Question:

The value of $\displaystyle \int_{0}^{\infty} e^{-x^2} \, dx$ is

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Use the identity $\int_0^\infty e^{-x^2} dx = \dfrac{\sqrt{\pi}}{2}$ and link it with $\Gamma(1/2) = \sqrt{\pi}$.
Updated On: Jun 24, 2025
  • $\dfrac{1}{3}\Gamma\left(\dfrac{3}{2}\right)$
  • $\dfrac{1}{2}\Gamma\left(\dfrac{1}{2}\right)$
  • $\dfrac{1}{2}\Gamma\left(\dfrac{3}{2}\right)$
  • $\Gamma\left(\dfrac{1}{5}\right)$
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The Correct Option is B

Solution and Explanation

Step 1: Recognize standard Gaussian integral
\[ \int_0^\infty e^{-x^2} dx = \frac{\sqrt{\pi}}{2} \] Step 2: Recall Gamma function relation
\[ \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} \Rightarrow \frac{1}{2} \Gamma\left(\frac{1}{2}\right) = \frac{\sqrt{\pi}}{2} \] Hence, \[ \boxed{\int_0^\infty e^{-x^2} dx = \frac{1}{2} \Gamma\left(\frac{1}{2}\right)} \]
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