Finding the area under e^-x^2 in the region x<=0
- Thread starter Trebor
- Start date
Here's one proof attributed to Stieltjes.
I will start it and you try to finish. Okay?.
\(\displaystyle \L\\\int_{-\infty}^{\infty}e^{-x^{2}}dx\)
Let \(\displaystyle \L\\I_{n}=\int_{0}^{\infty}x^{n}e^{-x^{2}}dx, \;\ n=0, 1, 2, ....\)
Establish a resursive relation:
\(\displaystyle \L\\I_{n}=\frac{n-1}{2}I_{n-2}\)
by integrating by parts.
Then we can deduce that:
\(\displaystyle \L\\I_{2k}=\frac{1\cdot{3}\cdot{5}....2k-1}{2^{k}}I_{0}\)
and
\(\displaystyle \L\\I_{2k+1}=\frac{1\cdot{2}\cdot{3}....k}{2}\)
Show that for all real values of \(\displaystyle \L\\{\lambda}\), the espression:
\(\displaystyle \L\\{\lambda}^{2}I_{n-1}+2{\lambda}I_{n}+I_{n+1}=\int_{0}^{\infty}x^{n-1}({\lambda}+x)^{2}e^{-x^{2}}dx\) is positive, and conclude that:
\(\displaystyle \L\\I_{n}^{2}<I_{n-1}I_{n+1}\)
By taking n=2k and n=2k+1, show that:
\(\displaystyle \L\\\frac{2\cdot{4}\cdot{6}...2k}{1\cdot{3}\cdot{5}...2k-1}\cdot\frac{1}{\sqrt{4k+2}} \;\ < \;\ I_{0} \;\ < \;\ \frac{2\cdot{4}\cdot{6}...2k}{1\cdot{3}\cdot{5}...2k-1}\cdot\frac{1}{\sqrt{4k}}\)
Now, apply Wallis's formula. See?. Easy :roll: :wink:
I will start it and you try to finish. Okay?.
\(\displaystyle \L\\\int_{-\infty}^{\infty}e^{-x^{2}}dx\)
Let \(\displaystyle \L\\I_{n}=\int_{0}^{\infty}x^{n}e^{-x^{2}}dx, \;\ n=0, 1, 2, ....\)
Establish a resursive relation:
\(\displaystyle \L\\I_{n}=\frac{n-1}{2}I_{n-2}\)
by integrating by parts.
Then we can deduce that:
\(\displaystyle \L\\I_{2k}=\frac{1\cdot{3}\cdot{5}....2k-1}{2^{k}}I_{0}\)
and
\(\displaystyle \L\\I_{2k+1}=\frac{1\cdot{2}\cdot{3}....k}{2}\)
Show that for all real values of \(\displaystyle \L\\{\lambda}\), the espression:
\(\displaystyle \L\\{\lambda}^{2}I_{n-1}+2{\lambda}I_{n}+I_{n+1}=\int_{0}^{\infty}x^{n-1}({\lambda}+x)^{2}e^{-x^{2}}dx\) is positive, and conclude that:
\(\displaystyle \L\\I_{n}^{2}<I_{n-1}I_{n+1}\)
By taking n=2k and n=2k+1, show that:
\(\displaystyle \L\\\frac{2\cdot{4}\cdot{6}...2k}{1\cdot{3}\cdot{5}...2k-1}\cdot\frac{1}{\sqrt{4k+2}} \;\ < \;\ I_{0} \;\ < \;\ \frac{2\cdot{4}\cdot{6}...2k}{1\cdot{3}\cdot{5}...2k-1}\cdot\frac{1}{\sqrt{4k}}\)
Now, apply Wallis's formula. See?. Easy :roll: :wink: