Tuesday, 28 June 2016

calculus - A closed form for a triple integral with sines and cosines



000sin(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



3000sin2(x)sin(y)cos(y)sin(z)cos(z)xyz(x+y+z)dxdydz



Now consider



F(a)=3000sin2(x)sin(y)cos(y)sin(z)cos(z)ea(x+y+z)xyz(x+y+z)dxdydz




Taking the derivative



F(a)=3000sin2(x)sin(y)cos(y)sin(z)cos(z)ea(x+y+z)xyzdxdydz



By symmetry we have



F(a)=3(0sin2(x)eaxxdx)(0sin(x)cos(x)eaxxdx)2



Using W|A I got




F(a)=316log(4a2+1)arctan2(2a)



By integeration we have



F(0)=3160log(4a2+1)arctan2(2a)da



Let x=2/a



I=380log(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+20arctan(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)}=b21b2{1b2+x211+x2}



We conclude that

b21b20log(1+x2)b2+x2log(1+x2)1+x2dx=b21b2{πblog(1+b)πlog(2)}



Where we used that



0log(a2+b2x2)c2+g2x2dx=πcglogag+bcg



By integration we deduce that



10πa21{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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