I am trying to prove this below.
$$
I:=\int_0^\infty \frac{\cos x}{x}\left(\int_0^x \frac{\sin t}{t}dt\right)^2dx=-\frac{7}{6}\zeta_3
$$
where
$$
\zeta_3=\sum_{n=1}^\infty \frac{1}{n^3}.
$$
I am not sure how to work with the integral over $t$ because it is from $0$ to $x$. If we can somehow write
$$
\int_0^\infty \frac{\cos x}{x} \left(\int_0^\infty \frac{\sin t}{t}dt-\int_x^\infty \frac{\sin t}{t}dt \right)^2dx=\int_0^\infty \frac{\cos x}{x}\left(\frac{\pi}{2}-\int_x^\infty \frac{\sin t}{t}dt\right)^2dx.
$$
I do not want to use an asymptotic expansion on the integral over $t$ from $x$ to $\infty$, I am looking for exact results. Note we can use $\int_0^\infty \frac{\sin t}{t}dt=\int_0^\infty \mathcal{L}[\sin t(s)]ds=\frac{\pi}{2}.$ Other than this approach I am not really sure how to go about this. Note by definition
$$
\int_0^x \frac{\sin t}{t}dt\equiv Si(x),
$$
but I'm not too sure what this definition can be used for in terms of a proof. Also note
$$
\int_0^\infty \frac{\cos x}{x}dx \to \infty.
$$
Answer
It's strange that you're unable to access that link. I do not understand that.
I accredit this to Shobit.
Anyway, what Shobit done was to write it as a triple integral.
$$I=\int_{0}^{\infty}\int_{0}^{1}\int_{0}^{1}\frac{\cos(x)\sin(xy)\sin(xz)}{xyz}dydzdx$$
$$I=1/4\int_{0}^{1}\int_{0}^{1}\int_{0}^{\infty}\frac{\cos(x(y-z+1))+\cos(x(y-z-1))-\cos(x(y+z+1))-\cos(x(y+z-1))}{xyz}dxdydz$$
Now, use the known result: $$\int_{0}^{\infty}\frac{\cos(bx)-\cos(ax)}{x}dx=\log(a/b)$$
Using this, it can now be written in terms of a log:
$$I=1/4\int_{0}^{1}\int_{0}^{1}\frac{1}{yz}\log\left(\frac{(y+z+1)(y+z-1)}{(y-z+1)(y-z-1)}\right)dydz$$
Now, integrating this w.r.t y is where the dilogs come into play:
$$I=1/4\int_{0}^{1}\frac{Li_{2}(\frac{1}{z+1})+Li_{2}(\frac{1}{z-1})-Li_{2}(\frac{-1}{z+1})-Li_{2}(\frac{1}{1-z})}{z}dz$$
Maybe you can finish it now on your own. It is now a matter of known dilog integrals. This is the bulk of it. But, if you need more I can write the rest later.
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