Metodi Matematici per l'energetica, Mostacci, esami e esercitazioni

∫

|z|=a

∂

|z|=a

∫ arg_(z) dz / z²

|z|=a

∫ |z| / z

|z|=a

• Re_(z)
• Im_(z)
• arg_(z)
• |z|
• (uco)

|z|=a

∫ Re_(z) / |z|

|z|=a

∫ arg² dz / z³

∂ |z| / z² dz

|z|=a

|z| dz / z^3/2

|z|=a

Solo Fa. NON OLOMORFE (non analitiche) PARAMETRIZZAZIONE (non si può usare Teo: dei Residui)

∫

|z-z_o|=R

∫ 1 / e^z-1

|z|=1

∫ 1 / e^z+1

∫ z / 1 - cos_(z)

|z|=1

∫^+∞₀ x sin(ax) dx / x²+b²

2 ∫^+∞₀ x e^ux dx / x²+1

∫^+∞₀ x e^ux dx / x²+9

∫^+∞₀ e^ux dx / x(x²+1)

∫^+∞_-∞ eu^x dx / x²-2x+3

∫^+∞₀ e^ux dx / (2x-π)²+1

∫^+∞₀ cos(ax) dx / x²+1

∫^+∞₀ cos(ax) dx / x⁴+1

∫^+∞₀ cos(ax) dx / (x⁴+1) TUTOR − FOGLIO BELOW

∫^+∞_-∞ e^i3x dx / x⁴+1

∫^+∞_-∞ x e^i2x dx / x²+1

∫^+∞_-∞ e^-i3x dx / x⁴+1

∫^+∞_-∞ x e^-i3x dx / x²+1

∫^+∞_-∞ x e^-iax dx / x²+b²+1

∫ log z dz

∫_|z|=a z log z dz

∫_|z|=1 z^{log z} dz

∫_|z|=1 z⁵log(3z) dz

∫_|z|=10 z^log(4z) dz

∫^+∞₀ x ^+au x dx / x⁴+1

∫^+∞₀ √x dx / x²+1

∫_|z|=10 z log (4z) dz / z²+25

∫ z³ log (7+3)dz / z⁴-1

∫_z+4|=0 z 3 ^{z log(z) dz} / (z-io)²

∫_|z+2|- z³log(7+3) dz / z⁴-1

∫_|z+2|=0 &frac;7z log(z)dz / z²+a²

∫^+∞_-∞ e^ux dx / √x(1+x)

DISPENSA

\[ \int_0^\infty \frac{x}{x^6+1} dx \quad \int_0^\infty \frac{x^{a-1}}{x^2+1} dx \quad \frac{1}{2\pi i} \int_{-i\infty}^{c+i\infty} \frac{e^{-\alpha i\pi z}}{\sqrt{z}} e^{-t z} dz \]

\[\frac{1}{2\pi i} \int_{-i\infty}^{c+i\infty} \frac{e^{tz}}{(1+z)^2(z-z)} dz, \quad t \in \mathbb{R}, -i\infty \lt c+i\infty \]

\[\int_{-\infty}^\infty \frac{x^2}{(x^2+1)(x^2+9)} dx \quad \int_{-\infty}^\infty \frac{\log(1+x^2)}{x^4+1} dx\]

\[\frac{1}{2\pi i} \int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty} \frac{e^{-zt}}{z^2 \cdot (e-z)} dz\]

\[I = \oint e^{-t|z|} (z^2+\Re(z)) dz\] \

\[|z| = 2, \quad \Imu_z(tz) > 0 \rightarrow in superior \]

PARAMETRI:

\[z=2e^{i\theta}, \quad \theta \in (0, 2\pi)\]

\[\Re(z)=-2\cos\theta=2\cos\theta\]

\[dz = 2ie^{i\theta}d\theta\]

\[ ... \]

∫ _|z|=a Re(z)/|z| dz

PARAMETRIZZO:z = ae ^iθ, θ ∈ [θ₀, θ₀ + 2π]Re(z) = a cosθdz = iae^iθdθ, |z| = a

= ∫_θ0^{θ₀ + 2π} a cosθ    / a^iθ iae^iθdθ = θ₀

∫_θ0^{θ₀ + 2π} iae^{iθ e iθ}α dθ + a/2 ∫_θ0^{θ₀ + 2π} ie^iθe^iθα dθ == a ∫_θ0^{θ₀ + 2π} [e^{i(θ + θ₀ + 2π)} - e^iθ] -

e^{i(θ₀ + 2π)} - e / α + ia/2 (θ + θ₀) - iaθ/ 2 _iαθ/2 + iαπiαθ² _{/ 2} =

"iαπ"

