Prove by induction of integration functions

Question

If $J_n= \int_{0}^{1}t^n(1-t)^ndt$, then $2(1+2n)J_n = nJ_{n-1}$ and $J_0 = 1$. [Hint: integration by parts]

Here is what I have done so far. Am I correct so far? How can I continue to prove it?

$J_n= \int_{0}^{1}t^n(1-t)^ndt$
= $[(1-t)^n\cdot \frac{t^{n+1}}{n+1}]\big|_0^1$ $- \int_{0}^{1}\frac{t^{n-1}}{n+1}\cdot n(1-t)^{n-1}(-1)dt $ $= \frac{n}{n+1}\int_{0}^{1}t^{n+1}(1-t)^{n-1}dt = \frac{n}{n+1}\int_{0}^{1}t^n(1-(1-t))\cdot(1-t)^{n-1}dt$ $= \frac{n}{n+1}\left(\int_{0}^{1}t^n(1-t)^{n-1}dt - \int_{0}^{1}t^n(1-t)^ndt\right) $ $ =\frac{n}{n+1}\left(\int_{0}^{1}t^{n-1}(1-(1-t))(1-t)^{n-1}dt - J_n\right)$

Answer 1

$$J_n = \int_0^1t^n(1-t)^ndt$$ Using Integration by parts, $$J_n = \frac{t^{n+1}(1-t)^n}{n+1}\vert_0^1+\frac{n}{n+1}\int_0^1t^{n+1}(1-t)^{n-1}dt$$ $$J_n = \frac{n}{n+1}\int_0^1t^{n}(1-t)^{n-1}(t)dt$$ $$J_n = \frac{n}{n+1}\int_0^1t^{n}(1-t)^{n-1}(t-1+1)dt$$ $$J_n = -\frac{n}{n+1}\int_0^1t^{n}(1-t)^{n}dt + \frac{n}{n+1}\int_0^1(1-t)^{n-1}t^ndt = -\frac{n}{n+1}J_n + \frac{n}{n+1}\int_0^1(1-t)^{n-1}t^ndt$$ $$\frac{2n+1}{n+1}J_n = \frac{n}{n+1}\int_0^1(1-t)^{n-1}t^ndt \cdots(1)$$ Using the property that $\int_0^af(x)dx=\int_0^af(a-x)dx$ for $a=1$, we get $$\frac{2n+1}{n+1}J_n = \frac{n}{n+1}\int_0^1(1-t)^nt^{n-1}dt \cdots(2)$$ Adding equations $(1)$ and $(2)$, $$2\frac{2n+1}{n+1}J_n = \frac{n}{n+1}\int_0^1(1-t)^{n-1}t^{n-1}dt = \frac{n}{n+1}J_{n-1}$$ $$\therefore 2(2n+1)J_n = nJ_{n-1}$$

which proves the desired result. Hope this helps.

Answer 2

What you have done is correct. If you multiply both sides by $n+1$ and transfer the last term to the left side you will almost get what you want except that you need the following: $\int_0^{1}t^{n-1}(1-t)^{n-1}(2t-1)dt=0$. This equation follows from the fact that the integral of an odd function from $-\frac 1 2 $ t0 $+\frac 1 2$ is $0$; to use this fact just make the substitution $s=t-\frac 1 2$ in $\int_0^{1}t^{n-1}(1-t)^{n-1}(2t-1)dt$.