Problem 108
Hard Special Techniques

Show that the Gamma function Gamma(p) = integral from 0

Show that the Gamma function $$ \Gamma(p) = \int_0^\infty x^{p-1} e^{-x} \, dx $$ converges for $p > 0$.

Explanation of Method

This problem investigates whether an improper integral converges, without evaluating it exactly. The domain is split at each point of impropriety, and near each one the integrand is compared to a simpler benchmark integral (typically a power $x^{-p}$ or an exponential) using a comparison test or limit comparison test; the integral converges precisely when every piece does.

The Gamma function is improper at both $x = 0$ and $x = \infty$. We must analyze convergence at both points.

Step 1: Split the integral

We split at $x = 1$ to separate the potential singularity at 0 from the behavior at infinity:

$$\Gamma(p) = \int_0^1 x^{p-1} e^{-x} \, dx + \int_1^\infty x^{p-1} e^{-x} \, dx$$

For convergence, BOTH integrals must converge.

Step 2: Analyze convergence near $x = 0$

Consider $\int_0^1 x^{p-1} e^{-x} \, dx$.

As $x \to 0^+$:

Therefore, near $x = 0$:

$$x^{p-1} e^{-x} \approx x^{p-1} \cdot 1 = x^{p-1}$$

Convergence criterion at $x = 0$:

The integral $\int_0^1 x^{p-1} \, dx$ is of the form $\int_0^a x^\alpha \, dx$.

We know this converges if and only if $\alpha > -1$.

In our case, $\alpha = p - 1$, so we need:

$$p - 1 > -1 \implies p > 0$$

Rigorous comparison:

For $x \in (0, 1]$, we have $e^{-x} \leq 1$, so:

$$0 \leq x^{p-1} e^{-x} \leq x^{p-1}$$

Since $\int_0^1 x^{p-1} \, dx$ converges for $p > 0$, by the comparison test:

$$\int_0^1 x^{p-1} e^{-x} \, dx \quad \text{converges for } p > 0$$

Therefore: The integral converges at $x = 0$ if and only if **$p > 0$**. ✓

Step 3: Analyze convergence at $x = \infty$

Consider $\int_1^\infty x^{p-1} e^{-x} \, dx$.

As $x \to \infty$:

The key question: Which dominates?

Fundamental fact: Exponential decay dominates polynomial growth!

For any $n > 0$, we have:

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

More precisely, for large $x$:

$$x^{p-1} e^{-x} < e^{-x/2}$$

for $x$ sufficiently large (independent of $p$).

Proof of exponential dominance:

For any $p$, choose $N$ large enough so that for $x \geq N$:

$$x^{p-1} < e^{x/2}$$

This is true for large enough $x$ since exponential growth beats polynomial growth.

Then for $x \geq N$:

$$x^{p-1} e^{-x} < e^{x/2} e^{-x} = e^{-x/2}$$

Since $\int_N^\infty e^{-x/2} \, dx = 2e^{-N/2}$ converges, by comparison:

$$\int_N^\infty x^{p-1} e^{-x} \, dx \quad \text{converges}$$

And since $\int_1^N x^{p-1} e^{-x} \, dx$ is a proper integral (finite), we conclude:

$$\int_1^\infty x^{p-1} e^{-x} \, dx \quad \text{converges for all } p$$

Therefore: The integral converges at $x = \infty$ **for all $p$** (no restriction needed). ✓

Step 4: Conclusion

Combining both conditions:

Convergence of Gamma Function

The Gamma function

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

converges if and only if $p > 0$.

$$\boxed{\text{For } p > 0, \text{ the Gamma function converges}}$$

Alternative proof for convergence at infinity

We can be more explicit. For $x \geq 1$:

$$x^{p-1} e^{-x} = \frac{x^{p-1}}{e^x}$$

By L'Hôpital's rule applied repeatedly (or by Taylor series), for any $n$:

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

Therefore, for large $x$:

$$\frac{x^{p-1}}{e^x} < \frac{1}{x^2}$$

Since $\int_1^\infty \frac{1}{x^2} \, dx = 1$ converges, our integral converges at infinity.

Special values and properties

Property 1: $\Gamma(1) = 1$

$$\Gamma(1) = \int_0^\infty e^{-x} \, dx = [-e^{-x}]_0^\infty = 0 - (-1) = 1$$

Property 2: Recurrence relation

For $p > 0$:

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

Proof by integration by parts:

Let $u = x^p$ and $dv = e^{-x} \, dx$.

Then $du = px^{p-1} \, dx$ and $v = -e^{-x}$.

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

The boundary term vanishes:

Therefore:

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

Property 3: For positive integers

$$\Gamma(n) = (n-1)! \quad \text{for } n \in \mathbb{N}$$

Proof by induction:

Base case: $\Gamma(1) = 1 = 0!$ ✓

Inductive step: If $\Gamma(n) = (n-1)!$, then:

$$\Gamma(n+1) = n\Gamma(n) = n \cdot (n-1)! = n!$$

Property 4: $\Gamma(1/2) = \sqrt{\pi}$

This can be proved using the substitution $x = t^2$ and the Gaussian integral.

Numerical examples

Example 1: $p = 1$

$$\Gamma(1) = 1$$

Example 2: $p = 2$

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

Example 3: $p = 3$

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

Example 4: $p = 1/2$

$$\Gamma(1/2) = \sqrt{\pi} \approx 1.772$$

Example 5: $p = 0.1$

$$\Gamma(0.1) \approx 9.513$$

(Note: $\Gamma(p)$ is very large for small positive $p$)

Why $p > 0$ is necessary

For $p \leq 0$:

Near $x = 0$, the integrand $x^{p-1} e^{-x} \approx x^{p-1}$.

If $p \leq 0$, then $p - 1 \leq -1$, and:

$$\int_0^1 x^{p-1} \, dx \quad \text{diverges}$$

Therefore, $\Gamma(p)$ does not converge for $p \leq 0$.

Applications

The Gamma function appears in:

Summary

  1. The Gamma function has potential issues at $x = 0$ and $x = \infty$
  2. Near $x = 0$: integrand $\sim x^{p-1}$, converges if $p > 0$
  3. At $x = \infty$: exponential decay $e^{-x}$ dominates any polynomial growth
  4. Combined: convergence requires only $p > 0$
  5. For positive integers: $\Gamma(n) = (n-1)!$
  6. Recurrence relation: $\Gamma(p+1) = p\Gamma(p)$

Key lessons:


Result

$$\boxed{\text{For } p > 0, \text{ the Gamma function converges}}$$