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I am trying to prove that these two integrals are equal $$ \int_0^\infty\int_0^\infty\dots\int_0^\infty \exp(-x z_1-z_1z_2-z_2z_3-\dots-z_{n-1}z_{n}-z_n) x^{c_1-1} z_1^{c_1+c_2-1} z_2^{c_2+c_3-1} \dots z_{n}^{c_n+c_{n+1}-1} dz_1 dz_2 \dots dz_{n} $$ $$ \int_0^\infty\int_0^\infty\dots\int_0^\infty \exp(-x ^{w_n} z_1-z_1z_2-z_2z_3-\dots-z_{n-1}z_{n}-z_n) x^{{w_n}\,c_{n+1}-1} z_n^{c_1+c_2-1} z_{n-1}^{c_2+c_3-1} \dots z_{1}^{c_n+c_{n+1}-1} dz_1 dz_2 \dots dz_{n} $$ where $x>0,c_n>0,\forall n$ and $$w_n = \begin{cases} 1 & \text{even}\,\,n \\ -1 & \text{odd}\,\,n \end{cases}.$$

Using Mathematica, I was able to prove up to $n=5$, but is this valid for all $n$?

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  • $\begingroup$ Just wondering: Why do you state your question with a $w$ only to restrict it to $w=1$? Isn't that a confusing waste of time? $\endgroup$ Commented Jul 26, 2022 at 19:42
  • $\begingroup$ @DavidG.Stork, $w=1$ only for even values of $n$ and $w=-1$ only for odd values of $n$ $\endgroup$ Commented Jul 26, 2022 at 19:45
  • $\begingroup$ I've rolled back your removal of the exponentials (resulting in diverging integrals). $\endgroup$ Commented Aug 1, 2022 at 9:37

1 Answer 1

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Basically $w_n=(-1)^n$. Just substitute $z_k=x^{(-1)^k}y_{n-k+1}$ in the first integral: $$ xz_1+\sum_{k=1}^{n-1}z_kz_{k+1}+z_n=x^{(-1)^n}y_1+\sum_{k=1}^{n-1}y_ky_{k+1}+y_n, \\ x^{c_1-1}\prod_{k=1}^{n}z_k^{c_k+c_{k+1}-1}\,dz_1\cdots dz_n=x^{c-1}\prod_{k=1}^{n}y_{n-k+1}^{c_k+c_{k+1}-1}\,dy_1\cdots dy_n,\\ c:=c_1+\sum_{k=1}^{n}(-1)^k(c_k+c_{k+1})=(-1)^nc_{n+1}, $$ and your equality follows immediately.

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