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I need to evaluate the following integral $$\int_0^\infty x^{\mu}\exp(-{\lambda}x^\kappa)\,dx$$ (where $\mu$, $\lambda$, and $\kappa$ are are all real) in terms of the Gamma function $\Gamma(t)=\int_0^\infty x^{t-1}e^{-x}\,dx$.

I am thinking that I need to make a substitution, but my previous attempts have failed, as after making the substitutions, I still have an $x$ term remaining in the integrand.

Since the parameters aren't given specific values, I am not sure how to proceed, since similar questions in which I have had to make a substitution, I have had $\mu=\kappa$.

If a substitution is required, what is it?

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    $\begingroup$ OK, so why not try making the substitution $u=\lambda x^\kappa$? $\endgroup$ – J. M. isn't a mathematician Jun 15 '16 at 0:49
  • $\begingroup$ @J.M. I still end up with x terms in the integrand, that I don't seem to be able to remove $\endgroup$ – randmath Jun 15 '16 at 0:58
  • $\begingroup$ What happens if you solve for $x$ in the substitution I gave? $\endgroup$ – J. M. isn't a mathematician Jun 15 '16 at 0:59
  • $\begingroup$ @J.M. I seem to have it in a more appropriate form now. I just need to figure out the correct restrictions on the parameters now. Thanks $\endgroup$ – randmath Jun 15 '16 at 1:07
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    $\begingroup$ Yes, although I did not say it explicitly, "solving for $x$" implies determining when this operation is valid. If you figure it out, you can write an answer to your question. $\endgroup$ – J. M. isn't a mathematician Jun 15 '16 at 1:09
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Assuming that $\mu,\kappa,\lambda$ have positive real parts, if you set $x=y^{1/\kappa}$ you are left with: $$I=\frac{1}{\kappa}\int_{0}^{+\infty}y^{\frac{\mu+1}{k}-1}e^{-\lambda y}\,dy $$ then you may set $y=\frac{z}{\lambda}$ to get:

$$ I=\frac{1}{\kappa\cdot\lambda^{\frac{\mu+1}{k}}}\int_{0}^{+\infty}z^{\frac{\mu+1}{k}-1}e^{-z}\,dz = \color{red}{\frac{1}{\kappa\cdot\lambda^{\frac{\mu+1}{k}}}\,\Gamma\left(\frac{\mu+1}{k}\right)}.$$

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