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Wolfram Alpha provides following answer for the integral $\displaystyle\int \frac{xe^{-x}}{(ax-c)^N}dx\,$:

$$-e^{-\frac{c}{a}}a^{-N-1}\left(c\Gamma(1-N,x-\frac{c}{a})+a\Gamma(2-N,x-\frac{c}{a})\right)+Constant$$

I do not understand how does incomplete Gamma function comes in the answer. As far as I know the definition of incomplete Gamma function is some definite integral but how does it appears in the indefinite integral. I will be very thankful if somebody guide me how to reach to the above answer and how does the limits appear to give Gamma function in the answer.

Thanks in advance.

1 Answers 1

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Substituting $u = ax-c$ gives us $$I = \frac {e^{-c/a}}{a^2} \int \frac {(u+c)e^{-u/a}}{u^N} \mathrm {d}u = \frac {e^{-c/a}}{a^2} I_1$$

Now $$I_1 = c\int \frac {e^{-u/a}}{u^N} \mathrm {d}u + \int u^{1-N} e^{-u/a} \mathrm {d}u = c I_{11} + I_{12} $$

In $I_{11}$, substituting $v = (u/a)^{1-N} $, gives us $$I _{11} = \frac {1}{1-N} \int e^{-v^{\frac {1}{1-N}}} \mathrm {d}v$$ Now this is an incomplete gamma function whose value comes out to be $$I_{11}= \frac {1}{1-N} [-(1-N)\Gamma (1-N, v^{\frac {1}{1-N}})] $$

Also, $I_{12} $ can be solved similarly writing $w= (u/a)^{2-N}$ giving us the final result as $$I_{12} = -a^{2-N} \Gamma (2-N, v^{\frac {1}{2-N}}) $$

ADD both of them and manipulate to get the final result. Hope it helps.

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    I am sorry but I do not understand how you used $I_{11}$ to gamma function. In your reference there are some limits however in this calculation there is no limit. Can you please include the definition of gamma function that you used in this answer. Thank you so much.2017-02-01
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    Many thanks for your comment and clarification. But I am extremely sorry I can not find this definition of incomplete gamma function on your provided link (https://en.m.wikipedia.org/wiki/Incomplete_gamma_function). The definitions on that link have integral with limits but the definition you wrote in your comment does not have any limits. I will be most thankful if you could provide some other reference where I can found this definition of gamma function. Sorry for bothering you again but I am novice. Thank you.2017-02-01