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)(x8−x4+1)=13[1x4+1+2x8−x4+1−x4x8−x4+1]
where
1x4+1=x−√22√2(−x2+√2x−1)+x+√22√2(x2+√2x+1)1x8−x4+1=2x−√62√2(2+√3)(2√3−3)(−2x2+√2x+√6x−2)+2x−√62√2(2+√3)(2√3−3)(−2x2−√2x+√6x−2)+2x+√62√2(2+√3)(2√3−3)(2x2+√2x+√6x+2)+2x+√62√2(2+√3)(2√3−3)(2x2−√2x+√6x+2)x4x8−x4+1=(1+√3)x2√2(2+√3)(2√3−3)(−2x2−√2x+√6x−2)+(1+√3)x2√2(2+√3)(2√3−3)(2x2−√2x+√6x+2)+(√3−3)x2√6(2+√3)(2√3−3)(−2x2+√2x+√6x−2)−(√3−1)x2√2(2+√3)(2√3−3)(2x2+√2x+√6x+2)
The rest evaluations are elementary but tedious.
No comments:
Post a Comment