关于证明积分∫₀^{(N+1/2)π}|sin(t)/t|dt > C log N的技术问询

Hey there! Let's work through this problem step by step—you're already halfway there by reducing the Dirichlet kernel integral to this sine integral, so great job getting that far.

First, the core idea here is to split the integral into intervals where we can bound (|\sin t|) and (1/t) easily, then use the asymptotic behavior of the harmonic series to connect the sum to (\log N).

Step 1: Split the integral into manageable intervals #

For any positive integer (N), we can split the interval ([0, (N+1/2)\pi]) into three types of segments:

• The initial segment ([0, \pi])
• Full period segments ([k\pi, (k+1)\pi]) for (k = 1, 2, ..., N-1)
• The final half-period segment ([N\pi, (N+1/2)\pi])

Step 2: Bound the integral on each segment #

Let's compute lower bounds for each part:

1. Initial segment ([0, \pi]):
   Since (\sin t \geq 0) here, the integral is just (\int_0^\pi \frac{\sin t}{t}dt = A), where (A) is a positive constant (approximately 1.8519—this is a fixed value, no dependence on (N)).

2. Full period segments ([k\pi, (k+1)\pi]):
   For any (k \geq 1), note that:

   • (t \leq (k+1)\pi), so (\frac{1}{t} \geq \frac{1}{(k+1)\pi})
   • (|\sin t|) has a period of (\pi), and (\int_{k\pi}^{(k+1)\pi} |\sin t|dt = 2) (this is true because integrating (|\sin t|) over any full (\pi)-length interval gives 2)

   Combining these, we get:
   [
   \int_{k\pi}^{(k+1)\pi} \frac{|\sin t|}{t}dt \geq \frac{2}{(k+1)\pi}
   ]

3. Final half-period segment ([N\pi, (N+1/2)\pi]):
   Here:

   • (t \leq (N+1/2)\pi), so (\frac{1}{t} \geq \frac{1}{(N+1/2)\pi})
   • Using substitution (u = t - N\pi), (|\sin t| = |\sin(u + N\pi)| = |\sin u| = \sin u) (since (u \in [0, \pi/2])), so (\int_{N\pi}^{(N+1/2)\pi} |\sin t|dt = \int_0^{\pi/2} \sin u du = 1)

   So this segment's integral is bounded below by:
   [
   \int_{N\pi}^{(N+1/2)\pi} \frac{|\sin t|}{t}dt \geq \frac{1}{(N+1/2)\pi}
   ]

Step 3: Combine the bounds and connect to (\log N) #

Now add up all these lower bounds:
[
f(N) \geq A + \sum_{k=1}^{N-1} \frac{2}{(k+1)\pi} + \frac{1}{(N+1/2)\pi}
]

Let's reindex the sum by letting (m = k+1)—this turns the sum into (\sum_{m=2}^N \frac{2}{m\pi}). We can rewrite this using the harmonic series (H_N = \sum_{m=1}^N \frac{1}{m}):
[
\sum_{m=2}^N \frac{2}{m\pi} = \frac{2}{\pi}\left(H_N - 1\right)
]

We know the harmonic series has the asymptotic expansion:
[
H_N = \log N + \gamma + o(1)
]
where (\gamma \approx 0.5772) is the Euler-Mascheroni constant, and (o(1)) goes to 0 as (N) grows large.

Substituting this back in, we get:
[
f(N) \geq A + \frac{2}{\pi}\left(\log N + \gamma - 1 + o(1)\right) + \frac{1}{(N+1/2)\pi}
]

Step 4: Finalize the constant (C) #

For large enough (N), the (o(1)) term and the last fraction (\frac{1}{(N+1/2)\pi}) become negligible. The remaining terms include (\frac{2}{\pi}\log N) plus a positive constant ((A + \frac{2}{\pi}(\gamma -1) \approx 1.58), which is positive).

For small (N) (like (N=1,2,3)), we can just compute (f(N)) directly and verify it's larger than some multiple of (\log N). For example:

• (N=1): (f(1) \approx 2.17 > 0 = C\log 1)
• (N=2): (f(2) \approx 2.33 > C\log 2) (even if (C = 1/\pi \approx 0.318), this holds)

So we can pick any constant (C < \frac{2}{\pi}) (e.g., (C = \frac{1}{\pi})) and there will exist some (N_0) such that for all (N \geq N_0), (f(N) > C\log N), and for (N < N_0), the inequality holds by direct computation.

This completes the proof—when you substitute this back into your Dirichlet kernel integral, you'll get the desired result (L_N > C'\log N) for some constant (C')!

备注：内容来源于stack exchange，提问作者Leaves