written 6.2 years ago by | • modified 6.0 years ago |
We can find above integral in two ways:
(i) |z|=1 is the circle with centre (0,0) & radius 1.
We use polar coordinates to evaluate above integral i.e. $z = re^{iθ}$, since r =1
Therefore $z = e^{iθ}$, $dz = ie^{iθ} \ dθ$ & θ varies from 0 to 2π, putting above values in the given integral
Therefore
$I =∫_0^{2π} \frac{e^{ke^{iθ}}}{e^{iθ} . ie^{iθ}} \ dθ$
= $ i ∫_0^{2π} e^{ke^{iθ}} \ dθ $
= $i ∫_0^{2π} e^{k(cosθ+isinθ)} \ dθ = i ∫_0^{2π} e^{kcosθ} .e^{iksinθ} \ dθ $
= $i ∫_0^{2π} e^{kcosθ} \ cos(ksinθ)+isin(k sinθ) \ dθ$…………………………(1)
(ii) Above integral can also evaluated by Cauchy’s Integral formula value of the integral formula
$I= ∫ \frac{e^{kz}}{z} =2πi [ e^{kz} ]_{z=0} = 2πi$ …………………………………..(2)
Equating (1) & (2),we get
$2πi = i ∫_0^{2π}e^{kcosθ} \ cos(ksinθ)+isin(k sinθ) \ dθ$
Equating imaginary parts,
$ 2π = ∫_0^{2π}e^{kcosθ} cos(ksinθ) \ dθ$
$ 2π = 2∫_0^π e^{kcosθ} \ cos(ksinθ) \ dθ$
$π = ∫_0^π e^{kcosθ} \ cos(ksinθ) \ dθ$