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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 $$ \int_0^\infty \frac{\cos x}{x}dx \to \infty. $$

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  • $\begingroup$ As quick hint, exand the integrant of the inernal integral into series, seems to me that will do the job. $\endgroup$ – TMS May 24 '14 at 23:44
  • $\begingroup$ I'd have written $\zeta(3)$ rather than $\zeta_3$. ${}\qquad{}$ $\endgroup$ – Michael Hardy May 24 '14 at 23:45
  • $\begingroup$ @MichaelHardy Okay that is great to know thanks a lot. $\endgroup$ – Jeff Faraci May 24 '14 at 23:47
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    $\begingroup$ See here for a nice solution: integralsandseries.prophpbb.com/… scroll to the bottom thread $\endgroup$ – Cody May 25 '14 at 0:00
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    $\begingroup$ @Integrals : Okay.. I wish Shobhit himself to post it here since he's the one who answered there. $\endgroup$ – gar May 25 '14 at 16:47
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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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    $\begingroup$ Thanks. This is fine to finish from here. I didn't see the Frullani like integral at all. And obviously writing it as a triple integral is very nice insight, I thought of neither. +1 $\endgroup$ – Jeff Faraci May 26 '14 at 22:09
  • $\begingroup$ Thanks Integral. If Shobit ever shows back up, please go ahead and give him a greenie because, afterall, it is his answer. I was just helping out because you could not access the link. $\endgroup$ – Cody May 27 '14 at 8:52

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