Saturday, 24 March 2018

calculus - Find intinftyinftyfrac1x12+1dx elementarily





Ho to find the following integral 1x12+1dxusing parts by substitution, partial fractions, etc. but not Cauchy's residue theorem?


Answer



Following Lucian's comment, by partial fractions we have
1x12+1=1(x4+1)(x8x4+1)=13[1x4+1+2x8x4+1x4x8x4+1]
where




1x4+1=x222(x2+2x1)+x+222(x2+2x+1)1x8x4+1=2x622(2+3)(233)(2x2+2x+6x2)+2x622(2+3)(233)(2x22x+6x2)+2x+622(2+3)(233)(2x2+2x+6x+2)+2x+622(2+3)(233)(2x22x+6x+2)x4x8x4+1=(1+3)x22(2+3)(233)(2x22x+6x2)+(1+3)x22(2+3)(233)(2x22x+6x+2)+(33)x26(2+3)(233)(2x2+2x+6x2)(31)x22(2+3)(233)(2x2+2x+6x+2)
The rest evaluations are elementary but tedious.


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