What is the value of Γ(5)

To find the value of Γ(5), we can use the formula for the gamma function. The gamma function is defined as the integral of t^(x-1) * e^(-t) from 0 to infinity. In this case, we want to find Γ(5), which means we need to evaluate the integral of t^(5-1) * e^(-t) from 0 to infinity.

Step 1: Write down the integral: ∫[0,∞] t^(5-1) * e^(-t) dt

Step 2: Simplify the integrand: t^(5-1) * e^(-t) = t^4 * e^(-t)

Step 3: Evaluate the integral: ∫[0,∞] t^4 * e^(-t) dt

Step 4: Use integration techniques to solve the integral. One common method is integration by parts.

Integration by parts: Let u = t^4 and dv = e^(-t) dt Then du = 4t^3 dt and v = -e^(-t)

Using the integration by parts formula: ∫ u dv = uv - ∫ v du

Applying this formula to our integral: ∫[0,∞] t^4 * e^(-t) dt = -t^4 * e^(-t) - ∫[0,∞] (-e^(-t) * 4t^3) dt

Step 5: Simplify the integral: ∫[0,∞] t^4 * e^(-t) dt = -t^4 * e^(-t) + 4∫[0,∞] t^3 * e^(-t) dt

Step 6: Repeat the integration by parts process for the remaining integral: Let u = t^3 and dv = e^(-t) dt Then du = 3t^2 dt and v = -e^(-t)

Using the integration by parts formula again: ∫[0,∞] t^3 * e^(-t) dt = -t^3 * e^(-t) - ∫[0,∞] (-e^(-t) * 3t^2) dt

Step 7: Simplify the integral: ∫[0,∞] t^3 * e^(-t) dt = -t^3 * e^(-t) + 3∫[0,∞] t^2 * e^(-t) dt

Step 8: Repeat the integration by parts process for the remaining integral: Let u = t^2 and dv = e^(-t) dt Then du = 2t dt and v = -e^(-t)

Using the integration by parts formula again: ∫[0,∞] t^2 * e^(-t) dt = -t^2 * e^(-t) - ∫[0,∞] (-e^(-t) * 2t) dt

Step 9: Simplify the integral: ∫[0,∞] t^2 * e^(-t) dt = -t^2 * e^(-t) + 2∫[0,∞] t * e^(-t) dt

Step 10: Repeat the integration by parts process for the remaining integral: Let u = t and dv = e^(-t) dt Then du = dt and v = -e^(-t)

Using the integration by parts formula again: ∫[0,∞] t * e^(-t) dt = -t * e^(-t) - ∫[0,∞] (-e^(-t) * 1) dt

Step 11: Simplify the integral: ∫[0,∞] t * e^(-t) dt = -t * e^(-t) + ∫[0,∞] e^(-t) dt

Step 12: Evaluate the remaining integrals: ∫[0,∞] e^(-t) dt = -e^(-t) evaluated from 0 to ∞ Since e^(-∞) is equal to 0, the integral becomes: ∫[0,∞] e^(-t) dt = -e^(-0) - (-e^(-∞)) = -1 - 0 = -1

Step 13: Substitute the values back into the original integral: ∫[0,∞] t * e^(-t) dt = -t * e^(-t) + ∫[0,∞] e^(-t) dt = -t * e^(-t) - 1

Step 14: Substitute the values back into the previous integrals: ∫[0,∞] t^2 * e^(-t) dt = -t^2 * e^(-t) + 2(-t * e^(-t) - 1) = -t^2 * e^(-t) - 2t * e^(-t) - 2

∫[0,∞] t^3 * e^(-t) dt = -t^3 * e^(-t) + 3(-t^2 * e^(-t) - 2t * e^(-t) - 2) = -t^3 * e^(-t) - 3t^2 * e^(-t) - 6t * e^(-t) - 6

∫[0,∞] t^4 * e^(-t) dt = -t^4 * e^(-t) + 4(-t^3 * e^(-t) - 3t^2 * e^(-t) - 6t * e^(-t) - 6) = -t^4 * e^(-t) - 4t^3 * e^(-t) - 12t^2 * e^(-t) - 24t * e^(-t) - 24

Step 15: Substitute the values back into the original integral: ∫[0,∞] t^4 * e^(-t) dt = -t^4 * e^(-t) - 4t^3 * e^(-t) - 12t^2 * e^(-t) - 24t * e^(-t) - 24

Step 16: Evaluate the limits of the integral: ∫[0,∞] t^4 * e^(-t) dt = lim[0→∞] (-t^4 * e^(-t) - 4t^3 * e^(-t) - 12t^2 * e^(-t) - 24t * e^(-t) - 24) - (-0^4 * e^(-0) - 4(0^3) * e^(-0) - 12(0^2) * e^(-0) - 24(0) * e^(-0) - 24)

Step 17: Simplify the limits: lim[0→∞] (-t^4 * e^(-t) - 4t^3 * e^(-t) - 12t^2 * e^(-t) - 24t * e^(-t) - 24) - (-0^4 * e^(-0) - 4(0^3) * e^(-0) - 12(0^2) * e^(-0) - 24(0) * e^(-0) - 24) = lim[0→∞] (-t^4 * e^(-t) - 4t^3 * e^(-t) - 12t^2 * e^(-t) - 24t * e^(-t) - 24) - (-0 - 0 - 0 - 0 - 24) = lim[0→∞] (-t^4 * e^(-t) - 4t^3 * e^(-t) - 12t^2 * e^(-t) - 24t * e^(-t) + 24)

Step 18: Take the limit as t approaches infinity: lim[0→∞] (-t^4 * e^(-t) - 4t^3 * e^(-t) - 12t^2 * e^(-t) - 24t * e^(-t) + 24) = -∞ * e^(-∞) - 4(∞^3) * e^(-∞) - 12(∞^2) * e^(-∞) - 24(∞) * e^(-∞) + 24 Since e^(-∞) is equal to 0, the expression becomes: = -∞ * 0 - 4(∞^3) * 0 - 12(∞^2) * 0 - 24(∞) * 0 + 24 = 0 - 0 - 0 - 0 + 24 = 24

Therefore, the value of Γ(5) is 24.