>>658
∫[0,∞] exp((2t+3)i) exp(-st) dt
= ∫[0,∞] exp((2i-s)t+3i) dt
= [exp((2i-s)t+3i)/(2i-s)]_[t=0,∞]
= e^(3i)/(s-2i)
= {cos(3)+i*sin(3)}(s+2i)/(s^2+4)

実数部分をとって
∫[0,∞] cos(2t+3)exp^(-st) dt
= {s cos(3) - 2sin(3)}/(s^2+4)