taking u = sin^2a(x) and v'=xcos(x) , I get:
-a\pi+(2a+1)(2a)\int\limits_0^{\pi/2} Sin^{2a}(x)x +Sin^{2a-1}(x)cos(x)dx
Sadly my math isn't great and can't seem to figure out how this would lead to the factorial form.
So I tried to integrate by parts again and try to simplify and it just...
\pi/2-(2a+1)\int\limits_0^{\pi/2} \sin^{2a}(x)cos(x)x
Sorry I still don't see how to finish the connection.
Edit: Didn't see you said twice IBP ill go back and retry this
Our integral
\int\limits_0^{\pi/2} \sin^{2a+1}(x)\,dx
Has a Factorial Form:
{(2^a a!)}^2 \over (2a+1)!
What is the process behind going from that integral to that factorial form?
My approach which is not very insightful:
I used mathematica to calculate the integral to return...
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Thanks Vanhees but I have not learned this notation yet which i believe will be the next chapter in the book. Once I get a little bit further ill revisit this calculation to see how this notation might simplify things.
Sorry about not posting the work it is a lot to write without knowing...
Homework Statement
Knowing the momentum operator -iħd/dx , the expectation value of momentum and the Fourier transforms how can I prove that <p> = ∫dk [mod square of ψ(k)] h/λ. From this, mod square of ψ(k) is defined to be equal to P(k) right?
Homework Equations
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