Here are complete solutions. 1) For α ≥ 0 let I(α) = ∫₀^∞ t^α e^{-t} dt. a) I(0) = ∫₀^∞ e^{-t} dt = [−e^{−t}]₀^∞ = 1. b) Convergence for every α ≥ 0. - On [0,1]: 0 ≤ t^α e^{-t} ≤ 1, hence ∫₀¹ t^α e^{-t} dt ≤ 1. - On [1,∞): consider g(t) = t^α e^{-t/2}. Then g'(t) = e^{-t/2} t^{α−1}(α − t/2), so g has a finite global maximum at t = 2α, with M := sup g(t) = (2α)^α e^{−α}. Thus for all t ≥ 1, t^α e^{-t} = g(t) e^{-t/2} ≤ M e^{-t/2}. The integral ∫₁^∞ M e^{-t/2} dt converges. Hence I(α) converges. (You can also split at any finite point such as 2α; the argument above gives a uniform bound with a constant M, which is enough for comparison with e^{-t/2}.) c) Recurrence via integration by parts (valid for α ≥ 1). Let u = t^α, dv = e^{-t} dt. Then du = α t^{α−1} dt, v = −e^{−t}. I(α) = [−t^α e^{−t}]₀^∞ + α ∫₀^∞ t^{α−1} e^{−t} dt. The boundary term is 0: at 0 it is 0 (α ≥ 1); as t → ∞, t^α e^{−t} → 0 by the bound in (b). Therefore I(α) = α I(α − 1). d) For a natural number n, I(n) = n I(n − 1) = n(n − 1) I(n − 2) = … = n! I(0) = n!. So I(n) = n!. 2) I(m,n) = ∫_{−π}^{π} cos(mx) cos(nx) dx, m,n ∈ N. Use cos A cos B = ½[cos(A − B) + cos(A + B)]. So I(m,n) = ½ ∫_{−π}^{π} cos((m − n)x) dx + ½ ∫_{−π}^{π} cos((m + n)x) dx. a) If m ≠ n, then both (m − n) and (m + n) are nonzero. For any nonzero k, ∫_{−π}^{π} cos(kx) dx = [sin(kx)/k]_{−π}^{π} = 0. Hence I(m,n) = 0. b) If m = n, then I(n,n) = ½ ∫_{−π}^{π} 1 dx + ½ ∫_{−π}^{π} cos(2nx) dx = ½(2π) + 0 = π. 3) Partial fraction decomposition of x^3/(1 + x^4). Factor 1 + x^4 over the reals: 1 + x^4 = (x^2 + √2 x + 1)(x^2 − √2 x + 1). Seek x^3/(1 + x^4) = (Ax + B)/(x^2 + √2 x + 1) + (Cx + D)/(x^2 − √2 x + 1). Equating coefficients after clearing denominators gives A + C = 1, −√2 A + B + √2 C + D = 0, (A + C) + √2(−B + D) = 0, B + D = 0. Solving: A = C = 1/2, B = 1/(2√2), D = −1/(2√2). Therefore x^3/(1 + x^4) = [ (1/2)x + 1/(2√2) ]/(x^2 + √2 x + 1) + [ (1/2)x − 1/(2√2) ]/(x^2 − √2 x + 1).

