1
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Frank Moses Feb 1 '17 at 13:43
  • 1
    $\begingroup$ The negative sign that comes in the RHS?? $\endgroup$ – Rohan Feb 1 '17 at 13:44
  • $\begingroup$ @Rohan which negative sign? $\endgroup$ – Frank Moses Feb 1 '17 at 13:46
  • 1
    $\begingroup$ @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. $\endgroup$ – Rohan Feb 1 '17 at 13:55
  • 1
    $\begingroup$ @FrankMoses any constant such that the integral converges. $\endgroup$ – Simply Beautiful Art Feb 2 '17 at 1:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.