📚 IB Mathematics: Using Parameter Differentiation to Evaluate Integrals | IB数学：利用对参数求导法计算积分
When a definite integral seems impossible to evaluate directly, one powerful strategy is to introduce an extra variable, called a parameter, into the integrand. By differentiating the resulting integral with respect to that parameter, we may obtain a much simpler expression. This method is often called differentiating under the integral sign, or Feynman's trick, after the physicist Richard Feynman.
当一个定积分看似无法直接计算时，一种强大的策略是在被积函数中引入一个额外的变量，称为参数。通过对所得积分关于该参数求导，我们可能得到一个简单得多的表达式。这种方法通常称为“积分号内求导”，或因物理学家理查德·费曼而得名的“费曼技巧”。
1. The Leibniz Integral Rule | 1. 莱布尼茨积分法则
The mathematical foundation of this technique is the Leibniz integral rule. For a function f(x,t) that is continuous and has a continuous partial derivative with respect to t, and for constant limits a and b, the derivative of the integral with respect to t is simply the integral of the partial derivative:
这一技巧的数学基础是莱布尼茨积分法则。若函数 f(x,t) 连续，且关于 t 的偏导数连续，上下限 a 与 b 为常数，则积分关于 t 的导数就等于偏导数的积分：
d/dt ∫ₐᵇ f(x,t) dx = ∫ₐᵇ (∂f/∂t) dx
If the limits themselves depend on t, we must add boundary terms. The general Leibniz rule is
如果积分上下限本身也依赖于 t，则必须加上边界项。一般形式的莱布尼茨法则为
d/dt ∫ₐ₍ₜ₎ᵇ₍ₜ₎ f(x,t) dx = f(b(t),t) b'(t) − f(a(t),t) a'(t) + ∫ₐ₍ₜ₎ᵇ₍ₜ₎ (∂f/∂t) dx
In most IB problems, we choose constant limits, so only the last term matters. The challenge is to choose the parameter so that the integrated derivative is easier to evaluate than the original integral.
在大多数IB题目中，我们选择常数为积分上下限，因此只需关注最后一项。关键挑战在于选择合适的参数，使得求导后的积分比原积分更容易计算。
2. Step-by-Step Strategy | 2. 分步策略
The method can be organised into a clear sequence of steps. Following these steps reduces the risk of mistakes.
该方法可以组织成清晰的步骤序列。按以下步骤操作可以降低犯错风险。
Define a parameterised integral: introduce a parameter a so that the desired integral becomes a special case of a family I(a).
定义含参数的积分：引入参数 a，使目标积分成为一族积分 I(a) 的特殊情形。
Differentiate with respect to the parameter: compute I'(a) by moving the derivative inside the integral, using the Leibniz rule.
对参数求导：利用莱布尼茨法则，将导数移入积分号内，计算 I'(a)。
Evaluate the new integral: this simpler integral should be computable by elementary methods.
计算新积分：这个更简单的积分应当可以通过基本方法求出。
Integrate back with respect to the parameter: solve for I(a) by indefinite integration, leaving an arbitrary constant.
对参数积分回去：对 I(a) 做不定积分，留下一个任意常数。
Determine the constant using a known value: usually I(0)=0, or by evaluating the integral at a convenient parameter value.
利用已知值确定常数：通常 I(0)=0，或选取一个便于计算的参数值求出积分。
Substitute the required value: evaluate I(a) at the parameter that reproduces the original problem.
代入所需参数值：计算能还原原问题的参数所对应的 I(a)。
3. Example 1: A Frullani-Type Integral | 3. 例一：Frullani型积分
Consider the integral
考虑积分
∫₀∞ (e⁻ᵃˣ − e⁻ˣ)/x dx = ? , a > 0
Define I(a) = ∫0∞ (e-ax - e-x)/xdx. Differentiating with respect to a gives
定义 I(a) = ∫0∞ (e-ax - e-x)/xdx。关于 a 求导得
I'(a) = ∫₀∞ (−x e⁻ᵃˣ)/x dx = −∫₀∞ e⁻ᵃˣ dx = −1/a
Integrating with respect to a, we get I(a) = -ln a + C. Since I(1)=0 (the two exponentials cancel), we have C=0. Therefore
对 a 积分得 I(a) = -ln a + C。因为 I(1)=0（两个指数函数相消），所以 C=0。因此
∫₀∞ (e⁻ᵃˣ − e⁻ˣ)/x dx = −ln a
For example, with a=2, the integral equals -ln 2. This example shows how differentiation removes a troublesome x in the denominator.
例如，当 a=2 时，积分等于 -ln 2。这个例子显示求导如何消去分母中麻烦的 x。
4. Example 2: A Logarithmic Integral | 4. 例二：含对数的积分
Another famous example is
另一个著名例子是
∫₀¹ (xᵃ − 1)/ln x dx = ln(a+1), a > −1
Let I(a)=∫01 (xa - 1)/ln xdx. Differentiating inside the integral, we use ∂ (xa)/∂ a = xa ln x. Hence
令 I(a)=∫01 (xa - 1)/ln xdx。在积分号内求导，利用 ∂ (xa)/∂ a = xa ln x。于是
