To solve these integrals using the Gamma function, we will use the following definitions and transformations:

1. Gamma Function: $\Gamma(s) = \int_0^{\infty} t^{s-1} e^{-t} \, dt$
2. Substitution: To transform the integrals into a form compatible with the Gamma function, we can use substitutions and properties of exponential expressions.

Let's go through each integral.

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1. Solving $\int_0^{\infty} 3^{-4z^2} \, dz$

Step 1: Rewrite the Integral

Notice that $3^{-4z^2} = e^{-(4 \ln 3) z^2}$. Therefore: $\int_0^{\infty} 3^{-4z^2} \, dz = \int_0^{\infty} e^{-(4 \ln 3) z^2} \, dz$

Step 2: Use a Substitution

To match the form of the Gamma function, let $u = (4 \ln 3) z^2$, so $z^2 = \frac{u}{4 \ln 3}$. Then, $dz = \frac{1}{2} \frac{1}{\sqrt{4 \ln 3 \cdot u}} \, du$, which simplifies to $dz = \frac{1}{2 \sqrt{4 \ln 3}} \frac{du}{\sqrt{u}}$.

This substitution gives: $\int_0^{\infty} e^{-(4 \ln 3) z^2} \, dz = \int_0^{\infty} e^{-u} \cdot \frac{1}{2 \sqrt{4 \ln 3}} \cdot \frac{1}{\sqrt{u}} \, du$

Step 3: Simplify

This expression becomes: $\frac{1}{2 \sqrt{4 \ln 3}} \int_0^{\infty} u^{-\frac{1}{2}} e^{-u} \, du$

Step 4: Recognize the Gamma Function

Notice that this integral now matches the form of the Gamma function $\Gamma(s) = \int_0^{\infty} u^{s-1} e^{-u} \, du$ with $s = \frac{1}{2}$: $= \frac{1}{2 \sqrt{4 \ln 3}} \Gamma\left(\frac{1}{2}\right)$

Step 5: Use the Known Value of $\Gamma\left(\frac{1}{2}\right)$

It is known that $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$. So: $\int_0^{\infty} 3^{-4z^2} \, dz = \frac{\sqrt{\pi}}{2 \sqrt{4 \ln 3}} = \frac{\sqrt{\pi}}{2 \cdot 2 \sqrt{\ln 3}} = \frac{\sqrt{\pi}}{4 \sqrt{\ln 3}}$

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2. Solving $\int_0^{\infty} e^{-h^2 x^2} \, dx$

Step 1: Use a Substitution

To simplify this integral, let $u = h^2 x^2$, so $x^2 = \frac{u}{h^2}$. Then, $dx = \frac{1}{2h \sqrt{u}} \, du$.

This gives: $\int_0^{\infty} e^{-h^2 x^2} \, dx = \int_0^{\infty} e^{-u} \cdot \frac{1}{2h \sqrt{u}} \, du$

Step 2: Simplify the Integral

This expression becomes: $\frac{1}{2h} \int_0^{\infty} u^{-\frac{1}{2}} e^{-u} \, du$

Step 3: Recognize the Gamma Function

Now the integral matches the form of the Gamma function with $s = \frac{1}{2}$: $= \frac{1}{2h} \Gamma\left(\frac{1}{2}\right)$

Step 4: Use the Known Value of $\Gamma\left(\frac{1}{2}\right)$

Since $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$, we have: $\int_0^{\infty} e^{-h^2 x^2} \, dx = \frac{\sqrt{\pi}}{2h}$

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Final Answers

1. $\int_0^{\infty} 3^{-4z^2} \, dz = \frac{\sqrt{\pi}}{4 \sqrt{\ln 3}}$
2. $\int_0^{\infty} e^{-h^2 x^2} \, dx = \frac{\sqrt{\pi}}{2h}$

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Would you like additional details on any of the steps?

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Related Questions

1. How can the Beta function be used to solve integrals with different forms?
2. What are the properties of the Gamma function that make it useful in evaluating integrals?
3. How is the value of $\Gamma\left(\frac{1}{2}\right)$ derived?
4. Are there alternative methods to solve these integrals without using the Gamma function?
5. What are some practical applications of the Gamma function in probability and statistics?

Tip:

The Gamma function $\Gamma(s)$ is closely related to factorials, as $\Gamma(n) = (n-1)!$ for a positive integer $n$, making it a powerful tool for handling continuous cases where factorials are used in discrete mathematics.