Analyse complexe Chapitre 2

Analyse complexe Chapitre 2

Lucie Le Briquer

Probl`eme.

1. 0<<(s)< d

|f(x)x^s−1|=|f(x)||x|^<(s)−1 ∼

x→0|f(0)||x|^<(s)−1 int´egrable car <(s)>0 Et en +∞:

|f(x)x^s−1| ∼

x→+∞λ|x|<(s)−1−d int´egrable puisque<(s)< d Th´eor`eme d’holomorphie sous le signeR

.

B ={z|0<<(z)< d} ∀x∈]0,+∞[, s7−→f(x)x^s−1est holomorphe surB K⊂B compact

∀s∈K,|f(x)x^s−1| ≤ |f(x)|x^a1_x≥1

| {z }

int´egrable

+|f(x)|a^b1_0≤x≤1

| {z }

int´egrable

o`ua= max

s∈U(<(s−1)) etb= min

s∈U(<(s−1)).

DoncF est holomorphe surB.

2. U =C\[0,+∞[ est un ouvert simplement connexe ne contenant pas 0−→on peut d´efinir logz

– exp logz=z=|z|e^iθ – log(z) = ln|z|+iarg(z) – arg(z)∈]0,2π[

– z^s:= exp(slogz) surU

1−iε 1 +iε

1•

1

On a :

(1 +iε)^s−−−−→

ε→0+ 1 (1−iε)^s−−−−→

ε→0− e^2iπs car log(1 +iε) = ln|1 +iε|

→0

+iarg(1 +iε)

→0

et log(1−iε) = ln|1−iε|

→0

+iarg(1−iε)

→2π

3.

R γ_ε,R

ε

Z

γ_ε,R

f(z)z^s−1dz= 2iπ X

α∈Pf

Res(f(z)z^s−1, α)

Attention. Res(f(z)z^s−1) 6= α^s−1Res(f, α) en g´en´eral, on a l’´egalit´e uniquement si des pˆoles simples.

Z

Cε

f(z)z^s−1dz= Z π/2

3π/2

f(εe^it)(εe^it)^s−1iεe^itdt γ(t) =εe^itaveec t∈[3π/2, π/2] et :

(εe^it)^s−1= exp((s−1) log(εe^it)) = exp((s−1)(lnε+it))

Z

Cε

f(z)z^s−1dz

≤

TCD

Z 3π/2

π/2

|f(εe^it)||ε^s−1||e^it(s−1)|dt −−−→

ε→0 0

4.

Z R

0

f(x+iε)(x+iε)^s−1dx−

Z R

0

f(x−iε)(x−iε)^s−1dx−−−→^TCD

ε→0

Z R

0

f(x)x^s−1dx−

Z R

0

f(x)x^s−1e^2iπ(s−1) Donc avecR−→+∞:

Z +∞

0

f(x)x^s−1dx

1−e^2iπ(s−1)

= 2iπX

Res(f(z)z^s−1, α)

2

5. Application.

Z +∞

0

x^s−1 dx

x²+ 1 = 1

1−e^2iπ(s−1) ×

Res x^s−1

x²+ 1, i

+ Res

x^s−1 x²+ 1,−i

= 2iπ

i^s−1

2i −⁽⁻ⁱ⁾_2i^s−1 1−e^2iπ(s−1)

=πe^iπ/2(s−1)−πe^{i3π/2(s−1)}

1−e^2iπ(s−1)i + angle moiti´e

3