I would like to ask for help with exercise I was given on QM. I am suppose to calculate: $$ \int_{0}^{L}|A|^2|x(L-x)|^2 dx = |A|^2\int_{0}^{L}(x(L-x))^2 $$ That is not a problem of course. I am suppose to use this formula to calculate the integral: $$ \int_{0}^{1}t^m(1-t)^n dt=\frac{m!n!}{(m+n+1)!} $$ Could anyone point my how to approach it. I have never seen such integral before.
Thank you for reeding. Any help will be appreciated.
I'm not sure how you're suppose to use "this formula", since your first equation is trivial; the only change is that $|A|^2$ is moved outside of the integral.
To calculate the value of $$I_{m,n}=\int_0^1 t^m(1-t)^n\ dt,$$ we can repeatedly integrate by parts to raise the degree of $t^m$ and lower the degree of $(1-t)^n$. First, note that if $n=0$ then the equation is true, since $$\int_0^1 t^m\ dt = \frac{1}{m+1} = \frac{m!}{(m+1)!}.$$ Integrating the genreal integral by parts once, we get $$\int_0^1 t^m(1-t)^n\ dt = \frac{t^{m+1}(1-t)^n}{m+1}\bigg|_0^1 + \frac{n}{m+1}\int_0^1 t^{m+1}(1-t)^{n-1}\ dt.$$ Simplifying, this gives us the relation $$I_{m,n} = \frac{n}{m+1}I_{m+1,n-1}.$$ Using this relation repeatedly until we hit $n=0$, we get \begin{align} I_{m,n} &= \frac{n!}{(m+1)(m+2)\cdots (m+n)}I_{m+n,0}\\ &= \frac{n!}{(m+1)(m+2)\cdots (m+n)(m+n+1)}\\ &= \frac{n!m!}{(n+m+1)!}. \end{align}