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Question Number 86643 by mathmax by abdo last updated on 29/Mar/20
calculate∫0∞cos(2ch(x))x2+9dx
Commented by mathmax by abdo last updated on 01/Apr/20
I=∫0∞cos(2chx)x2+9dx⇒2I=∫−∞+∞cos(2chx)x2+9dx=Re(∫−∞+∞ei2chxx2+9dx)letφ(z)=e2ichzz2+9⇒φ(z)=e2ich(x)(z−3i)(z+3i)residustheoremgive∫−∞+∞φ(z)dz=2iπRes(φ,3i)=3iπ×e2ich(3i)6i=π2e2ich(3i)wehavech(3i)=e3i+e−3i2=cos3⇒e2ich(3i)=e2icos(3)=cos(2cos3)+isin(2cos3)⇒∫−∞+∞φ(z)dz=π2e2icos(3)⇒2I=π2cos(2cos3)⇒I=π4cos(2cos(3))
Commented by Ar Brandon last updated on 02/Apr/20
What′sresidustheoremeplease?
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