✔ 最佳答案
∫x cos x dx = ?
∫x cos x dx
= ∫x d(sin x)
= x sin x - ∫sin x dx
= x sin x + cos x + C
∫((sin x)/x) dx = ?
Method 1:
∫((sin x)/x) dx
= ∫Σ(n = 0 to ∞)[((- 1)ⁿ x^(2n + 1))/(x(2n + 1)!)] dx（use sin x = Σ(n = 0 to ∞)[((- 1)ⁿ x^(2n + 1))/(2n + 1)!]）
= ∫Σ(n = 0 to ∞)[((- 1)ⁿ x²ⁿ)/((2n + 1)!)] dx
= Σ(n = 0 to ∞)[((- 1)ⁿ x^(2n + 1))/((2n + 1)(2n + 1)!)] + C
Method 2:
∫((sin x)/x) dx
= ∫(1/x) d(- cos x)
= - (cos x)/x - ∫- cos x d(1/x)
= - (cos x)/x - ∫((cos x)/x²) dx
= - (cos x)/x - ∫(1/x²) d(sin x)
= - (cos x)/x - (sin x)/x² + ∫sin x d(1/x²)
= - (cos x)/x - (sin x)/x² + ∫((- 2 sin x)/x³) dx
= - (cos x)/x - (sin x)/x² + ∫((2!/x³) d(cos x)
= - (cos x)/x - (sin x)/x² + (2! cos x)/x³ - ∫cos x d(2!/x³)
= - (cos x)/x - (sin x)/x² + (2! cos x)/x³ - ∫((- 3! cos x)/(x^4)) dx
= - (cos x)/x - (sin x)/x² + (2! cos x)/x³ + ∫(3!/(x^4)) d(sin x)
= - (cos x)/x - (sin x)/x² + (2! cos x)/x³ + (3! sin x)/(x^4) - ∫sin x d(3!/(x^4))
= - (cos x)/x - (sin x)/x² + (2! cos x)/x³ + (3! sin x)/(x^4) - ∫((- 4! sin x)/(x^5)) dx
= - (cos x)/x - (sin x)/x² + (2! cos x)/x³ + (3! sin x)/(x^4) - ∫(4!/(x^5)) d(cos x)
= - (cos x)/x - (sin x)/x² + (2! cos x)/x³ + (3! sin x)/(x^4) - (4! cos x)/(x^5) + ∫cos x d(4!/(x^5))
= - (cos x)/x - (sin x)/x² + (2! cos x)/x³ + (3! sin x)/(x^4) - (4! cos x)/(x^5) + ∫((- 5! cos x)/(x^6)) dx
= - (cos x)/x - (sin x)/x² + (2! cos x)/x³ + (3! sin x)/(x^4) - (4! cos x)/(x^5) - ∫((5!/(x^6)) d(sin x)
= - (cos x)/x - (sin x)/x² + (2! cos x)/x³ + (3! sin x)/(x^4) - (4! cos x)/(x^5) - (5! sin x)/(x^6) + ∫sin x d(5!/(x^6))
=
．
．
．
= [- (0! cos x)/x - (1! sin x)/x² + (2! cos x)/x³ + (3! sin x)/(x^4) - (4! cos x)/(x^5) - (5! sin x)/(x^6) + ……] + C
= Σ(n = 0 to ∞)[((- 1)^(n + 1) (2n)! cos x)/(x^(2n + 1)) + ((- 1)^(n + 1) (2n + 1)! sin x)/(x^(2n + 2))] + C
Note that Method 1 is suitable for small values of x, whereas Method 2 is suitable for large values of x.
∫((sin x)/x) dx is still undefined at x = 0, although by putting x = 0 in the answer in Method 1 still gives the result Σ(n = 0 to ∞)[(- 1)ⁿ/((2n + 1)(2n + 1)!)] + C, which the value is exist.
靈感來自
http://en.wikipedia.org/wiki/Sine_integral#Asymptotic_expansion
