Differentiation under the integral sign

Differentiation under the integral sign

1. Idea — Leibniz with Fixed Limits (Preliminary)

Consider $F(α) = ∫_{a}^{b} f(x,α) dx$ where $a,b$ are constants (do not depend on $α$) and $f$ and $f_α = ∂f/∂α$ are continuous on $[a,b]×[α_1,α_2]$. Then you may differentiate through the integral:

i.e. differentiate the integrand partially with $x$ treated as constant, then integrate. This follows from uniform continuity and lets you swap $d/dα$ with $∫dx$ when conditions hold.

2. Why Useful

Converts a hard integral depending on a parameter into an easier differential equation, or evaluates $F(α)$ by integrating $dF/dα$.

Classic: $F(α)=∫_{0}^{π/2} log(1+α cos x) dx$ → $F'(α)=∫_{0}^{π/2} cos x/(1+α cos x) dx$ easier to handle at $α=0$.

3. Conditions (when you may swap)

• $f(x,α)$ and $∂f/∂α$ continuous in both variables on closed rectangle containing $(x,α)$ range, and $a,b$ finite constants. For infinite limits, need uniform convergence (covered later).

4. Worked Examples

(a) F(α)= ∫_{0}^{1} x^α dx = 1/(α+1) (α>-1): differentiate w.rt α?
    Direct: F'(α)= ∫_{0}^{1} x^α log x dx = -1/(α+1)^2 matches direct derivative -1/(α+1)^2 ✓
    Conversely, evaluate ∫_{0}^{1} x^2 log x dx = F'(2)= -1/9 ✓

(b) F(a)= ∫_{0}^{π} log(1+ a cos x) dx : F(0)=0, F'(a)= ∫_{0}^{π} cos x/(1+a cos x) dx
    = (π/a) - (1/a)∫_{0}^{π} dx/(1+a cos x) → after integration gives F(a)=π log((1+√(1-a^2))/2) for |a|<1 ✓

(c) Evaluate I = ∫_{0}^{∞} e^{-α x} sin x / x dx? Fixed? Not fixed limit? Needs general rule next topic.

5. Relation to Next Topic

This fixed-limit case is the special case of Leibniz rule where $a,b$ constant. Next topic adds variable limits $a(α), b(α)$.

6. Exam Points

• State $d/dα ∫_{a}^{b} f(x,α)dx = ∫_{a}^{b} f_α dx$ with conditions — 3 marks, frequent.
• Evaluate $∫_{0}^{1} x^α log x dx$ or $∫_{0}^{π} x sin x/(1+...)$ via this — 5 marks.
• Distinguish from general Leibniz — 3 marks.