My Math Forum Wallis and Stirling and Poisson: a relation.

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 January 31st, 2012, 02:54 PM #1 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Wallis and Stirling and Poisson: a relation. It isn't hard to prove that $\int\limits_0^{\frac{\pi }{2}} {{{\sin }^m}x} dx= \frac{{m - 1}}{m}\int\limits_0^{\frac{\pi }{2}} {{{\sin }^{m - 2}}x} dx$ and thus that $\int\limits_0^{\frac{\pi }{2}} {{{\sin }^m}x} dx= \frac{{\left( {m - 1} \right)!!}}{{m!!}}$ for odd $m$ $\int\limits_0^{\frac{\pi }{2}} {{{\sin }^m}x} dx= \frac{{\left( {m - 1} \right)!!}}{{m!!}}\frac{\pi }{2}$ for even $m$ Morever, since we are in the first quadrant, the following inequation holds. $\int\limits_0^{\frac{\pi }{2}} {{{\sin }^{2n + 1}}x} dx < \int\limits_0^{\frac{\pi }{2}} {{{\sin }^{2n}}x} dx < \int\limits_0^{\frac{\pi }{2}} {{{\sin }^{2n - 1}}x} dx$ Thus: $\frac{{\left( {2n} \right)!!}}{{\left( {2n + 1} \right)!!}} < \frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}}\frac{\pi }{2} < \frac{{\left( {2n - 2} \right)!!}}{{\left( {2n - 1} \right)!!}}$ Arranging the inequation properly produces: ${\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{{2n + 1}} < \frac{\pi }{2} < {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{{2n}}$ ${\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{{2n}}{{2n + 1}}\frac{1}{n} < \frac{\pi }{2} < {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{n}$ Thus we conclude that ${\lim }\limits_{n \to \infty } {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{n}= \frac{\pi }{2}$ and that ${\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{{2n + \theta }}= \frac{\pi }{2}$ with $0 < \theta < 1$ which is a variation of Wallis infinite product. Let's focus on Poisson's integral $\int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}} dx= \sqrt \pi$ It is easily proven by Taylor series that $1-x^2 < e^{-x^2} < \frac{1}{1+x^2}$ Thus $\left(1-x^2 \right)^n< e^{-nx^2} < \frac{1}{\left(1+x^2 \right)^n}$ Upon integration we have $\int\limits_0^1 {{{\left( {1 - {x^2}} \right)}^n}dx < \int\limits_0^\infty {{e^{ - n{x^2}}}dx} < } \int\limits_0^\infty {\frac{{dx}}{{{{\left( {1 + {x^2}} \right)}^n}}}}$ Similar to the integration in Wallis' identity we can prove that $\int\limits_0^{\frac{\pi }{2}} {{{\cos}^m}x} dx= \frac{{\left( {m - 1} \right)!!}}{{m!!}}$ for odd $m$ $\int\limits_0^{\frac{\pi }{2}} {{{\cos}^m}x} dx= \frac{{\left( {m - 1} \right)!!}}{{m!!}}\frac{\pi }{2}$ for even $m$ and with appropiate change of variables, namely$x= \sin t$ and $x= \tan t$ we conclude that $\int\limits_0^1 {{{\left( {1 - {x^2}} \right)}^n}dx}= \frac{{\left( {2n} \right)!!}}{{\left( {2n + 1} \right)!!}}$ $\int\limits_0^\infty {\frac{{dx}}{{{{\left( {1 + {x^2}} \right)}^n}}}}= \frac{{\left( {2n - 3} \right)!!}}{{\left( {2n-2} \right)!!}}\frac{\pi }{2}$ Thus: $\frac{{\left( {2n} \right)!!}}{{\left( {2n + 1} \right)!!}} < \int\limits_0^\infty {{e^{ - n{x^2}}}dx} < \frac{{\left( {2n - 3} \right)!!}}{{\left( {2n-2} \right)!!}}\frac{\pi }{2}$ And with $\sqrt{n}x= u$ (but keep the $x$) $\frac{{\left( {2n} \right)!!}}{{\left( {2n + 1} \right)!!}} < \frac{1}{{\sqrt n }}\int\limits_0^\infty {{e^{ - {x^2}}}dx} < \frac{{\left( {2n - 3} \right)!!}}{{\left( {2n-2} \right)!!}}\frac{\pi }{2}$ Let's set $\int\limits_0^\infty {{e^{ - {x^2}}}} dx= \Phi$ and thus we have, using Wallis identity: $\frac{n}{{{{\left( {2n + 1} \right)}^2}}}{\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n + 1} \right)!!}}} \right]^2} < {\Phi ^2} < n{\left[ {\frac{{\left( {2n - 3} \right)!!}}{{\left( {2n - 2} \right)!!}}} \right]^2}\frac{{{\pi ^2}}}{4}$ $\frac{n}{{{{\left( {2n + 1} \right)}^2}}}\left( {2n + \theta } \right)\frac{\pi }{2} < {\Phi ^2} < \frac{n}{{2\left( {n - 1} \right) + \theta }}\frac{\pi }{2}$ But if $n \to \infty$ we get $\frac{\pi }{4} < {\Phi ^2} < \frac{\pi }{4}$ and thus $\int\limits_0^\infty {{e^{ - {x^2}}}} dx= \frac{{\sqrt \pi }}{2}$ which yields $\int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}} dx= \sqrt \pi$ I'll wait for some feedback untill I get to Stirling's approximation. I learned this proofs from a Spanish Analisis book, by Rey Pastor (a true diamond if you have to ask).

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