Esercitazione guidata Analisi Matematica 1 - decima parte

1)

∫₀^{π/4 2}/_(1+tgx)² dxtgx = tx = arctg tdx = ¹/_1+t² dt= ∫₀^{1 2}/_{(1+t)² ∙ 1/(1+t²)} dt= 2 ∫₀^{1 1}/_{(1+t)² (1+t²)} dt(1+t)² (1+t²) = 0⇒ 1+t=0        1+t²=0t=-1m=2A,BC,D¹/_{(1+t)² (1+t²)} = ^A/_1+t + ^B/_(1+t)² + ^Ct+D/_1+t²A(1+t)(1+t²) + B(1+t²) + (Ct+D)(1+t)²=(1+t)² (1+t²)= A(1+t²+t+t³) + B(1+t²) + (Ct+D)(1+t²+2t)= (1+t)² (1+t²)= A+At²+At+At³ + B+Bt² + Ct³ + Ct + Dt² + 2Ct=(1+t)² (1+t²)

1)∫₀^π/4 2/(1+tgx)² dxtgx = tx = arctg tdx = 1/(1+t²) dt= ∫₀¹ 2/(1+t)² · 1/(1+t²) dt= 2 ∫₀¹ 1/((1+t)²(1+t²)) dt(1+t)²(1+t²) = 0⇔ 1+t = 0t = -1m = 2↑ A,B1+t² = 0↑ C,D1/((1+t)²(1+t²)) = A/(1+t) + B/(1+t)² + (Ct+D)/(1+t²)= A(1+t)(1+t²) + B(1+t²) + (Ct+D)(1+t)²= (1+t)²(1+t²)= A(1+t²+t+t³) + B(1+t²) + (Ct+D)(1+t²+2t)= (1+t)²(1+t²)= A+A t²+t A+t A³ + B t² + B+ Ct+Ct²+Dt+2 Ct+ D + Dt² + 2Dt= t³(A + C) + t²(A + B + 2C + D) ++ t (A + C + 2D) + (A + B + D){A + C = 0A + B + 2C + D = 0A + C + 2D = 0A + B + D = 1

2)

∫₀¹ sinh x / (e^x - 1) dx= lim_{ε -> 0⁺} ∫_ε¹ sinh x / (e^x - 1) dx∫_ε¹ sinh x / (e^x - 1) dx =sinh x = (e^x - e^-x) / 2sinh x = 1/2 (e^x - 1/e^x) = 1/2 (e^2x - 1)/e^x= 1/2 (e^x - 1)(e^x + 1) / e^xsinh x / (e^x-1) = 1/2 (e^x - 1)(e^x + 1) / e^x (e^x - 1)= ¹⁄₂ ^{ex + 1}⁄_e^x= ¹⁄₂ (1 + ¹⁄_e^x)= ¹⁄₂ (1 + e^-x)

∫_ε^{1 sinh x}⁄_{e^x - 1} dx = ¹⁄₂ ∫_ε¹ (1 + e^-x) dx= ¹⁄₂ ( ∫_ε¹ dx + ∫_ε¹ e^-x dx )= ¹⁄₂ ( [ x ]_ε¹ - [ e^-x ]_ε¹ )= ¹⁄₂ ( (1 - ε) - (e^-1 - e^-ε) )= ¹⁄₂ (1 - ε - ¹⁄_e + ¹⁄_eε )

∫01 sinh x⁄ex - 1 dx = limε->0+ ∫ε1 sinh x⁄ex - 1 dx= limε->0+ 1⁄2 (1 - ε - 1⁄e + ( 1⁄eε ) )= 1⁄2 (1 - 1⁄e + 1) = 1⁄2 (2 - 1⁄e)

3)

\(\int_{1}^{+\infty}x^2e^{-x} \, dx = \lim_{b \to +\infty} \int_{1}^{b} x^2e^{-x} \, dx\)\(f \in \mathcal{E}([1, +\infty))\)\(b > 1\)\([1, b] \subseteq [1, +\infty)\)\(f \in \mathcal{E}([1, b])\)

\(\int_{1}^{b} x^2e^{-x}\, dx = \left[-e^{-x}x^2\right]_{1}^{b} - \int_{1}^{b} -e^{-x} \cdot 2x \, dx\)= \(\left[-e^{-x}x^2\right]_1^b + 2\int_{1}^{b} xe^{-x} \, dx\)= \([-e^{-b} b^2 + e^{-1}\] + 2 \left\{\left[-e^{-x}x\right]_{1}^{b} - \int_{1}^{b} -e^{-x} \, dx\right\}\)= \([-e^{-b} b^2 + e^{-1}\] + 2 \{-e^{-b}b + e^{-1} - [e^{-x}]_{1}^{b} \}\)= \(-e^{-b} b^2 + e^{-1} - 2e^{-b} + 2e^{-1} - 2(-e^{-b} - e^{-1})\)= \(-e^{-b} b^2 + 3e^{-1} - 2e^{-b} - 2e^{-b} + 2e^{-1}\)= \(e^{-b} b^2 + 5e^{-1} - 2e^{-b} - 2e^{-b}\)\(\to 5e^{-1}\)

abab(a, c](c, b]c ∈ (a, b)∫_a^b f(x) dx = ∫_a^c f(x) dx + ∫_c^b f(x) dx= lim_{ε → a⁺ ∫ε^c f(x) dx + limt → b^- ∫c^t f(x) dx}

