How would you integrate $\frac{Si(x)}{x}$? The function $Si(x)$ can be obtained when we integrate 
$\frac{\sin(x)}{x}$. But how would we go about integrating $\frac{Si(x)}{x}$?
More information about the function $Si(x)$ can be found here https://en.wikipedia.org/wiki/Trigonometric_integral
Edit: Just checked wolframalpha and even it did not have any answer.
 A: There is no reason for suspecting that the antiderivative of $\operatorname{Si}(x)/x$ can be expressed in terms of “known” functions.
The power series expansion of $\operatorname{Si}(x)$ is
$$
\operatorname{Si}(x)=\sum_{n\ge0}\frac{(-1)^nx^{2n+1}}{(2n+1)^2\cdot(2n)!}
$$
Therefore the power series expansion of $\operatorname{Si}(x)/x$ is
$$
\sum_{n\ge0}\frac{(-1)^nx^{2n}}{(2n+1)^2\cdot(2n)!}
$$
and therefore the antiderivatives are
$$
c+\sum_{n\ge0}\frac{(-1)^nx^{2n+1}}{(2n+1)^3\cdot(2n)!}
$$
Note that there is a pattern here: if you start with the function
$$
f_0(x)=x\cos x
$$
then its series expansion is
$$
\sum_{n\ge0}\frac{(-1)^nx^{2n+1}}{(2n)!}
$$
If we integrate $f_0(x)/x$, we get
$$
f_1(x)=\sum_{n\ge0}\frac{(-1)^nx^{2n+1}}{(2n+1)\cdot(2n)!}=\sin x
$$
(using here and below the antiderivative that evaluates $0$ at $x=0$).
If we integrate $f_1(x)/x$, we get
$$
f_2(x)=\sum_{n\ge0}\frac{(-1)^nx^{2n+1}}{(2n+1)^2\cdot(2n)!}=\operatorname{Si}x
$$
and so on.
Repeating the process yields
$$
f_k(x)=\sum_{n\ge0}\frac{(-1)^nx^{2n+1}}{(2n+1)^k\cdot(2n)!}
$$
and
$$
Df_{k+1}(x)=\frac{f_k(x)}{x}
$$
A: so, if by definition:
$$\text{Si}(x)=\int_0^x\frac{\sin(t)}{t}dt$$
then can we say:
$$I=\int\frac{\text{Si}(x)}{x}dx=\int\frac{\int_0^x\frac{\sin(t)}{t}dt}{x}dx=\int\int_0^x\frac{\sin(t)}{tx}dtdx$$
now if we insert a variable:
$$I(y)=\int_0^y\int_0^x\frac{\sin(t)}{tx}dtdx$$
and this may be solvable through change of variables
