Prove that Gamma(p+1) = integral from 0 to () infty x^p

Prove that $$ \Gamma(p+1) = \int_0^\infty x^p e^{-x} \, dx = p! $$ where $p$ is a natural number.

We will prove this by establishing the recurrence relation for the Gamma function and then using mathematical induction.

Step 1: Prove the recurrence relation $\Gamma(p+1) = p\Gamma(p)$

We use integration by parts on:

$$\Gamma(p+1) = \int_0^\infty x^p e^{-x} \, dx$$

Let:

• $u = x^p$, so $du = px^{p-1} \, dx$
• $dv = e^{-x} \, dx$, so $v = -e^{-x}$

By integration by parts:

$$\int_0^\infty x^p e^{-x} \, dx = \left[x^p \cdot (-e^{-x})\right]_0^\infty - \int_0^\infty (-e^{-x}) \cdot px^{p-1} \, dx$$

$$= \left[-x^p e^{-x}\right]_0^\infty + p\int_0^\infty x^{p-1} e^{-x} \, dx$$

Evaluate the boundary term:

At $x = \infty$:

$$\lim_{x \to \infty} x^p e^{-x} = 0$$

This is because exponential decay dominates polynomial growth. More rigorously, by L'Hôpital's rule applied $p$ times:

$$\lim_{x \to \infty} \frac{x^p}{e^x} = \lim_{x \to \infty} \frac{p!}{e^x} = 0$$

At $x = 0$:

$$\lim_{x \to 0^+} x^p e^{-x} = 0 \cdot 1 = 0 \quad \text{(for } p \geq 1\text{)}$$

Therefore:

$$\left[-x^p e^{-x}\right]_0^\infty = 0 - 0 = 0$$

Conclusion:

$$\Gamma(p+1) = 0 + p\int_0^\infty x^{p-1} e^{-x} \, dx = p\Gamma(p)$$

Recurrence Relation

For $p > 0$:

$$\Gamma(p+1) = p\Gamma(p)$$

Step 2: Compute $\Gamma(1)$

$$\Gamma(1) = \int_0^\infty x^0 e^{-x} \, dx = \int_0^\infty e^{-x} \, dx$$

$$= \left[-e^{-x}\right]_0^\infty = 0 - (-1) = 1$$

Therefore:

$$\boxed{\Gamma(1) = 1 = 0!}$$

Step 3: Prove $\Gamma(p+1) = p!$ by induction

Gamma Function equals Factorial

For all natural numbers $p \geq 0$:

$$\Gamma(p+1) = p!$$

Proof by mathematical induction:

Base case: $p = 0$

$$\Gamma(0+1) = \Gamma(1) = 1 = 0!$$

The base case holds. ✓

Inductive hypothesis:

Assume that for some $p \geq 0$:

$$\Gamma(p+1) = p!$$

Inductive step:

We need to prove:

$$\Gamma((p+1)+1) = (p+1)!$$

Using the recurrence relation with $p+1$ in place of $p$:

$$\Gamma(p+2) = (p+1) \cdot \Gamma(p+1)$$

By the inductive hypothesis, $\Gamma(p+1) = p!$, so:

$$\Gamma(p+2) = (p+1) \cdot p! = (p+1)!$$

This completes the inductive step. ✓

Conclusion:

By the principle of mathematical induction, for all natural numbers $p \geq 0$:

$$\boxed{\Gamma(p+1) = p!}$$

Step 4: Verification with examples

Example 1: $p = 0$

$$\Gamma(1) = \int_0^\infty e^{-x} \, dx = 1 = 0!$$

Example 2: $p = 1$

$$\Gamma(2) = 1 \cdot \Gamma(1) = 1 \cdot 1 = 1 = 1!$$

Example 3: $p = 2$

$$\Gamma(3) = 2 \cdot \Gamma(2) = 2 \cdot 1 = 2 = 2!$$

Example 4: $p = 3$

$$\Gamma(4) = 3 \cdot \Gamma(3) = 3 \cdot 2 = 6 = 3!$$

Example 5: $p = 4$

$$\Gamma(5) = 4 \cdot \Gamma(4) = 4 \cdot 6 = 24 = 4!$$

All examples confirm our result! ✓

Complete table

$$\begin{array}{c|c|c} p & p! & \Gamma(p+1) \\ \hline 0 & 1 & 1 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 6 & 6 \\ 4 & 24 & 24 \\ 5 & 120 & 120 \end{array}$$

Summary

1. Recurrence relation: Using integration by parts, we proved $\Gamma(p+1) = p\Gamma(p)$
2. Base case: Direct integration gives $\Gamma(1) = 1 = 0!$
3. Induction: We proved $\Gamma(p+1) = p!$ for all natural numbers $p \geq 0$
4. Verification: Multiple examples confirm the result

Key insights:

• The Gamma function extends factorial to all positive real numbers
• For natural numbers: $\Gamma(n+1) = n!$ exactly
• The recurrence $\Gamma(p+1) = p\Gamma(p)$ mirrors $n! = n \cdot (n-1)!$
• Integration by parts is the key technique
• Exponential decay ensures boundary terms vanish