Absolutely awesome evaluation. This integral had so many things going. Really enjoyed the evaluation - and the grand finale is the famous Gaussian integral. Makes everything so satisfying.
I might be becoming an old curmudgeon, but I don't like a binomial coefficient of a continuous variable. But I guess if we think of it as a shorthand for a ratio of gamma functions it's ok.
Hi Alex. I don’t think so. The problem was written that way but I don’t think it’s required unless I’m forgetting some place where that was used in the video
oh wait i think it really does need to be a positive integer. The trouble is that x can be negative and so x to a non-integer exponent is not defined in most cases.
@@owl3math yes but when you expanded the binomial you got (x^()-2)! and because the interval is minus infinity to infinity then x can be 0 thus end up with (-2)! Which is clearly imaginary so you assumed the integral can get imaginary values.( When i said x^() then () is 2k+1). So it doesn't really matter if 2k is not an integer because we allowed an imaginary answer to exist. Otherwise when you expanded the binomial you should have said x^()≥2. Thats my claim I haven't went to a university yet i am last year high school so i dont know everything thats why i asked
@@alexkaralekas4060 it's interesting...I was thinking the real trouble is at 3:15 where I set the lower bound to minus infinity. That only works if 2k+1 is odd which implies k as an integer.
Absolutely awesome evaluation. This integral had so many things going. Really enjoyed the evaluation - and the grand finale is the famous Gaussian integral. Makes everything so satisfying.
Thanks!!! Yes I thought this one was really fun and interesting. And plus throw in a little bit of tricky algebra 😀👍
That was great! It had something for everyone.
ha! Yeah i think so too but I'm optimistic that way. But i really do like it because its a unique and interesting problem. thanks!
This one is definitely a nice combo.
Who's Sid? His name is all over the place on math topics.
Sid Sivakumar. He provided me a lot of really good original problems so many of them became videos. 😀
I might be becoming an old curmudgeon, but I don't like a binomial coefficient of a continuous variable. But I guess if we think of it as a shorthand for a ratio of gamma functions it's ok.
🤣 yes good point. Yeah i think as shorthand its ok :)
x^(2k+1) choose 2 = (x^(2k+1))!/2(x^(2k+1)-2)!
=x^(2k+1)•(x^(2k+1)-1)/2
=(x^(4k+2)-x^(2k+1))/2
z=x^(2k+1)
dz=(2k+1)x^2k•dx
I=1/(2k+1)•int[-♾️,♾️](e^((z-z^2)/2))dz
z-z^2=1/4-(z-1/2)^2
I=e^(1/8)/(2k+1)•int[-♾️,♾️](e^(-(z-1/2)^2/2))dz
p=z-1/2
dp=dz
I=2e^(1/8)/(2k+1)•int[0,♾️](e^(-p^2/2))dp
r=p/sqrt(2)
dr=dp/sqrt(2)
I=2sqrt(2)e^(1/8)/(2k+1)•int[0,♾️](e^-r^2)dr
s=r^2
r=s^(1/2)
dr=1/2•s^(-1/2)•ds
I=sqrt(2)e^(1/8)/(2k+1)•int[0,♾️](s^(-1/2)•e^(-s))ds
I=sqrt(2)e^(1/8)•gamma(1/2)/(2k+1)
gamma(1/2)=sqrt(pi)
I=sqrt(2pi)e^(1/8)/(2k+1)
Does k have to be an integer?
Hi Alex. I don’t think so. The problem was written that way but I don’t think it’s required unless I’m forgetting some place where that was used in the video
oh wait i think it really does need to be a positive integer. The trouble is that x can be negative and so x to a non-integer exponent is not defined in most cases.
@@owl3math yes but when you expanded the binomial you got (x^()-2)! and because the interval is minus infinity to infinity then x can be 0 thus end up with (-2)! Which is clearly imaginary so you assumed the integral can get imaginary values.( When i said x^() then () is 2k+1). So it doesn't really matter if 2k is not an integer because we allowed an imaginary answer to exist. Otherwise when you expanded the binomial you should have said x^()≥2. Thats my claim I haven't went to a university yet i am last year high school so i dont know everything thats why i asked
@@alexkaralekas4060 it's interesting...I was thinking the real trouble is at 3:15 where I set the lower bound to minus infinity. That only works if 2k+1 is odd which implies k as an integer.
@@owl3math still if k is for example 1/2 there isn't an issue so k doesn't always needs to be an integer