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If we have $\theta(x) = (\log x)^n$, and I want to integrate $S = \int_0^\infty \theta(x)f(x)\,dx$, where $$\large {f(x) = \left[ \theta^{(1 + 1/\alpha + \beta/\alpha)x} \right]^{\alpha + \beta} e^{-\theta x^{\alpha}}}.$$ By putting value in the above integral $$S = \int_0^{\infty}\left[ (\log x)^n \theta^{(1 + 1/\alpha + \beta/\alpha)x} \right]^{\alpha + \beta} e^{-\theta x^{\alpha}} \,dx$$ By substitution $\theta x^{\alpha} = t$, then $\alpha \theta x^{\alpha - 1}\, dx = dt$ and $0 < t < \infty$. After simplification, this becomes $$S = \int_0^{\infty}\left[\log \left(\dfrac{t}{\theta}\right)^{1/\alpha}\right]^n \theta^{1/\alpha} t^{1 + \beta/\alpha} e^{-t} \,dt\\ S = \theta^{1/\alpha} \int_0^{\infty} \left[\log \left(\dfrac{t}{\theta}\right)^{1/\alpha}\right]^n t^{1 + \beta/\alpha} e^{-t} \,dt$$

Here three variables are involved in the above integral, now how I can integrate this?

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    $\begingroup$ You need to put this in LaTeX form for it to be understandable. This site uses MathJax; see meta.math.stackexchange.com/questions/5020/… for a tutorial. $\endgroup$ – rogerl Jun 26 '14 at 17:00
  • $\begingroup$ I'm guessing the power on $\theta$ is a mistake? $\endgroup$ – John Fernley Jun 26 '14 at 19:56

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