How to evaluate the integral <msubsup> &#x222B;<!-- ∫ --> a <mi mathvariant="normal">

Jorge Lawson

Jorge Lawson

Answered question

2022-05-29

How to evaluate the integral a x ( 1 exp ( x a ) ) d x

Answer & Explanation

concludirgt

concludirgt

Beginner2022-05-30Added 11 answers

Denoting the integral as I = a x ( 1 e x a ) d x, we can take the substitution t = x a . We get d t = a x a + 1 d x which gives
I = 1 a 0 a a 1 e t t 1 + 2 a d t = 1 a 0 a a ( 1 e t ) ( a 2 t 2 a ) d t
The above integral is in perfect form to apply integration by parts. We thus get
I = 1 a [ a 2 ( e t 1 t 2 a ) | 0 a a + a 2 0 a a t ( 1 2 a ) 1 e t d t ]
Recalling that the lower incomplete Gamma function is given by γ ( α , x ) := 0 x t α 1 e t d t we get
I = a 2 2 ( e a a 1 ) + 1 2 lim t 0 + 1 e t t 2 a L + 1 2 γ ( 1 2 a , a a )
We just need to evaluate the limit L to conclude the problem, and this is the part that will give us the restriction on the possible values of a. By applying L'Hôpital's rule we get
L = a 2 lim t 0 + e t t 2 a 1
And here the value of the limit entirely depends on if 2 a 1 is positive, negative or 0. If 2 a 1 > 0, then we have a 1 0 situation and the limit is infinity, so I diverges. If we have 2 a 1 = 0 then γ ( 1 2 a , a a ) = γ ( 0 , 2 2 ) diverges, so here I again diverges. If 2 a 1 < 0 then 1 t 2 a 1 0 as t 0 + and thus L=0. We can finally conclude that
I = a 2 2 ( e a a 1 ) + 1 2 γ ( 1 2 a , a a ) , a > 2

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