含参数a的0到1积分求解及双对数函数求值问询
Hey there! Let's work through this integral step by step, and I'll break down exactly how to handle the dilogarithm part so you feel confident evaluating it.
First, let's simplify the integral with a substitution you already considered: let $t = \frac{y}{a}$. That means $y = at$, $dy = a dt$, and our limits shift from $y=0$ to $t=0$, and $y=1$ to $t=\frac{1}{a}$. Plugging this in, the integral simplifies nicely:
$$\int_{0}^{1} \frac{\ln{|1-\frac{y}{a}|}}{y} dy = \int_{0}^{1/a} \frac{\ln{|1 - t|}}{at} \cdot a dt = \int_{0}^{1/a} \frac{\ln{|1 - t|}}{t} dt$$
Now, this integral is directly tied to the definition of the dilogarithm function ($\text{Li}_2(z)$). Remember, the standard definition is:
$$\text{Li}2(z) = -\int{0}^{z} \frac{\ln(1 - t)}{t} dt$$
But we have an absolute value, so we need to split this into cases based on the value of $a$ (since that determines whether $\frac{1}{a}$ is less than or greater than 1, which affects the sign inside the log).
Case 1: $a > 1$
When $a > 1$, $\frac{1}{a} < 1$, so $1 - t > 0$ across the entire interval $[0, \frac{1}{a}]$. We can drop the absolute value, and using the dilogarithm definition directly:
$$\int_{0}^{1/a} \frac{\ln(1 - t)}{t} dt = -\text{Li}_2\left(\frac{1}{a}\right)$$
So your original integral evaluates to $\boldsymbol{-\text{Li}_2\left(\frac{1}{a}\right)}$.
Case 2: $0 < a < 1$
Here, $\frac{1}{a} > 1$, so we need to split the integral at $t=1$ to handle the absolute value (since for $t > 1$, $|1 - t| = t - 1$):
$$\int_{0}^{1/a} \frac{\ln{|1 - t|}}{t} dt = \int_{0}^{1} \frac{\ln(1 - t)}{t} dt + \int_{1}^{1/a} \frac{\ln(t - 1)}{t} dt$$
First, the first integral is a known special value: $\int_{0}^{1} \frac{\ln(1 - t)}{t} dt = -\text{Li}_2(1) = -\frac{\pi^2}{6}$ (since $\text{Li}2(1) = \sum{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$, a classic result from p-series).
For the second integral, use substitution $u = \frac{1}{t}$ (so $t = \frac{1}{u}$, $dt = -\frac{1}{u^2} du$). The limits shift from $t=1$ to $u=1$, and $t=\frac{1}{a}$ to $u=a$. Plugging in:
$$\int_{1}^{1/a} \frac{\ln(t - 1)}{t} dt = \int_{1}^{a} \frac{\ln\left(\frac{1}{u} - 1\right)}{\frac{1}{u}} \cdot \left(-\frac{1}{u^2}\right) du = \int_{a}^{1} \frac{\ln\left(\frac{1 - u}{u}\right)}{u} du$$
Split the log term:
$$\int_{a}^{1} \frac{\ln(1 - u) - \ln u}{u} du = \int_{a}^{1} \frac{\ln(1 - u)}{u} du - \int_{a}^{1} \frac{\ln u}{u} du$$
- The first part here is $\text{Li}2(a) - \text{Li}2(1)$ (using the dilogarithm definition: $\int{a}^{1} \frac{\ln(1-u)}{u} du = \int{0}^{1} \frac{\ln(1-u)}{u} du - \int_{0}^{a} \frac{\ln(1-u)}{u} du = -\text{Li}_2(1) + \text{Li}_2(a)$).
- The second integral is straightforward: let $v = \ln u$, $dv = \frac{1}{u} du$, so $\int \frac{\ln u}{u} du = \frac{1}{2}(\ln u)^2 + C$. Evaluating from $a$ to 1 gives $0 - \frac{1}{2}(\ln a)^2 = -\frac{1}{2}(\ln a)^2$.
Putting all these pieces together for $0 < a < 1$:
$$-\frac{\pi^2}{6} + \left(\text{Li}_2(a) - \frac{\pi^2}{6}\right) + \frac{1}{2}(\ln a)^2 = \boldsymbol{\text{Li}_2(a) - \frac{\pi^2}{3} + \frac{1}{2}(\ln a)^2}$$
Case 3: $a < 0$
Let $a = -b$ where $b > 0$. Then $|1 - \frac{y}{a}| = |1 + \frac{y}{b}| = 1 + \frac{y}{b}$ (since $y \geq 0$, $b > 0$). Using substitution $t = \frac{y}{b}$, the integral becomes:
$$\int_{0}^{1} \frac{\ln\left(1 + \frac{y}{b}\right)}{y} dy = \int_{0}^{1/b} \frac{\ln(1 + t)}{t} dt$$
This ties to the dilogarithm's negative argument property: $\int_{0}^{z} \frac{\ln(1 + t)}{t} dt = -\text{Li}_2(-z)$. Here $z = \frac{1}{b} = -\frac{1}{a}$, so the integral evaluates to $\boldsymbol{-\text{Li}_2\left(\frac{1}{a}\right)}$ (same form as the $a > 1$ case).
How to Evaluate the Dilogarithm
You don't need to be intimidated by $\text{Li}_2(z)$—here are practical ways to compute it:
- Special Values: Memorize or look up common results like:
- $\text{Li}_2(1) = \frac{\pi^2}{6}$
- $\text{Li}_2(-1) = -\frac{\pi^2}{12}$
- $\text{Li}_2\left(\frac{1}{2}\right) = \frac{\pi^2}{12} - \frac{(\ln 2)^2}{2}$
- Taylor Series: For $|z| \leq 1$, use the expansion $\text{Li}2(z) = \sum{n=1}^\infty \frac{zn}{n2}$. This converges quickly for $z$ near 0.
- Euler's Reflection Formula: For $z > 1$, use $\text{Li}_2(z) = -\text{Li}_2\left(\frac{1}{z}\right) - \frac{\pi^2}{6} - \frac{1}{2}(\ln z)^2$ to convert it to a value with $|1/z| < 1$, which you can compute with the Taylor series.
备注：内容来源于stack exchange，提问作者Kraken
