Prove by mathematical induction

f'(r)=∫(e^(rcosθ)cos(rsinθ))'dθ

=∫[e^(rcosθ)cosθcos(rsinθ)-e^(rcosθ)sin(rsinθ)cosθ]dθ

=∫e^(rcosθ)[cosθcos(rsinθ)-sin(rsinθ)sinθ]dθ

=∫e^(rcosθ)cos(θ+rsinθ)dθ

Suppose f (k) (r) = ∫ e (rcos θ) cos (k θ+rsin θ) d θ.

f^(k+ 1)(r)=∫[e^(rcosθ)cos(kθ+rsinθ)]'dθ

=∫[e^(rcosθ)cosθcos(kθ+rsinθ)-e^(rcosθ)sin(kθ+rsinθ)sinθ]dθ

=∫e^(rcosθ)cos((k+ 1)θ+rsinθ)dθ

So when n=k+ 1, it also holds. To sum up, n∈N* is true.

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We solve F(r) by calculating the integral ∫e[r(cosθ+isθ)dθ, and let z = e (i θ) and dz=izdθ.

∫e[r(cosθ+isθ)dθ=∮e(rz)/(iz)dz, and the integral curve is |z|= 1, counterclockwise.

E (rz)/(iz) is the first pole in |z|= 1, and the remainder at z=0 is1/i =-i.

According to residue theorem, ∮ e (rz)/(iz) dz = 2π i *-i = 2π.

However, ∫ e [r (cos θ+isin θ) d θ = ∫ e (rcos θ) [cos (rsin θ)+isin θ] d θ = ∫ e (rcos θ) cos (rsin θ) d θ+i.

Comparison shows that ∫ e (rcos θ) cos (rsin θ) dθ = 2π, ∫ e (rcos θ) sin (rsin θ) dθ = 0.