In my work, I keep coming across the term

$$ f(x) = \frac{1}{\sqrt{2\pi}} 2^{\frac{x-1}{2}} \Gamma(\frac{x+1}{2}), $$

in particular for $x \in [0,1]$. I have been working on bounding the density function of a product of powers of standard Gaussians, i.e.

$$ Y = X_1^{e_1}\cdots X_k^{e_k} $$

for $X_i \sim \mathcal{N}(0,1)$ i.i.d., and have been applying those bounds to bound e.g.

$$ E[\lvert Y \rvert^{\delta}], $$

for $\delta \in [0,1]$.

Has anyone else come across this kind of term before? And could you help me understand/prove why, in the range of $ x \in [0,1]$, it is bounded by $\frac{1}{2}$?

Thank you!

Edit: Added more detail, clarified the question, added the missing constant. Thank you for the feedback.

  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community
    Sep 14 at 16:29
  • $\begingroup$ This should be pretty close to $n!!$, the double factorial. $\endgroup$
    – user44191
    Sep 14 at 17:23
  • $\begingroup$ I have edited the question to be more specific. Thank you for your feedback. $\endgroup$
    – MathsIsFun
    Sep 14 at 23:23
  • $\begingroup$ I cannot help you understand why $f(x)$ is bounded by 1/2 for $x\in [0,1]$. The problem is it does not seem to be so. For instance, I find $f(0)=\sqrt{\pi/2}$. $\endgroup$ Sep 15 at 2:36
  • $\begingroup$ @MichaelRenardy The usual problem, I forgot that I always carry the constant $\frac{1}{\sqrt{2\pi}}$. Sloppiness on my part. $\endgroup$
    – MathsIsFun
    Sep 15 at 11:09

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