计算∫_{-π}^πcosⁿ(a-x)cos((n-2r)x)dx并证明其等于λcos((n-2r)a)（无需递推公式）

Alright, let's break down this integral problem and prove the result without leaning on recurrence relations. Here's a step-by-step approach that uses complex exponentials (Euler's formula) and orthogonality of trigonometric functions—no recursive tricks needed.

Step 1: Rewrite cosines using Euler's Formula #

First, recall that any cosine term can be expressed using complex exponentials:
cosθ = (e^(iθ) + e^(-iθ))/2

Let's apply this to cosⁿ(a-x) first. Expanding it with the binomial theorem:

cosⁿ(a-x) = [(e^(i(a-x)) + e^(-i(a-x)))/2]^n
          = (1/2ⁿ) Σ_{k=0}^n C(n,k) e^(i(n-k)(a-x)) e^(-ik(a-x))
          = (1/2ⁿ) Σ_{k=0}^n C(n,k) e^(i(n-2k)(a-x))

Next, rewrite the other cosine term in the integral the same way:
cos((n-2r)x) = (e^(i(n-2r)x) + e^(-i(n-2r)x))/2

Multiplying these together gives us the full integrand in terms of complex exponentials:

cosⁿ(a-x)cos((n-2r)x) = (1/(2^(n+1))) Σ_{k=0}^n C(n,k) [e^(i(n-2k)(a-x)) + e^(-i(n-2k)(a-x))] * [e^(i(n-2r)x) + e^(-i(n-2r)x)]

Step 2: Filter terms that contribute to the integral #

The key here is using the orthogonality of complex exponentials: for any integer m ≠ 0,
∫_{-π}^π e^(imx)dx = 0
Only when m=0 does this integral equal 2π.

Let's expand the product inside the sum and look for terms where the exponent of x is zero (these are the only ones that won't vanish when integrated):

• For e^(i(n-2k)(a-x)) * e^(i(n-2r)x):
  The exponent of x is -(n-2k) + (n-2r) = 2(k-r). This equals zero only when k=r.
• For e^(i(n-2k)(a-x)) * e^(-i(n-2r)x):
  The exponent of x is -(n-2k) - (n-2r) = -2n + 2k + 2r. This equals zero only when k = n - r (since r < n/2, n-r is a valid index between 0 and n).
• For e^(-i(n-2k)(a-x)) * e^(i(n-2r)x):
  The exponent of x is (n-2k) + (n-2r) = 2n - 2k - 2r. This equals zero only when k = n - r.
• For e^(-i(n-2k)(a-x)) * e^(-i(n-2r)x):
  The exponent of x is (n-2k) - (n-2r) = 2(r - k). This equals zero only when k=r.

All other values of k will lead to integrals that evaluate to zero, so we can ignore them entirely.

Step 3: Calculate the non-zero integral terms #

First, handle the k=r case. The non-zero terms here simplify to:

(1/(2^(n+1))) C(n,r) [e^(i(n-2r)a) + e^(-i(n-2r)a)]

Integrating this over [-π, π] gives:

(1/(2^(n+1))) C(n,r) * 2π * [e^(i(n-2r)a) + e^(-i(n-2r)a)]

Using (e^(iθ) + e^(-iθ)) = 2cosθ, this reduces to:
(π C(n,r)/2^(n-1)) cos((n-2r)a)

Next, handle the k=n-r case. Since C(n, n-r) = C(n,r) (symmetry of binomial coefficients), the non-zero terms here simplify to:

(1/(2^(n+1))) C(n,r) [e^(-i(n-2r)a) + e^(i(n-2r)a)]

Integrating this gives the same result as the k=r case:
(π C(n,r)/2^(n-1)) cos((n-2r)a)

Step 4: Combine results to find λ #

Adding the two non-zero results together gives the final integral value:

∫_{-π}^πcosⁿ(a-x)cos((n-2r)x)dx = (π C(n,r)/2^(n-1) + π C(n,r)/2^(n-1)) cos((n-2r)a)
                                = (π C(n,r)/2^(n-2)) cos((n-2r)a)

So the parameter λ is exactly π C(n,r)/2^(n-2) (where C(n,r) is the binomial coefficient n!/(r!(n-r)!)).

内容的提问来源于stack exchange，提问作者Alex