$\int_{0}^{\infty}\sqrt{x}e^{-x^{3}}dx=$

Question

Accepted Answer

To solve the integral $\int_{0}^{\infty}\sqrt{x}e^{-x^{3}}dx$, we need to evaluate it using an appropriate substitution and transformation technique.

### Explanation for Option 1:

1. **Substitution**: Let $ u = x^{3} $. Then, $ du = 3x^{2}dx $, which implies $ dx = \frac{du}{3x^{2}} $. Since $ x = u^{1/3} $, we have $ x^{2} = u^{2/3} $. Therefore, $ dx = \frac{du}{3u^{2/3}} $.

2. **Transform the Integral**: Substitute these into the integral:
   $$
   \int_{0}^{\infty} \sqrt{x} e^{-x^{3}} dx = \int_{0}^{\infty} (u^{1/6}) e^{-u} \frac{du}{3u^{2/3}}
   $$
   Simplifying, we get:
   $$
   \frac{1}{3} \int_{0}^{\infty} u^{1/6 - 2/3} e^{-u} du = \frac{1}{3} \int_{0}^{\infty} u^{-1/2} e^{-u} du
   $$

3. **Gamma Function**: The integral $\int_{0}^{\infty} u^{-1/2} e^{-u} du$ is a standard Gamma function $\Gamma(s)$ where $s = \frac{1}{2}$. The Gamma function $\Gamma(\frac{1}{2})$ is known to be $\sqrt{\pi}$.

4. **Final Calculation**:
   $$
   \frac{1}{3} \Gamma\left(\frac{1}{2}\right) = \frac{1}{3} \sqrt{\pi}
   $$

Thus, the value of the integral is $\frac{1}{3} \sqrt{\pi}$, which corresponds to Option 1.

### Explanation for Incorrect Options:

- **Option 2: $\frac{1}{2}\sqrt{\pi}$**: This option is incorrect because the factor derived from the substitution and transformation process is $\frac{1}{3}$, not $\frac{1}{2}$. The correct factor is crucial in determining the final result of the integral.

- **Option 3: $\sqrt{\pi}$**: This option is incorrect because it does not account for the factor of $\frac{1}{3}$ that arises from the substitution process. The integral evaluates to $\frac{1}{3} \sqrt{\pi}$, not $\sqrt{\pi}$.

- **Option 4: 0**: This option is incorrect because the integral of a positive function over a positive domain cannot be zero unless the function itself is zero, which is not the case here. The function $\sqrt{x} e^{-x^{3}}$ is positive for $x > 0$.

In conclusion, the correct answer is Option 1: $\frac{1}{3} \sqrt{\pi}$, as derived through the appropriate substitution and evaluation using the Gamma function.