We know the indefinite integral with d being the differential operator (e.g., by substitution t=ez and partial fraction)
dzcoshz+cosθ=−isinθdlnez+eiθez+e−iθ.
Can we find a nice contour in the complex plane and using the Cauchy residue theorem to directly show its definite form?
∫∞0dzcoshz+cosθ=θsinθ
Other methods are welcome too.
The upper (neither the lower) hemispherical contour does not work because the segment integral close to the imaginary axis at infinity does not vanish.
Answer
A more direct approach than the one I mentioned in the comments is to integrate f(z)=zcoshz+cosθ,0<θ<π, around a rectangle in the upper half-plane of height 2πi.
Doing so, we get ∫∞−∞xcoshx+cosθdx−∫∞−∞x+2πicoshx+cosθdx=2πi(Res[f(z),i(π−θ)]+Res[f(z),i(π+θ)])=2πi(i(π−θ)sinh(i(π−θ))+i(π+θ)sinh(i(π+θ)))=2πi(π−θsinθ−π+θsinθ)=−4πiθsinθ.
(The integrals along the sides of the rectangle vanish in the limit since |coshz| grows exponentially as Re(z)→±∞.)
Therefore, 2πi∫∞−∞1coshx+cosθdx=4πiθsinθ, and the result follows since the integrand is even.
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