 a}+\frac{\pi \text{Li}_2\left(-\frac{3}{a-1}\right)}{4 a}+\frac{\pi \text{Li}_2\left(-\frac{1}{a}\right)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{a+2}{2 (a+1)}\right)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{a (a+2)}{(a+1)^2}\right)}{4 a}-\frac{\pi \text{Li}_2\left(-\frac{1}{2}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{1}{1-a}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a}{a-1}\right)}{4 a}-\frac{\pi \text{Li}_2\left(-\frac{1}{a-1}\right)}{4 a}-\frac{\pi \text{Li}_2(-a)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{1}{a+1}\right)}{4 a}-\frac{3 \pi \text{Li}_2\left(\frac{a}{a+1}\right)}{4 a}-\frac{7 \pi ^3}{24 a}+\frac{3 \pi \log ^2(2)}{8 a}+\frac{\pi \log ^2(a)}{8 a}-\frac{\pi \log ^2(a+1)}{2 a}+\frac{\pi \log (2) \log (a-1)}{4 a}+\frac{\pi \log (2) \log (a+1)}{4 a}-\frac{\pi \log (2) \log (a)}{2 a}-\frac{\pi \log (2) \log (a+2)}{2 a}+\frac{\pi \log \left(\frac{a+2}{a+1}\right) \log \left(\frac{1}{(a+1)^2}\right)}{4 a}+\frac{\pi \log \left(-\frac{1}{a+1}\right) \log \left(\frac{a (a+2)}{(a+1)^2}\right)}{4 a}+\frac{\pi \log (3) \log (a+2)}{4 a}+\frac{\pi \log (a) \log (a+2)}{2 a}-\frac{\pi \log (a) \log (a+1)}{2 a}-\frac{i \pi ^2 \log \left(\frac{1}{1-a}\right)}{4 a}-\frac{\pi \log (3) \log (a-1)}{4 a}-\frac{\pi \log \left(-\frac{1}{a+1}\right) \log \left(\frac{a+2}{a+1}\right)}{4 a}-\frac{\pi \log \left(-\frac{1}{a+1}\right) \log \left(-\frac{a}{a+1}\right)}{4 a}-\frac{\pi \log (a-1) \log (a+2)}{4 a}-\frac{\pi \log (a+1) \log (a+2)}{4 a} & (1\le a\le3), \\ \color{Red}{-\frac{\pi \text{Li}_2\left(\frac{a^2}{a^2-1}\right)}{4 a}}-\frac{\pi \log \left(-\frac{a}{a+1}\right) \log \left(\frac{1}{1-a^2}\right)}{4 a}+\frac{\pi \text{Li}_2(-2)}{4 a}+\frac{\pi \text{Li}_2(2)}{4 a}+\frac{\pi \text{Li}_2\left(-\frac{1}{2}\right)}{2 a}+\frac{\pi \text{Li}_2\left(\frac{1}{1-a}\right)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{1}{a-1}\right)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{a}{a-1}\right)}{4 a}+\frac{\pi \text{Li}_2\left(-\frac{1}{a-1}\right)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{1}{a+1}\right)}{2 a}+\frac{\pi \text{Li}_2\left(\frac{a}{a+1}\right)}{2 a}-\frac{\pi \text{Li}_2\left(-\frac{a}{2}\right)}{2 a}-\frac{\pi \text{Li}_2(1-a)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a+2}{2 (1-a)}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a-2}{a-1}\right)}{4 a}-\frac{\pi \text{Li}_2\left(-\frac{3}{a-1}\right)}{4 a}-\frac{\pi \text{Li}_2(a-1)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a+2}{2 (a+1)}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a (a+2)}{(a+1)^2}\right)}{4 a}+\frac{\pi ^3}{3 a}-\frac{\pi \log ^2(2)}{4 a}+\frac{\pi \log ^2(a+1)}{2 a}+\frac{i \pi ^2 \log (2)}{4 a}+\frac{\pi \log (2) \log (a)}{2 a}+\frac{\pi \log (2) \log (a+2)}{2 a}-\frac{\pi \log (2) \log (a-1)}{4 a}-\frac{\pi \log (2) \log (a+1)}{4 a}+\frac{i \pi ^2 \log \left(\frac{1}{1-a}\right)}{4 a}+\frac{\pi \log (3) \log (a-2)}{4 a}+\frac{\pi \log (a-2) \log (a-1)}{4 a}+\frac{\pi \log \left(\frac{a+2}{a+1}\right) \log \left(-\frac{1}{a+1}\right)}{4 a}+\frac{\pi \log \left(-\frac{1}{a+1}\right) \log \left(-\frac{a}{a+1}\right)}{4 a}+\frac{\pi \log (a-1) \log (a+2)}{4 a}+\frac{\pi \log (a+1) \log (a+2)}{4 a}-\frac{\pi \log (a) \log (a+2)}{2 a}-\frac{\pi \log (a-2) \log \left(\frac{1}{a-1}\right)}{4 a}-\frac{\pi \log (3) \log \left(\frac{a+2}{a-1}\right)}{4 a}-\frac{\pi \log (2-a) \log (a-1)}{4 a}-\frac{\pi \log (a) \log \left(\frac{1}{a+1}\right)}{4 a}-\frac{\pi \log (3) \log \left(\frac{a-2}{a+1}\right)}{4 a}-\frac{\pi \log \left(\frac{1}{(a+1)^2}\right) \log \left(\frac{a+2}{a+1}\right)}{4 a}-\frac{\pi \log (3) \log (a+1)}{4 a}-\frac{\pi \log \left(-\frac{1}{a+1}\right) \log \left(\frac{a (a+2)}{(a+1)^2}\right)}{4 a}& (a\ge3). \end{cases}$$
Update 3: Common fourier transforms $$\begin{aligned} &1.\int_{0}^{\infty}\operatorname{Ci}(x)\cos(\omega x)\text{d}x= \begin{cases} 0 &(0\le\omega<1), \\ \displaystyle{ -\frac{\pi}{4} }&(\omega=1), \\ \displaystyle{ -\frac{\pi}{2\omega} }&(\omega>1). \end{cases}\\ &2.\int_{0}^{\infty}\operatorname{Ci}(x)\sin(\omega x)\text{d}x= \begin{cases} \displaystyle{-\frac{\ln(1-\omega^2)}{2\omega}} &(0\le\omega<1), \\ \displaystyle{ +\infty }&(\omega=1), \\ \displaystyle{-\frac{\ln(\omega^2-1)}{2\omega} }&(\omega>1). \end{cases}\\ &3.\int_{0}^{\infty}\operatorname{Ci}(x)^2\cos(\omega x)\text{d}x= \begin{cases} \displaystyle{ \frac{\pi\ln(1+\omega)}{2\omega} }&(0\le\omega\le2), \\ \displaystyle{ \frac{\pi\ln(\omega^2-1)}{2\omega} }&(\omega\ge2). \end{cases}\\ &4.\int_{0}^{\infty}\operatorname{si}(x)\sin(\omega x)\text{d}x= \begin{cases} 0 &(0\le\omega<1), \\ \displaystyle{ \frac{\pi}{4} }&(\omega=1), \\ \displaystyle{ \frac{\pi}{2\omega} }&(\omega>1). \end{cases}\\ &5.\int_{0}^{\infty}\operatorname{si}(x)\cos(\omega x)\text{d}x= \begin{cases} \displaystyle{\frac{1}{2\omega}\ln\left ( \frac{1+\omega}{1-\omega} \right ) } &(0\le\omega<1), \\ \displaystyle{ +\infty }&(\omega=1), \\ \displaystyle{\frac{1}{2\omega}\ln\left ( \frac{\omega+1}{\omega-1} \right ) }&(\omega>1). \end{cases}\\ &6.\int_{0}^{\infty}\operatorname{si}(x)^2\cos(\omega x)\text{d}x= \begin{cases} \displaystyle{ \frac{\pi\ln(1+\omega)}{2\omega} }&(0\le\omega\le2), \\ \displaystyle{ \frac{\pi}{2\omega}\ln\left ( \frac{\omega+1}{\omega-1} \right ) }&(\omega\ge2). \end{cases}\\ &7.\int_{0}^{\infty}\frac{\operatorname{Si}(x)}{x}\cos(\omega x)\text{d}x= \begin{cases} \displaystyle{-\frac{\pi}{2}\ln(\omega)} &(0<\omega\le1), \\ \displaystyle{0 }&(\omega\ge1). \end{cases}\\ \end{aligned}$$
Definition: Functions $\operatorname{Si}_n(x)$ are defined by $$\operatorname{Si}_0(x)=\sin(x),\operatorname{Si}_n(x) =\int_{0}^{x} \frac{\operatorname{Si}_{n-1}(t)}{t}\text{d}t.$$ And we are able to get $$ \int_{0}^{\infty}\frac{\operatorname{Si}_2(x)\operatorname{si}(x)^2}{x} \text{d} x=\frac{7\pi^5}{1440}. $$
