0
1.3kviews
If C is the circle $|z|=1$, using the integral $\int \frac{e^{kz}}{z}dz$, where K is real, show that $\int_0^{\pi} e^{k cos \theta} cos (ksin \theta) d \theta = \pi$
0
14views

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,

$2π = ∫_0^{2π}e^{kcosθ} cos(ksinθ) \ dθ$

$2π = 2∫_0^π e^{kcosθ} \ cos(ksinθ) \ dθ$

$π = ∫_0^π e^{kcosθ} \ cos(ksinθ) \ dθ$