∫∞0∫∞0∫∞0sin(x)sin(y)sin(z)xyz(x+y+z)(sin(x)cos(y)cos(z)+sin(y)cos(z)cos(x)+sin(z)cos(x)cos(y))dxdydz
I saw this integral I posted on a page on Facebook . The author claims that there is a closed form for it.
My Attempt
This can be rewritten as
3∫∞0∫∞0∫∞0sin2(x)sin(y)cos(y)sin(z)cos(z)xyz(x+y+z)dxdydz
Now consider
F(a)=3∫∞0∫∞0∫∞0sin2(x)sin(y)cos(y)sin(z)cos(z)e−a(x+y+z)xyz(x+y+z)dxdydz
Taking the derivative
F′(a)=−3∫∞0∫∞0∫∞0sin2(x)sin(y)cos(y)sin(z)cos(z)e−a(x+y+z)xyzdxdydz
By symmetry we have
F′(a)=−3(∫∞0sin2(x)e−axxdx)(∫∞0sin(x)cos(x)e−axxdx)2
Using W|A I got
F′(a)=−316log(4a2+1)arctan2(2a)
By integeration we have
F(0)=316∫∞0log(4a2+1)arctan2(2a)da
Let x=2/a
I=38∫∞0log(x2+1)arctan2(x)x2dx
Question
I seem not be able to verify (1) is correct nor find a closed form for it, any ideas ?
Answer
Ok I was able to find the integral
∫∞0log(x2+1)arctan2(x)x2dx
First note that
∫log(1+x2)x2dx=2arctan(x)−log(1+x2)x+C
Using integration by parts
I=π312+2∫∞0arctan(x)log(1+x2)(1+x2)xdx
For the integral let
F(a)=∫∞0arctan(ax)log(1+x2)(1+x2)xdx
By differentiation we have
F′(a)=∫∞0log(1+x2)(1+a2x2)(1+x2)dx
Letting 1/a=b we get
1(1+a2x2)(1+x2)=1a2{1((1/a)2+x)(1+x2)}=b21−b2{1b2+x2−11+x2}
We conclude that
b21−b2∫∞0log(1+x2)b2+x2−log(1+x2)1+x2dx=b21−b2{πblog(1+b)−πlog(2)}
Where we used that
∫∞0log(a2+b2x2)c2+g2x2dx=πcglogag+bcg
By integration we deduce that
∫10πa2−1{alog(1+1a)−log(2)}da=π2log2(2)
For the last one I used wolfram alpha, however it shouldn't be difficult to prove.
Finally we have
∫∞0log(x2+1)arctan2(x)x2dx=π312+πlog2(2)
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