Problem

Show that $xF(1, 1, 2; -x) = \ln(1+x)$.

Solution

Step 1: Expand the Hypergeometric Series

Here $\alpha=1$, $\beta=1$, $\gamma=2$. Replace $x$ with $-x$.

$F(1, 1, 2; -x) = 1 + \frac{1 \cdot 1}{1 \cdot 2}(-x) + \frac{1(2) \cdot 1(2)}{1 \cdot 2 \cdot 2(3)}(-x)^2 + \cdots$

Step 2: Simplify terms

• Term 1: $1$
• Term 2: $\frac{1}{2}(-x) = -\frac{x}{2}$
• Term 3: $\frac{2 \cdot 2}{2 \cdot 6} x^2 = \frac{4}{12} x^2 = \frac{x^2}{3}$
• Term 4: $\frac{1(2)(3) \cdot 1(2)(3)}{6 \cdot 2(3)(4)} (-x)^3 = -\frac{x^3}{4}$

So:

$F(1, 1, 2; -x) = 1 - \frac{x}{2} + \frac{x^2}{3} - \frac{x^3}{4} + \cdots$

Step 3: Multiply by $x$

$xF(1, 1, 2; -x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$

Step 4: Identify the Maclaurin Series

This series is the standard expansion for $\ln(1+x)$.

Therefore:

$xF(1, 1, 2; -x) = \ln(1+x)$

Q.E.D.