Math Problem Statement

give step by step solution

Using Beta and Gamma functions, prove the following: \( \int_0^{\infty} 3^{-4z^2} \, dz \int_0^{\infty} e^{-h x^2} \, dx \)

Solution

To solve this integral using the Beta and Gamma functions, we need to break it down and relate it to these special functions. Let’s go through it step by step.

The expression we are given is:

034z2dz0ehx2dx\int_0^{\infty} 3^{-4z^2} \, dz \int_0^{\infty} e^{-h x^2} \, dx

We’ll approach each part separately and see if we can simplify it using known forms of Beta and Gamma integrals.

Step 1: Evaluate the Inner Integral with respect to xx

Consider the inner integral:

0ehx2dx\int_0^{\infty} e^{-h x^2} \, dx

This is a Gaussian integral of the form:

0eax2dx=12πa\int_0^{\infty} e^{-a x^2} \, dx = \frac{1}{2} \sqrt{\frac{\pi}{a}}

In our case, a=ha = h, so:

0ehx2dx=12πh\int_0^{\infty} e^{-h x^2} \, dx = \frac{1}{2} \sqrt{\frac{\pi}{h}}

Step 2: Substitute Result into the Outer Integral

Now that we have evaluated the inner integral, we can rewrite the expression as:

12πh034z2dz\frac{1}{2} \sqrt{\frac{\pi}{h}} \int_0^{\infty} 3^{-4z^2} \, dz

Step 3: Evaluate the Outer Integral with respect to zz

Now, consider the remaining integral:

034z2dz\int_0^{\infty} 3^{-4z^2} \, dz

Rewrite 34z23^{-4z^2} as e4z2ln3e^{-4z^2 \ln 3}. Then the integral becomes:

0e4z2ln3dz\int_0^{\infty} e^{-4z^2 \ln 3} \, dz

This is again a Gaussian integral, similar to the one we saw earlier, where now the coefficient of z2z^2 is 4ln34 \ln 3. So, we use the Gaussian integral formula:

0eaz2dz=12πa\int_0^{\infty} e^{-a z^2} \, dz = \frac{1}{2} \sqrt{\frac{\pi}{a}}

Here, a=4ln3a = 4 \ln 3, so:

0e4z2ln3dz=12π4ln3=12π2ln3=π4ln3\int_0^{\infty} e^{-4z^2 \ln 3} \, dz = \frac{1}{2} \sqrt{\frac{\pi}{4 \ln 3}} = \frac{1}{2} \cdot \frac{\sqrt{\pi}}{2 \sqrt{\ln 3}} = \frac{\sqrt{\pi}}{4 \sqrt{\ln 3}}

Step 4: Combine Results

Now we combine the results from both integrals:

12πhπ4ln3=π8hln3\frac{1}{2} \sqrt{\frac{\pi}{h}} \cdot \frac{\sqrt{\pi}}{4 \sqrt{\ln 3}} = \frac{\pi}{8 \sqrt{h \ln 3}}

Final Answer

Thus, the evaluated expression is:

034z2dz0ehx2dx=π8hln3\int_0^{\infty} 3^{-4z^2} \, dz \int_0^{\infty} e^{-h x^2} \, dx = \frac{\pi}{8 \sqrt{h \ln 3}}

Would you like a deeper explanation of any step?


Additional Questions:

  1. How does the Gamma function relate to integrals involving factorials?
  2. Can we use the Beta function to solve similar types of integrals?
  3. What other methods could we use to evaluate double integrals involving exponentials?
  4. How would the solution change if the limits of integration were different?
  5. How is the Gaussian integral formula derived?

Tip:

When encountering nested integrals, solve the inner integral first to simplify the expression before tackling the outer integral.

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Math Problem Analysis

Mathematical Concepts

Calculus
Special Functions
Integration
Beta and Gamma Functions
Gaussian Integral

Formulas

\( \int_0^{\infty} e^{-a x^2} \, dx = \frac{1}{2} \sqrt{\frac{\pi}{a}} \) (Gaussian integral formula)
Gamma function: \( \Gamma(n) = \int_0^{\infty} t^{n-1} e^{-t} \, dt \)

Theorems

Gamma Function Properties
Gaussian Integral Theorem

Suitable Grade Level

Undergraduate