1

I'm wondering how can we evaluate $$\int\cdots\int_{\mathbb R^n}\prod_{i=1}^{n}\frac{x_i^me^{-x_i^2}}{e^{\frac{x_i}2}\cosh(\frac{x_i}2)}\mathbb dx_1 \cdots \mathbb dx_n$$ using Gamma/Beta functions? Thanks in advance.

Infiniticism
  • 8,644
  • 1
  • 22
  • 67
Costas
  • 123
  • I am new to the community and I post some old exams problems I found while studying for mathematical methods of physics. The truth is I have made almost 0 progress on the ones I post – Costas Nov 17 '20 at 13:12
  • I will try not to post such questions and to contribute to the community with effort! – Costas Nov 17 '20 at 13:13
  • Thanks for the edit!! – Costas Nov 17 '20 at 13:14
  • I also wanted to ask you, if a problem I posted received 0 answers, shall I post it again? – Costas Nov 17 '20 at 13:16
  • 3
    When $m$ is even, the integral reduces to $$\left(\int_{-\infty}^{\infty}x^me^{-x^2},\mathrm{d}x\right)^n=\Gamma\left(\frac{m+1}{2}\right)^m.$$ When $m$ is odd, even the simplest case $n=1$ and $m=1$ yields $$-2\int_{0}^{\infty}xe^{-x^2}\tanh(x/2),\mathrm{d}x=-\frac{1}{2}\int_{0}^{\infty}e^{-x^2}\operatorname{sech}^2(x/2),\mathrm{d}x,$$ which I am not sure how to tackle. – Sangchul Lee Nov 17 '20 at 13:52
  • Can you tell me how you get this integral ? It seems right thought – Costas Nov 17 '20 at 14:31
  • 2
    When $m$ is even, then note that \begin{align}\int_{-\infty}^{\infty}\frac{x^me^{-x^2}}{e^{x/2}\cosh(x/2)},\mathrm{d}x&=\int_{0}^{\infty}\left(\frac{x^me^{-x^2}}{e^{x/2}\cosh(x/2)}+\frac{(-x)^me^{-(-x)^2}}{e^{-x/2}\cosh(-x/2)}\right),\mathrm{d}x\&=2\int_{0}^{\infty}x^me^{-x^2},\mathrm{d}x.\end{align} The last integral reduces to the gamma function via the substitution $t=x^2$. – Sangchul Lee Nov 17 '20 at 15:14
  • Thanks, I will post if i find the answer for m=odd – Costas Nov 17 '20 at 15:20
  • I couldnt find anything for m=odd.. – Costas Nov 18 '20 at 23:21

0 Answers0