Prove that: integral from 0 to infinity x^(2n+1) e^(-x²)

Prove that: $$ \int_0^{\infty} x^{2n+1} e^{-x^2} \, dx = \frac{n!}{2} $$ where $n$ is a natural number.

We will prove this by using substitution to reduce the integral to the Gamma function.

Step 1: Apply substitution $u = x^2$

Let $u = x^2$, so:

• $du = 2x \, dx$, which gives $x \, dx = \frac{1}{2} du$
• $x^{2n+1} = x^{2n} \cdot x = (x^2)^n \cdot x = u^n \cdot x$
• When $x = 0$: $u = 0$
• When $x = \infty$: $u = \infty$

Substituting into the integral:

$$\int_0^{\infty} x^{2n+1} e^{-x^2} \, dx = \int_0^{\infty} x^{2n} \cdot x \cdot e^{-x^2} \, dx$$

$$= \int_0^{\infty} u^n \cdot x \cdot e^{-u} \, dx$$

Since $x \, dx = \frac{1}{2} du$, we have:

$$= \int_0^{\infty} u^n e^{-u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int_0^{\infty} u^n e^{-u} \, du$$

Step 2: Recognize the Gamma function

By definition of the Gamma function:

$$\Gamma(n+1) = \int_0^{\infty} u^n e^{-u} \, du$$

And from Problem 109, we know that for natural numbers $n \geq 0$:

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

Step 3: Complete the proof

Therefore:

$$\int_0^{\infty} x^{2n+1} e^{-x^2} \, dx = \frac{1}{2} \int_0^{\infty} u^n e^{-u} \, du = \frac{1}{2} \Gamma(n+1) = \frac{n!}{2}$$

Odd Power Gaussian Integral

For any natural number $n \geq 0$:

$$\int_0^{\infty} x^{2n+1} e^{-x^2} \, dx = \frac{n!}{2}$$

$$\boxed{\int_0^{\infty} x^{2n+1} e^{-x^2} \, dx = \frac{n!}{2}}$$

Step 4: Verification with examples

Example 1: $n = 0$

$$\int_0^{\infty} x^1 e^{-x^2} \, dx = \int_0^{\infty} x e^{-x^2} \, dx = \frac{0!}{2} = \frac{1}{2}$$

Direct verification: Let $u = x^2$, $du = 2x \, dx$:

$$\int_0^{\infty} x e^{-x^2} \, dx = \frac{1}{2}\int_0^{\infty} e^{-u} \, du = \frac{1}{2}[-e^{-u}]_0^{\infty} = \frac{1}{2}(0-(-1)) = \frac{1}{2}$$

Confirmed! ✓

Example 2: $n = 1$

$$\int_0^{\infty} x^3 e^{-x^2} \, dx = \frac{1!}{2} = \frac{1}{2}$$

Using our result:

$$\frac{1}{2}\int_0^{\infty} u e^{-u} \, du = \frac{1}{2}\Gamma(2) = \frac{1}{2} \cdot 1 = \frac{1}{2}$$

Confirmed! ✓

Example 3: $n = 2$

$$\int_0^{\infty} x^5 e^{-x^2} \, dx = \frac{2!}{2} = \frac{2}{2} = 1$$

Using our result:

$$\frac{1}{2}\int_0^{\infty} u^2 e^{-u} \, du = \frac{1}{2}\Gamma(3) = \frac{1}{2} \cdot 2 = 1$$

Confirmed! ✓

Example 4: $n = 3$

$$\int_0^{\infty} x^7 e^{-x^2} \, dx = \frac{3!}{2} = \frac{6}{2} = 3$$

Example 5: $n = 4$

$$\int_0^{\infty} x^9 e^{-x^2} \, dx = \frac{4!}{2} = \frac{24}{2} = 12$$

Complete table

$$\begin{array}{c|c|c} n & 2n+1 & \int_0^{\infty} x^{2n+1} e^{-x^2} \, dx \\ \hline 0 & 1 & \frac{1}{2} \\ 1 & 3 & \frac{1}{2} \\ 2 & 5 & 1 \\ 3 & 7 & 3 \\ 4 & 9 & 12 \end{array}$$

Summary

1. Substitution: Using $u = x^2$ transforms the integral to $\frac{1}{2}\int_0^{\infty} u^n e^{-u} \, du$
2. Gamma function: Recognizing $\int_0^{\infty} u^n e^{-u} \, du = \Gamma(n+1) = n!$
3. Conclusion: Therefore $\int_0^{\infty} x^{2n+1} e^{-x^2} \, dx = \frac{n!}{2}$
4. Verification: Multiple examples confirm the result

Key insights:

• The substitution $u = x^2$ converts Gaussian integrals to exponential integrals
• Only odd powers $x^{2n+1}$ give simple factorial results
• The factor $\frac{1}{2}$ comes from the Jacobian of the substitution
• This connects Gaussian integrals to the Gamma function
• These integrals appear frequently in probability (moments of normal distribution)