prove that ∫∞−∞x41+x8dx=π√2sinπ8
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 π√2sinπ8. 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+e−iz2, 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(z−1+z)(z−2+z2)=2πi(−18)ei5π22cos(5π8)2cos(5π4)=πi(−1)icos(π2+π8)(−1√2)=π√2sin(π8)
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