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Using the substitution $u_i=x_i^p$ and letting $r:=q/p$ and $a:=1-1/p\in[0,1)$, we see that the integral in question is \begin{align} I&=np^{-n}\int_0^1\cdots\int_0^1\frac{du_1\cdots du_n}{u_1^a\cdots u_n^a} u_1^r\int_0^\infty dt\,e^{-t(u_1+\cdots+u_n)} \\ &=np^{-n}\int_0^\infty dt\,\int_0^1\frac{du_1}{u_1^{a-r}}\,e^{-tu_1} \Big(\int_0^1\frac{du_2}{u_2^a}\,e^{-tu_2}\Big)^{n-1} \\ &=np^{-n}\int_0^\infty \frac{dt}{t^{(n-1)(1-a)}}\gamma(1+r-a,t)\, \gamma(1-a,t)^{n-1}, \end{align} where $\gamma$ is the lower incomplete gamma function.

Using the substitution $u_i=x_i^p$ and letting $r:=q/p$ and $a:=1-1/p\in[0,1)$, we see that the integral in question is \begin{align} I&=np^{-n}\int_0^1\cdots\int_0^1\frac{du_1\cdots du_n}{u_1^a\cdots u_n^a} u_1^r\int_0^\infty dt\,e^{-t(u_1+\cdots+u_n)} \\ &=np^{-n}\int_0^\infty dt\,\int_0^1\frac{du_1}{u_1^{a-r}}\,e^{-tu_1} \Big(\int_0^1\frac{du_2}{u_2^a}\,e^{-tu_2}\Big)^{n-1} \\ &=np^{-n}\int_0^\infty \frac{dt}{t^{r+n(1-a)}}\gamma(1+r-a,t)\, \gamma(1-a,t)^{n-1}, \end{align} where $\gamma$ is the lower incomplete gamma function.

Using the substitution $u_i=x_i^p$ and letting $r:=q/p$ and $a:=1-1/p\in[0,1)$, we see that the integral in question is \begin{align} I&=np^{-n}\int_0^1\cdots\int_0^1\frac{du_1\cdots du_n}{u_1^a\cdots u_n^a} u_1^r\int_0^\infty dt\,e^{-t(u_1+\cdots+u_n)} \\ &=np^{-n}\int_0^\infty dt\,\int_0^1\frac{du_1}{u_1^{a-r}}\,e^{-tu_1} \Big(\int_0^1\frac{du_2}{u_2^a}\,e^{-tu_2}\Big)^{n-1} \\ &=np^{-n}\int_0^\infty \frac{dt}{t^{(n-1)(1-a)}}\int_0^1\frac{du_1}{u_1^{a-r}}\,e^{-tu_1}\, \gamma(1-a,t-1)^{n-1}, \end{align} where $\gamma$ is the lower incomplete gamma function.

Using the substitution $u_i=x_i^p$ and letting $r:=q/p$ and $a:=1-1/p\in[0,1)$, we see that the integral in question is \begin{align} I&=np^{-n}\int_0^1\cdots\int_0^1\frac{du_1\cdots du_n}{u_1^a\cdots u_n^a} u_1^r\int_0^\infty dt\,e^{-t(u_1+\cdots+u_n)} \\ &=np^{-n}\int_0^\infty dt\,\int_0^1\frac{du_1}{u_1^{a-r}}\,e^{-tu_1} \Big(\int_0^1\frac{du_2}{u_2^a}\,e^{-tu_2}\Big)^{n-1} \\ &=np^{-n}\int_0^\infty \frac{dt}{t^{(n-1)(1-a)}}\gamma(1+r-a,t)\, \gamma(1-a,t)^{n-1}, \end{align} where $\gamma$ is the lower incomplete gamma function.

Using the substitution $u_i=x_i^p$ and letting $r:=q/p$ and $a:=1-1/p\in[0,1)$, we see that the integral in question is \begin{align} I&=np^{-n}\int_0^1\cdots\int_0^1\frac{du_1\cdots du_n}{u_1^a\cdots u_n^a} u_1^r\int_0^\infty dt\,e^{-t(u_1+\cdots+u_n)} \\ &=np^{-n}\int_0^\infty dt\,\int_0^1\frac{du_1}{u_1^{a-r}}\,e^{-tu_1} \Big(\int_0^1\frac{du_2}{u_2^a}\,e^{-tu_2}\Big)^{n-1} \\ &=np^{-n}\int_0^\infty dt\,t^{(n-1)(a-1)}\int_0^1\frac{du_1}{u_1^{a-r}}\,e^{-tu_1}\, \gamma(1-a,t-1)^{n-1}, \end{align} where $\gamma$ is the lower incomplete gamma function.

Using the substitution $u_i=x_i^p$ and letting $r:=q/p$ and $a:=1-1/p\in[0,1)$, we see that the integral in question is \begin{align} I&=np^{-n}\int_0^1\cdots\int_0^1\frac{du_1\cdots du_n}{u_1^a\cdots u_n^a} u_1^r\int_0^\infty dt\,e^{-t(u_1+\cdots+u_n)} \\ &=np^{-n}\int_0^\infty dt\,\int_0^1\frac{du_1}{u_1^{a-r}}\,e^{-tu_1} \Big(\int_0^1\frac{du_2}{u_2^a}\,e^{-tu_2}\Big)^{n-1} \\ &=np^{-n}\int_0^\infty \frac{dt}{t^{(n-1)(1-a)}}\int_0^1\frac{du_1}{u_1^{a-r}}\,e^{-tu_1}\, \gamma(1-a,t-1)^{n-1}, \end{align} where $\gamma$ is the lower incomplete gamma function.