PARAMETRIZZO:z = ae^iθdz = aie^iθ

    ,

z = a, θ ε [θ₀, θ₀ + 2π]

∫_|z|=a 1/z dz

= ∫_θ0^{θ₀ + 2π} ia dθ = ia [θ]

= iαθ + iα2π- iαθ= iaπ

PARAMETRIZZO:

z = az^-iθ z = a (z = a

z²

* z²dz = aie^iθdθ

, ∫_|z|=a 1/z² dz

= ∫_θ0^{θ₀ + 2π} -ie ^iθ / ze * (e^{-iθ /} θ= [e^{-iθ + θ}]

0

PARAMETRIZZO:z = ae^iθ θ ε [θ₀, θ₀+2π]dz = aie^iθ "θ",

z

= a * e

= a^3/2]

ia ∫_θ0^3/2 e^-iθ/2

= ia,

= ia ∫ q

∫_-∞^{+∞ eu x}/_{x² - 2x + 2} dx

∫_R ^euz/_{z² - 2z + 2} dz = ∫_R' ^{eiz - e-iz}/_2i 1/_{z² - 2z + 2} dz

= 1/_2i ∫_R e^-iz/_{z² - 2z + 2} dz - 1/_2i ∫_R e^-iz/_{z² + 2} dz

• lim_|z|→+∞ z·F(z) = lim_|z|→+∞ z·e^-iz/_{z² - 2z + 2} = 0
• lim_|z|→+∞ z·F(z) = lim_|z|→+∞ z·e^-iz/_{z² + 2} = 0

∫_R F(z) dz = 0

= ∫_Γ+ e^-iz/_{z² - 2z + 2} dz - 1/_2i ∫_Γ'- e^-iz/_{z² + 2} dz

= 1/_2i (2πi Res(1+i)) - 1/_2i (-2πi Res(1-i))

Res (z_k) = lim_z→zk

• Res(1+i) = eⁱ⁽¹⁺ⁱ⁾/_{(z² + 2z - 2)²} = e^-1/2i
• Res(1-i) = e^-i(1-i)/_{2(1-i) - 2} = e^-1/2i

I = 1/2i (±2πi) e^-i(1-i)/2i = πe^-1 Reu (1)

∑ −∞_

-☹z+1 v. 7- se

...leeemi σIOORDANN

f ...iv -7fe

liv di(- i") ≡ 7-9e-3i

I = RZO

efsin

&χ≡

∣(z+i)(-i)=...-7≡

≡≡

πe -

πe -3

∮ log (z) dz|z| = a

∮ f(x)dx = ∫ f(x)dx + ∫ f(x)dx + + ∫ f(x)dx - ∫ f(x)dx

f(z)dz = ∫ f(z)dz = ∫ f(z)dz + ∫ f(z)dz - - ∫ f(z)dz

z = ρ e^iθ

∫_r0 = 0

d z = e^{i θ} d ρ

log (z) = log (ρ e^{i θ}) == i ϑ ρ + i π

d z = e^{-i θ} d ρ

- ∫ π d ϑ + ∫ - i π d ϑ == - 2π i ∫ d ϑ == - 2 π i a

∫_Γ -7eu(ζ7) / ζ² + 25 dζ

Poli : z = ±5i

Br06610 : 10

∫_Γ f(ζ)dζ = ∫_A f(ζ)dζ + ∫_A' f(ζ)dζ + ∫_B' f(ζ) dζ = 2πi Σ Res(f_zk)

Verifico lemma di Jordan (∫_R0 )

Res(z_k) = lim_z→zk (z - z_k) z eu(ζ7) / ζ² + 25 = -7k eu(ζ4z_k) / Z_k

Res(5i) = eu(4 · 5i) = 1/2 [eu(-20) + iπ/2]

Res(-5i) = eu(-20i) = 1/2 [eu(20) - iπ/2]

Segmento BB' = (0, -π)

z = ρ e^-iπ dz = e^-iπ dφ = iρ dφ

log(z+3) log(4ρ e^iπ) = eu(4ρ) + iπ

Segmento AA' = (θ = -π)

log(z+3) = eu (ιρ) - iπ

I = 2πi Σ Res =

= xπi M / k² [2eu(20) ± iπ / 2 - iπ / 2] = πi eu(20)

= [πi eu(400) + πi / 0∫_Γ 2ρ / ρ² + 25 dφ = πi eu(ρ² +25)]_0¹⁰

= πi [eu(4000) + eu(125) - eu(257)]

= πi [eu(1000) + eu(5) ]