How would you integrate $\frac{Si(x)}{x}$?

Question

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.

Answer 1

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 $$ Df_{k+1}(x)=\frac{f_k(x)}{x} $$

Answer 2

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