∫_-∞² 1/(x-1) dxDf : x ≠ 1-∞2(-∞, 1)(1, 2]c < 1(-∞, c](c, 1) (1, 2](-∞, c] (c, +∞)x ≠ 0(-∞, 0) (0, +∞)(-∞, c] (c, 0) (0, d] [d, +∞)

4)

∫₀¹ (1 - cos x) / (√x³ log(1+x)) dxf ∈ (0,1) Df: (0,+∞) { √x³ log(1+x) ≠ 0 1+x > 0 x³ > 0 } ⇒ { x ≠ 0 x > -1 x > 0 } (0,+∞)√x³ ≠ 0 ∧ log(1+x) ≠ 0 x ≠ 0 ∧ 1+x ≠ 1 ⇒ x ≠ 0

x → 0⁺ f(x) → ? 1 - cos x ∼ ½ x²√x³ log(1+x) ∼ √x³ = x√x f(x) ∼ ½ / x²√x = ½ / x√x

∫₀¹ f(x) dx ∼ ½ ∫₀¹ 1/√x dx CONVERGE⇒ ∫₀¹ f(x) dx converge.

5)

₂^+∞ ∫ (π - 2 arctg x) / (x² log x) dxDF: (0, 1) ∪ (1, +∞){ x² log x ≠ 0 { x > 0 ⇒{ x ≠ 0, 1 { x > 0 ∫ ∈ Θ ([2, +∞))x² log x ≠ 0 ⟺ x² ≠ 0 ∧ log x ≠ 0⟺ x ≠ 0 ∧ x ≠ 1

x → +∞π - 2 arctg x =arctg x + arctg (1/x) = π/2 ∀ x > 0π - 2 arctg x = 2 (π/2 - arctg x)= 2 arctg (1/x) ~ (2/x)arctg x ~ x , x → ∞arctg (1/x) ~ 1/x , x → +∞

₂^+∞ ∫ (π - 2 arctg x) / (x² log x) dx ~₂^+∞ ∫ 2 / (x³ log x) dx converge

6)

∫₀¹ √x / (e^αx - √(1 + 2x)) dx   α ≥ 1x -> 0⁺   √x ~ √xe^αx - √(1 + 2x) = (√(1 + αx + o(x)) - (1 + 1/2 . 2x + o(x))= αx - x + o(x)= (α - 1)x + o(x)

se   α = 1 = o(x)e^αx - √(1 + 2x) == (1 + αx + 1/2 (αx)² + o(x²)) - (1 + 1/2 . 2x - 1/8 (2x)² + o(x²))= αx + α²/2 x² + o(x²) - x + 1/2 x² + o(x²)= (α - 1)x + 1/2 (α² + 1) x² + o(x²)

α - 1 = 0   => α = 1f(x) = √x / (x² + o(x²)) ~ √x / x² = 1 / x^{2 - 1/2} = 1 / x^3/2∫₀¹ f(x) dx ~ ∫₀¹ 1 / x^3/2 dx   DIVERGE

α - 1 ≠ 0   α > 1f(x) = √x / (α - 1)x + o(x)) ~ √x / (α - 1)x = 1 / x^{1 - 1/2} = 1 / √x∫₀¹f(x)dx ∼ ∫₀¹¹/_√x dx CONVERGE

7)

∫₀¹(^x/_sinx - 1)^α dx α ∈ ℝ^x/_sinx - 1 = ^x-sinx/_sinx= ^{x3(- 1/₆x3 + o(x3))}/_{x - ¹/6x³ + o(x³)} = ^{1/₆x3 + o(x3)}/_{x + o(x)}∼ ¹/₆ x³/_x = ¹/₆ x²

∫₀¹(f(x))^α dx ∼ ∫₀¹x^2α dx = ∫₀^{1 1}/_x^-2α dxConverge se e solo se -2α < 1=> α > -¹/₂

8)

∫₁^{∞ sin x}/_(logx+x)² dx f ∈ ℓ (ℝ(1,+∞))ma non ha segno costante│^{sin x}/_{(log x + x)²}│ = ^{│sin x│}/_{(log x + x)²} ≤ ¹/_{(log x + x)²}(│sin x│ ≤ 1)