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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:

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

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 \(x\)

Consider the inner integral:

\[
\int_0^{\infty} e^{-h x^2} \, dx
\]

This is a Gaussian integral of the form:

\[
\int_0^{\infty} e^{-a x^2} \, dx = \frac{1}{2} \sqrt{\frac{\pi}{a}}
\]

In our case, \( a = h \), so:

\[
\int_0^{\infty} e^{-h x^2} \, dx = \frac{1}{2} \sqrt{\frac{\pi}{h}}
\]

### Step 2: Substitute Result into the Outer Integral

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

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

### Step 3: Evaluate the Outer Integral with respect to \(z\)

Now, consider the remaining integral:

\[
\int_0^{\infty} 3^{-4z^2} \, dz
\]

Rewrite \( 3^{-4z^2} \) as \( e^{-4z^2 \ln 3} \). Then the integral becomes:

\[
\int_0^{\infty} e^{-4z^2 \ln 3} \, dz
\]

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

\[
\int_0^{\infty} e^{-a z^2} \, dz = \frac{1}{2} \sqrt{\frac{\pi}{a}}
\]

Here, \( a = 4 \ln 3 \), so:

\[
\int_0^{\infty} e^{-4z^2 \ln 3} \, dz = \frac{1}{2} \sqrt{\frac{\pi}{4 \ln 3}} = \frac{1}{2} \cdot \frac{\sqrt{\pi}}{2 \sqrt{\ln 3}} = \frac{\sqrt{\pi}}{4 \sqrt{\ln 3}}
\]

### Step 4: Combine Results

Now we combine the results from both integrals:

\[
\frac{1}{2} \sqrt{\frac{\pi}{h}} \cdot \frac{\sqrt{\pi}}{4 \sqrt{\ln 3}} = \frac{\pi}{8 \sqrt{h \ln 3}}
\]

### Final Answer

Thus, the evaluated expression is:

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

Would you like a deeper explanation of any step?

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### 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.

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

Step-by-step solution for evaluating the integral \( \int_0^{\infty} 3^{-4z^2} \, dz \int_0^{\infty} e^{-h x^2} \, dx \) using Beta and Gamma functions. Explore Gaussian integrals and special functions with a detailed breakdown suitable for undergraduate-level calculus.

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