Let's solve the **first part of Q15** step-by-step using Beta and Gamma functions. --- ## **Q15: Using Beta or Gamma functions, Prove that** ### **(i) \(\Gamma(1) = 1\)** #### **Step 1: Definition of Gamma Function** \[ \Gamma(n) = \int_0^\infty x^{n-1} e^{-x} dx \] #### **Step 2: Substitute \( n = 1 \)** \[ \Gamma(1) = \int_0^\infty x^{1-1} e^{-x} dx = \int_0^\infty x^0 e^{-x} dx = \int_0^\infty e^{-x} dx \] #### **Step 3: Integrate** \[ \int_0^\infty e^{-x} dx = \left[ -e^{-x} \right]_0^\infty = 0 - (-1) = 1 \] #### **Final Answer** \[ \boxed{\Gamma(1) = 1} \] --- ### **(ii) \(\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}\)** #### **Step 1: Substitute \( n = \frac{1}{2} \) in Gamma function** \[ \Gamma\left(\frac{1}{2}\right) = \int_0^\infty x^{-\frac{1}{2}} e^{-x} dx \] #### **Step 2: Use Substitution** Let \( x = t^2 \implies dx = 2t dt \) and as \( x \to 0, t \to 0 \); as \( x \to \infty, t \to \infty \): \[ x^{-\frac{1}{2}} = (t^2)^{-\frac{1}{2}} = t^{-1} \] Now substitute: \[ \int_0^\infty x^{-\frac{1}{2}} e^{-x} dx = \int_0^\infty t^{-1} e^{-t^2} \cdot 2t dt = 2\int_0^\infty e^{-t^2} dt \] #### **Step 3: Value of the Gaussian Integral** \[ \int_0^\infty e^{-t^2} dt = \frac{\sqrt{\pi}}{2} \] #### **Step 4: Substitute the Value** \[ \Gamma\left(\frac{1}{2}\right) = 2 \cdot \frac{\sqrt{\pi}}{2} = \sqrt{\pi} \] #### **Final Answer** \[ \boxed{\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}} \] --- ## **Summary:** 1. \(\Gamma(1) = 1\) 2. \(\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}\) --- If you want the solutions for other questions, let me know!