0

The function $g(t)$ givenby $$ g(t)=\frac{n(n-1)}{2}\left[\frac{t^{n-2}(2-t)^{n+1}-t^{2n-1}}{n+1}-\frac{t^{n}(2-t)^{n-1}-t^{2n-1}}{n-1}\right]$$ integrates to 1 form $t=0$ to $t=1$.From numerical simulations one can see that limit of $\int_0^1 tg(t) dt$ as n goes to infinity approaches one i.e $ \lim_{n \to \infty}\int_0^1 tg(t) dt$ is expected to be 1.However, i am not able to prove it analytically.Any help/hint in this regard is greatly appreciated

AgnostMystic
  • 1,654
  • The $t^{2 n - 1}$ terms can be dropped. The difference $$t \frac {n (n - 1)} 2 \frac {t^n (2 - t)^{n - 1}} {n - 1} - t \frac {n (n - 1)} 2 \frac {t^n (2 - t)^{n - 1}} {n + 1}$$ is bounded and converges to zero pointwise on $[0, 1)$, so the limit of its integral is zero. Therefore we can replace $t^n (2 - t)^{n - 1}/(n - 1)$ in $g(t)$ with $t^n (2 - t)^{n - 1}/(n + 1)$ to reduce the problem to $$\lim_{n \to \infty} \frac {2 n (n - 1)} {n + 1} \int_0^1 (1 - t) (t (2 - t))^{n - 1} dt.$$ The indefinite integral $\int (2 t - t^2)^{n - 1} d(2 t - t^2)$ is elementary. – Maxim Mar 06 '21 at 14:28

1 Answers1

1

The only thing to use is $$I(p,q)=\int_0^1 t^p\,(2-t)^q\,dt=2^{p+q+1} B_{\frac{1}{2}}(p+1,q+1)$$ where appears the incomplete Beta function.

Multiplying by $t$ we then have

$$\int_0^1\Big[t^{n-1}(2-t)^{n+1}-t^{2n}\Big]\,dt=2^{2 n+1} B_{\frac{1}{2}}(n,n+2)-\frac{1}{2 n+1}$$ $$\int_0^1\Big[t^{n+1}(2-t)^{n-1}-t^{2n}\Big]\,dt=2^{2 n+1} B_{\frac{1}{2}}(n+2,n)-\frac{1}{2 n+1}$$ All of this gives $$\int_0^1 t\,g(t)\, dt=$$ $$\frac{n \left(4^n \left(2 n (2 n+1) B_{\frac{1}{2}}(n,n+1)-(2 n (n+1)+1) B_{\frac{1}{2}}(n,n)\right)+1\right)}{(n+1) (2 n+1)}$$ But, hoping that I am not mistaken, this tends to $1$ and not $0$.

Could you check if there is or not a typo in the definition of $g(t)$ ?

Claude Leibovici
  • 260,315