Saturday, June 29, 2019

integration - prove that intinftyinftyfracx41+x8dx=fracpisqrt2sinfracpi8






prove that x41+x8dx=π2sinπ8




My attempt:

C is semicircle in upper half complex plane



Simple poles = eiπ8,ei3π8,ei5π8,ei7π8 lie in upper semi-circle C and real axis



Given integral value =2πi(sum of residues)=2πi(18)[ei5π8+ei15π8+ei25π8+ei35π8]=0.27059π



This is numerically equal to π2sinπ8. But without using calculator, how to get this expression.


Answer



Just extending what you've got so far. Let's note z=ei5π8 and recall that cosz=eiz+eiz2, then:
2πi(18)[ei5π8+ei15π8+ei25π8+ei35π8]=2πi(18)[z+z3+z5+z7]=2πi(18)z[1+z2+z4+z6]=2πi(18)z[1+z2+z4(1+z2)]=2πi(18)z(1+z2)(1+z4)=2πi(18)z4(z1+z)(z2+z2)=2πi(18)ei5π22cos(5π8)2cos(5π4)=πi(1)icos(π2+π8)(12)=π2sin(π8)


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