Gamma Integral

Gamma Integral

In this section, we study about a special improper integral of the form ^∞∫₀ e ^−x xⁿ⁻¹dx , where n is a positive integer. Here, we have

By L’Hoˆpital’s  rule, for every positive integer m , we get,

Example 9.43

Prove that ^∞∫₀ e^−x xⁿ dx = n!, where n is a positive integer.

Solution

Applying integration by parts, we get

Let In = ^∞∫₀ e^−x xⁿ dx .Then, I_n = nI_n−1 .

So, we get I_n = n ( n −1)I_n−2 .

Proceeding in this way, we get ultimately,

I_n = n ( n − 1)( n − 2) ( 2)(1)I₀ .

But, I₀ = ^∞∫₀ e ^−x x⁰dx = ( −e^−x)^∞₀ = 0 +1 = 1 . So, we get I_n = n (n − 1)(n − 2) (2)(1) = n!.

Hence, we get

Result

^∞∫₀ e ^−x xⁿ dx = n!, where n is a nonnegative integer.

Note

The integral ^∞∫₀ e^−x xⁿ⁻¹dx defines a unique positive integer for every positive integer n ≥ 1.

Definition 9.1

^∞∫₀ e ^−x xⁿ⁻¹dx is called the gamma integral. It is denoted by Γ(n) and is read as “gamma of n ”.

Note

Example 9.44

Evaluate ^∞∫₀ e ^−ax xⁿ dx ,where a > 0 .

Solution

Making the substitution t = ax , we get dt = adx and x = 0 ⇒ t = 0 and x = ∞ ⇒ t = ∞ .

Hence, we get