Integration of x n e − x d xThe question is a definite integral ∫ 0 ∞ x n e x d xSo, I'm integrating it by parts, and going by the LAITE principle, I get: I n = − x n e x + ∫ 0 ∞ n x n − 1 e x d xThe value inside the integral is n I n − 1 , so all I'm left is to get the limit of the first term as x ranges from 0 to infinity. ∑ x n e x Somehow, I'm not exactly sure why I feel like the limit of the first term would go to zero (or n!), given that formula, but I'm not exactly sure how to substitute the n! formula for e x . How do I proceed from here?The options are n ! − n I n − 1 n ! + n I n − 1 n I n − 1 none of these

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Integration of       x    n        e          −      x        d  xThe question is a definite integral      ∫          0              ∞                  x      n              e      x          d  xSo, I&#039;m integrating it by parts, and going by the LAITE principle, I get:      I    n    =            −              x        n                    e      x        +      ∫          0              ∞                  n              x                  n          −          1                            e      x          d  xThe value inside the integral is  n      I          n      −      1      , so all I&#039;m left is to get the limit of the first term as x ranges from 0 to infinity.  ∑            x      n              e      x      Somehow, I&#039;m not exactly sure why I feel like the limit of the first term would go to zero (or n!), given that formula, but I&#039;m not exactly sure how to substitute the n! formula for       e    x  . How do I proceed from here?The options are  n  !  −  n      I          n      −      1        n  !  +  n      I          n      −      1        n      I          n      −      1      none of these

Alma Garner · Accepted Answer

By parts,      ∫          x      =      0        ∞        x    n        e          −      x          d  x  =  −      x    n        e          −      x                          |                    x      =      0        ∞    +  n      ∫          x      =      0        ∞        x          n      −      1            e          −      x          d  x  .The first term vanishes so that      I    n    =  n      I          n      −      1        .As the base case is      I    0    =      ∫          x      =      0        ∞        e          −      x          d  x  =  −      e          −      x                          |                    x      =      0        ∞    =  1  ,we get the superb formula      I    n    =  n  !Note that as       e    x   contains all powers of x,      e    x    =      ∑          k      =      0        ∞              x      k              k      !        ≥            x              n        +        1                    (      n      +      1      )      !        ,and  0  ≤      x    n        e          −      x        ≤      1          (      n      +      1      )      !      x        ,allowing the limit at   ∞ to cancel.

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