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I have read following definition of incomplete gamma function $$\Gamma(a,x)=\int_x^{\infty} t^{a-1}e^{-t}dt$$ where $Re(a)>0$. This definition is according to DLMF website (http://dlmf.nist.gov/8.2). I have come across another definition of incomplete Gamma function which is given below $$\int e^{-x^{k}}dx=-\frac{1}{k}\Gamma\left(\frac{1}{k},x^k\right)$$ Now this defintion can be verified from wolfram alpha (https://www.wolframalpha.com/input/?i=int+e%5E(-x%5Ek)). Although I some other knowledgeable person (MSE user with high score) has also used this definition but I have not seen this definition in the literature. I have also tried to find this definition on the internet but I was unsuccessful. I will be very thankful if somebody help me in getting the literature where I can found this second definition. Thanks in advance.

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Consider the u-substitution $x=u^{1/k}$.

$$\int e^{-x^k}\ dx=\int e^{-u}\ \frac{du}{ku^{(k-1)/k}}$$

Solving this with the Gamma function gives

$$I=-\frac1k\Gamma\left(\frac1k,u\right)+c=-\frac1k\Gamma\left(\frac1k,x^k\right)+c$$

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    Many thanks for your answer. Please add more steps I cannot get the answer from the hint. I just have no idea how to include the limits when actually there are no limits involved. Your help is much appreciated. – 2017-02-01
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    The negative sign that comes in the RHS?? – 2017-02-01
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    @Rohan which negative sign? – 2017-02-01
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    @Rohan I think the expression on the right side is correct according to substitution. Can you add remaining steps until we get the answer in incomplete gamma function form. Please note that I have not seen any definition in the literature where incomplete gamma function is defined as an integral without limits. Thank you so much for your help. – 2017-02-01
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    @FrankMoses Frankly speaking in my answer for that particular integral, I myself used WolframAlpha. But I also now wonder how they have used it in an indefinite integral form. If I get the answer, I will post here. – 2017-02-01
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    @Rohan thank you for help. I am wondering about the limits. The problem is as follows. Suppose we solve some indefinite integral (of similar kind as discussed in this post and the one discussed in http://math.stackexchange.com/questions/2123777/intermediate-steps-needed-in-integral-int-fracxe-xax-cndx/2123819#2123819) through wolfram alpha and after getting the answer we put the limits. Will it be right answer? Irrespective of the limits used? Your thoughts on this are much appreciated. Thank you, – 2017-02-01
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    @FrankMoses I updated my answer. :-) – 2017-02-01
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    @SimplyBeautifulArt I have two questions for your answer. (i) there is no negative sign present on the right side of your second equation. (ii) I am very confused about how you have converted a indefinite integral into Gamma function. Did you also used wolfram alpha to get this answer? If this is not the case then please show me the reference where I can find a formula that shows indefinite integral definition of gamma function. I am very thankful to your help. – 2017-02-02
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    @FrankMoses (i) Oh right, my bad. The negative comes from the fact that $x$ is in the lower bound of the integral of the gamma function, so we have to multiply be $-1$ to flip it to the top. (ii) You may note that the integral is equal to the gamma function, with exception of bounds and a constant $1/k$ Indeed, the integral is of the form $\int x^ae^{-x}\ dx$, which is what I used. One may also note that:$$\int f(x)\ dx=\int_a^xf(t)\ dt+c$$ – 2017-02-02
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    @SimplyBeautifulArt Thanks alot. At last I got a general formula $$\int f(x)dx=\int_{a}^{x}f(t)dt+c$$. Do you really mean the lower limit to be $a$ or should it be $-\infty$ or $0$? – 2017-02-02
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    @FrankMoses any constant such that the integral converges. – 2017-02-02
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    @Rohan I think the formula used by wolfram alpha is $$\int f(x)dx=\int_{a}^{x}f(t)dt+c$$ as mentioned by Simply Beautiful Art. – 2017-02-02