# Evaluating an integral in terms of the gamma function

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?

• OK, so why not try making the substitution $u=\lambda x^\kappa$? – J. M. isn't a mathematician Jun 15 '16 at 0:49
• @J.M. I still end up with x terms in the integrand, that I don't seem to be able to remove – randmath Jun 15 '16 at 0:58
• What happens if you solve for $x$ in the substitution I gave? – J. M. isn't a mathematician Jun 15 '16 at 0:59
• @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 – randmath Jun 15 '16 at 1:07
• 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. – J. M. isn't a mathematician Jun 15 '16 at 1:09

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