Problem 2.

Evaluate (∫_-∞^∞)(cos(b*x)/(x^2+d^2)*(d^)(x)), d>0,b>0 by residues.

substituting in the integral,

(∫_-∞^∞)(cos(b*x)/(x^2+d^2)*(d^)(x))=1/2*(∫_-∞^∞)((ℇ^(𝑖*b*x))/(x^2+d^2)*(d^)(x))+(∫_-∞^∞)((ℇ^(-𝑖*b*x))/(x^2+d^2)*(d^)(x))

these 2 integrals are symmetrical, so we will only take one of them. Let's consider a ƒ(𝓏) :
ƒ(𝓏)=(ℇ^(𝑖*b*z))/(z^2+d^2), where z is a complex value. This function has simple poles at z=±𝑖*d. The residue of ƒ(𝓏) at z=𝑖*d is

Res(ƒ(𝓏),z=𝑖*d)=(lim_z*->*𝑖*d)((ℇ^(𝑖*b*z))/((z-𝑖*d)*(z+𝑖*d)))=

(ℇ^(𝑖*b(𝑖*d)))/((𝑖*d-𝑖*d)*(𝑖*d+𝑖*d))=(ℇ^(-d*b))/(2*𝑖*d)

By the residue theorem, the integral over the closed contour is

(∮_)(ƒ(𝓏)*(d^)(z))=2*π*𝑖*Res(ƒ(𝓏),z=-𝑖*d) so (∫_-∞^∞)((ℇ^(𝑖*b*x))/(x^2+d^2)*(d^)(x))=(2*π*𝑖*ℇ^(-d*b))/(2*𝑖*d) or π/d*ℇ^(-d*b) .

The original integral is I=Re((∫_-∞^∞)((ℇ^(𝑖*b*x))/(x^2+d^2)*(d^)(x)))=π/d*ℇ^(-d*b), so the final answer is (∫_-∞^∞)(cos(b*x)/(x^2+d^2)*(d^)(x))=π/d*ℇ^(-d*b) .