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this integral is proving to be very challenging so help is required: $$\int_{0}^{\infty}\beta x^{\beta -1} e^{-x^{\beta}} dx$$

This has been my approach so far: $$\beta \int_{0}^{\infty} e^{-x^{\beta}} x^{\beta -1} dx$$ The next logical step in this was to perform a $u$ substitution having $u = x^{\beta}$ and $du = \beta x^{\beta}$ but that didnt not help simplify the intergral and neither would if i use $u$ to be $(tx-x^{\beta})$ so I am pretty much stuck. Also I am worried about what would happen to the limits of the Integrals if I use $u = x^{\beta}$ help please?

Raul Gonzales
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  • Are $t$ and $\beta$ fixed? – Maximilian Janisch Nov 14 '19 at 16:29
  • is a dx or dt ? Note that $du =\beta x^{\beta -1}$ for $u=x^{\beta}$ – user577215664 Nov 14 '19 at 16:35
  • Wolfram Alpha cannot solve this, so you need to explain why you think there is a closed solution. – Klaus Nov 14 '19 at 16:37
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    @isham it is dx mate sorry – Raul Gonzales Nov 14 '19 at 16:48
  • @Klaus Wolfram Alpha was my first port of call Klaus. this is why i am saying that this integration seems to be more complex than usual. there is a closed solution because i need to proof that $$E(X^r) = \Gamma(\frac{r}{\beta}+1)$$ – Raul Gonzales Nov 14 '19 at 17:14
  • $$\int_0^{\infty } \exp (t x) \beta x^{\beta -1} \exp \left(-x^{\beta }\right) , dx=\sum {j=0}^{\infty } \frac{t^j \Gamma \left(1+\frac{j}{\beta }\right)}{j!}=\Psi{1,1} \left[ \begin{array}{l} (1,\frac{1}{\beta }) \ (1,0)\end{array} ; t\right]$$ See:https://math.stackexchange.com/questions/1880261/zeros-of-fox-wright-function – Mariusz Iwaniuk Nov 14 '19 at 17:51
  • I think this can be solved similar to this https://math.stackexchange.com/questions/2586898/how-to-evaluate-this-notoriously-hard-exponential-integral – stocha Nov 14 '19 at 19:53

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