能否在未知ζ(2)值时用复分析方法计算∫₀¹(log(x)/(x-1))dx？

Absolutely! You can compute this integral using complex analysis without relying on the known value of ζ(2). Let me walk you through a concrete, step-by-step approach that uses the residue theorem and careful handling of logarithmic branch cuts and infinite sums.

Step 1: Rewrite the integral for convenience #

First, note that the integrand has a removable singularity at x=1 (since $\lim_{x \to 1} \frac{\log x}{x-1} = 1$), so the integral converges. Let's rewrite it to simplify upcoming steps:
I = ∫₀¹ (logx)/(x-1) dx = ∫₀¹ (-logx)/(1 - x) dx

If we expand $\frac{1}{1-x}$ as a geometric series (valid for $|x|<1$), this integral becomes the sum $\sum_{n=0}^\infty \int₀¹ (-logx)x^n dx = \sum_{n=0}^\infty \frac{1}{(n+1)^2} = ζ(2)$. But since we don't want to use the known value of ζ(2), we'll use complex analysis to compute this sum directly.

Step 2: Use residue theorem to compute $\sum_{n=1}^\infty \frac{1}{n^2}$ #

We'll use the function f(z) = π cot(πz) / z² and a square contour (centered at the origin, with vertices at $(N+\frac{1}{2})(\pm1\pm i)$ for integer $N \to \infty$). Here's why this works:

• $\cot(πz)$ has simple poles at all integer $z = n$, with residue $\frac{1}{π}$ at each pole.
• The contour encloses all poles $z = \pm1, \pm2, ..., \pm N$, and avoids the pole at $z=0$ (we'll handle that separately).

Compute residues: #

• At $z = n$ (non-zero integer): Res(f, n) = \lim_{z \to n} (z - n) * \frac{π cot(πz)}{z²} = π*\frac{1}{π}/n² = \frac{1}{n²}
• At $z = 0$: We need the Laurent expansion of $\cot(πz)$ around $z=0$:
  cot(πz) = \frac{1}{πz} - \frac{πz}{3} - \frac{π³z³}{45} - ...
  Multiply by $π$ and divide by $z²$:
  f(z) = \frac{π cot(πz)}{z²} = \frac{1}{z³} - \frac{π²}{3z} - \frac{π⁴z}{45} - ...
  The residue at $z=0$ is the coefficient of $\frac{1}{z}$, which is -\frac{π²}{3}.

Apply residue theorem: #

As $N \to \infty$, the integral of $f(z)$ around the square contour tends to 0 (since $|\cot(πz)|$ is bounded on the contour when $N$ is large). By residue theorem:
0 = \sum_{n=-N, n≠0}^N Res(f, n) + Res(f, 0)
Substitute the residues:
0 = 2\sum_{n=1}^N \frac{1}{n²} - \frac{π²}{3}
Taking $N \to \infty$, we get:
2ζ(2) = \frac{π²}{3} ⇒ ζ(2) = \frac{π²}{6}

Step 3: Connect back to the original integral #

Since we've shown via complex analysis that $\sum_{n=0}^\infty \frac{1}{(n+1)^2} = \frac{π²}{6}$, and this sum equals our original integral $I$, we conclude:
∫₀¹ (logx)/(x-1) dx = \frac{π²}{6}

Alternative direct contour integral approach #

If you prefer to compute the integral directly without linking it to the zeta function, you can use a keyhole contour with the function f(z) = \frac{(log z)²}{z² - 1} (handling the logarithmic branch cut along the positive real axis). By evaluating the contour integral, subtracting contributions from large/small circles (which vanish), and simplifying the integrals along the branch cut, you'll also arrive at the result $\frac{π²}{6}$ without invoking ζ(2)'s known value.

内容的提问来源于stack exchange，提问作者Alex