Evaluate Integral of 1/(x(x+1)(x+2)...(x+m))

Question

Evaluate the following integral:

$$\int\frac{\text{d}x}{x(x+1)(x+2)\cdots(x+m)}$$

where $m\in\mathbb{N}$.

$$\int\frac{\text{0.5}}{x} = 0.5ln(x)$$ + $$\int\frac{\text{-1}}{x+1}= -ln(x+1) $$ + $$\int\frac{\text{0.5}}{x+2}= 0.5ln(x+2)$$

Answer 1

The integral is easily writable as

$$\int \frac{\text{d}x}{\prod_{k = 0}^m (x+k)}$$

The productory can be evaluated, and the result is called a Pochammer symbol:

$$\prod_{k = 0}^m (x+k) = (x+1)_m$$

Which can be expressed in terms of the Gamma function:

$$(x+1)_m = \frac{\Gamma (m+x+1)}{\Gamma (x+1)}$$

In your case then, the integral becomes

$$\int \frac{\Gamma(x+1)}{\Gamma(m+x+1)}\ \text{d}x$$

Unless you have to evaluate a definite integral, I cannot help you more than this.