# Detailed Solution in Tutor Mode ## 1. Integral Definition and Properties Let \( I(a) = \int_0^\infty t^a e^{-t} dt \) for \( a \geq 0 \). ### (a) Show \( I(0) = 1 \) To find \( I(0) \): \[ I(0) = \int_0^\infty e^{-t} dt \] Calculating the integral: \[ I(0) = \left[-e^{-t}\right]_0^\infty = 0 - (-1) = 1 \] Thus, \( I(0) = 1 \). ### (b) Show \( I(a) \) converges for \( a > 0 \) **Convergence on \( [0, 1] \)**: - Since \( 0 \leq t^a e^{-t} \leq 1 \) for \( t \in [0, 1] \): \[ \int_0^1 t^a e^{-t} dt \leq \int_0^1 1 dt = 1 \] **Convergence on \( [1, \infty) \)**: - Using the hint, for \( t > 2a \): \[ t^a e^{-t} < e^{-t/2} \] - Define \( g(t) = t^a e^{-t/2} \); find its maximum at \( t = 2a \): \[ g(2a) = (2a)^a e^{-a} \] - For \( t \geq 1 \): \[ t^a e^{-t} \leq g(t)e^{-t/2} \leq M e^{-t/2} \] - The integral converges: \[ \int_1^\infty M e^{-t/2} dt \text{ converges.} \] Thus, \( I(a) \) converges for \( a > 0 \). ### (c) Show \( I(a) = a I(a-1) \) for \( a > 1 \) using Integration by Parts Let: - \( u = t^a \) - \( dv = e^{-t} dt \) - Then, \( du = a t^{a-1} dt \) and \( v = -e^{-t} \) Applying integration by parts: \[ I(a) = \left[-t^a e^{-t}\right]_0^\infty + a \int_0^\infty t^{a-1} e^{-t} dt \] - Boundary term is 0 (as shown previously). Thus, \[ I(a) = a I(a-1) \] ### (d) Conclude \( I(n) = n! \) for natural numbers \( n \) Using the recurrence relation: \[ I(n) = n I(n-1) = n(n-1) I(n-2) = \ldots = n! I(0) \] Since \( I(0) = 1 \): \[ I(n) = n! \] ## 2. Integral of Cosine Functions Define: \[ I(m,n) = \int_{-\pi}^{\pi} \cos(mx) \cos(nx) dx \] ### (a) Show \( I(m,n) = 0 \) whenever \( m \neq n \) Using the identity: \[ \cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)] \] Thus, \[ I(m,n) = \frac{1}{2} \int_{-\pi}^{\pi} \cos((m-n)x) dx + \frac{1}{2} \int_{-\pi}^{\pi} \cos((m+n)x) dx \] - If \( m \neq n \), both terms integrate to 0: \[ \int_{-\pi}^{\pi} \cos(kx) dx = 0 \text{ for any nonzero } k \] Thus, \( I(m,n) = 0 \). ### (b) Show \( I(m,n) = \pi \) whenever \( m = n \) If \( m = n \): \[ I(n,n) = \frac{1}{2} \int_{-\pi}^{\pi} 1 dx + \frac{1}{2} \int_{-\pi}^{\pi} \cos(2nx) dx \] - The first integral evaluates to \( 2\pi \): \[ I(n,n) = \frac{1}{2} (2\pi) + 0 = \pi \] ## 3. Partial Fraction Decomposition To find the partial fraction decomposition of: \[ \frac{x^3}{1 + x^4} \] ### Step 1: Factor \( 1 + x^4 \) Factoring over the reals: \[ 1 + x^4 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1) \] ### Step 2: Set up the Partial Fractions Assume: \[ \frac{x^3}{1 + x^4} = \frac{Ax + B}{x^2 + \sqrt{2}x + 1} + \frac{Cx + D}{x^2 - \sqrt{2}x + 1} \] ### Step 3: Clear Denominators Multiply through by \( 1 + x^4 \): \[ x^3 = (Ax + B)(x^2 - \sqrt{2}x + 1) + (Cx + D)(x^2 + \sqrt{2}x + 1) \] ### Step 4: Equate Coefficients Expand and equate coefficients: 1. \( A + C = 1 \) 2. \( -\sqrt{2}A + B + \sqrt{2}C + D = 0 \) 3. \( A + C - \sqrt{2}B + \sqrt{2}D = 0 \) 4. \( B + D = 0 \) ### Step 5: Solve the System From \( B + D = 0 \): - Let \( D = -B \). Substituting into other equations yields: - After solving, we find: \[ A = C = \frac{1}{2}, \quad B = \frac{1}{2\sqrt{2}}, \quad D = -\frac{1}{2\sqrt{2}} \] ### Final Decomposition Thus, the partial fraction decomposition is: \[ \frac{x^3}{1 + x^4} = \frac{(1/2)x + (1/(2\sqrt{2}))}{x^2 + \sqrt{2}x + 1} + \frac{(1/2)x - (1/(2\sqrt{2}))}{x^2 - \sqrt{2}x + 1} \]