I am trying to prove this below.
I:=∫∞0cosxx(∫x0sinttdt)2dx=−76ζ3
where
ζ3=∞∑n=11n3.
I am not sure how to work with the integral over t because it is from 0 to x. If we can somehow write
∫∞0cosxx(∫∞0sinttdt−∫∞xsinttdt)2dx=∫∞0cosxx(π2−∫∞xsinttdt)2dx.
I do not want to use an asymptotic expansion on the integral over t from x to ∞, I am looking for exact results. Note we can use ∫∞0sinttdt=∫∞0L[sint(s)]ds=π2. Other than this approach I am not really sure how to go about this. Note by definition
∫x0sinttdt≡Si(x),
but I'm not too sure what this definition can be used for in terms of a proof. Also note
∫∞0cosxxdx→∞.
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=∫∞0∫10∫10cos(x)sin(xy)sin(xz)xyzdydzdx
I=1/4∫10∫10∫∞0cos(x(y−z+1))+cos(x(y−z−1))−cos(x(y+z+1))−cos(x(y+z−1))xyzdxdydz
Now, use the known result: ∫∞0cos(bx)−cos(ax)xdx=log(a/b)
Using this, it can now be written in terms of a log:
I=1/4∫10∫101yzlog((y+z+1)(y+z−1)(y−z+1)(y−z−1))dydz
Now, integrating this w.r.t y is where the dilogs come into play:
I=1/4∫10Li2(1z+1)+Li2(1z−1)−Li2(−1z+1)−Li2(11−z)zdz
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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