To prove the following integral identity:
\[
\left(\int_0^\infty e^{-x^2} x^{\frac{1}{2}} \, dx \right) \left(\int_0^\infty x^2 e^{-x^4} \, dx \right) = \frac{\pi^{\frac{3}{2}}}{4},
\]
we will evaluate each integral separately and then multiply the results.

1. Evaluate the Integral \( I_1 = \int_0^\infty e^{-x^2} x^{\frac{1}{2}} \, dx \)

Let’s start by evaluating:
\[
I_1 = \int_0^\infty x^{\frac{1}{2}} e^{-x^2} \, dx
\]
Use the substitution \( u = x^2 \), hence \( du = 2x \, dx \) or \( dx = \frac{du}{2x} \). Also, \( x = u^{\frac{1}{2}} \), so \( dx = \frac{du}{2u^{\frac{1}{2}}} \).

Substitute \( x \) in the integral:
\[
I_1 = \frac{1}{2} \int_0^\infty e^{-u} u^{\frac{1}{4}} \cdot \frac{du}{2u^{\frac{1}{2}}} = \frac{1}{2} \int_0^\infty e^{-u} u^{-\frac{3}{4}} \, du
\]
This integral is a Gamma function \( \Gamma(z) \) with \( z = \frac{1}{4} \):
\[
I_1 = \frac{1}{2} \Gamma\left(\frac{1}{4}\right)
\]

\section*{2. Evaluate the Integral \( I_2 = \int_0^\infty x^2 e^{-x^4} \, dx \)}

Next, evaluate:
\[
I_2 = \int_0^\infty x^2 e^{-x^4} \, dx
\]
Use the substitution \( v = x^4 \), hence \( dv = 4x^3 \, dx \) or \( dx = \frac{dv}{4x^3} \). Also, \( x^3 = v^{\frac{3}{4}} \), so \( dx = \frac{dv}{4v^{\frac{3}{4}}} \).

Substitute \( x \) in the integral:
\[
I_2 = \frac{1}{4} \int_0^\infty v^{\frac{1}{4}} e^{-v} \cdot \frac{dv}{4v^{\frac{3}{4}}} = \frac{1}{4} \int_0^\infty v^{-\frac{1}{2}} e^{-v} \, dv
\]
This is also a Gamma function \( \Gamma(z) \) with \( z = \frac{1}{2} \):
\[
I_2 = \frac{1}{4} \Gamma\left(\frac{1}{2}\right)
\]
Using \( \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} \):
\[
I_2 = \frac{1}{4} \sqrt{\pi}
\]

\section*{3. Multiply the Results}

Multiply the two integrals:
\[
I_1 \cdot I_2 = \left(\frac{1}{2} \Gamma\left(\frac{1}{4}\right)\right) \left(\frac{1}{4} \sqrt{\pi}\right)
= \frac{1}{8} \Gamma\left(\frac{1}{4}\right) \sqrt{\pi}
\]
Using the known result:
\[
\Gamma\left(\frac{1}{4}\right) = 2^{\frac{3}{2}} \pi^{\frac{1}{4}},
\]
\[
\frac{1}{8} \Gamma\left(\frac{1}{4}\right) \sqrt{\pi} = \frac{1}{8} \cdot 2^{\frac{3}{2}} \pi^{\frac{1}{4}} \cdot \sqrt{\pi} = \frac{2 \pi^{\frac{3}{2}}}{8} = \frac{\pi^{\frac{3}{2}}}{4}
\]

Therefore:
\[
\left(\int_0^\infty e^{-x^2} x^{\frac{1}{2}} \, dx \right) \left(\int_0^\infty x^2 e^{-x^4} \, dx \right) = \frac{\pi^{\frac{3}{2}}}{4}
\]
This completes the proof.