y entonces

Question

Si $\displaystyle I = \int_{0}^{1}x^{1004}\cdot (1-x)^{1004}dx$ y $\displaystyle J=\int_{0}^{1}x^{1004}\cdot \left(1-x^{2010}\right)^{1004}dx\;,$

Entonces la relación entre $I$y $J.$

$\bf{My\; Try::}$ Da $$\displaystyle J = \int_{0}^{1}x^{1004}\cdot (1-x^{2010})^{1004}dx\;,$$ Now Put $\displaystyle x^{1005}=t\;,$ Then $1005x^{1004}dx = dt$

Y límite de cambio, $$\displaystyle J=\frac{1}{1005}\int_{0}^{1}(1-t^2)^{1004}dt = \frac{1}{1005}\int_{0}^{1}\left[1-(1-t)^2\right]^{1004}dt

Así conseguimos %#% $ #%

Ahora cómo puedo resolver después de eso, ayudarme

Gracias

Answer 1

Usando su sustitución para $J$ tenemos

$$J=\frac{1}{1005}\int_{0}^{1}(1-t)^{1004}(1+t)^{1004}dt=\frac{1}{1005}\int_{0}^{1}t^{1004}(2-t)^{1004}dt$$

Ahora uso sustitución $t=2y$ obtenemos

$$J=\frac{1}{1005}\int_{0}^{\frac{1}{2}}2^{2009}y^{1004}(1-y)^{1004}dy Por lo tanto

$$J=\frac{2^{2009}}{1005}\int_{0}^{\frac{1}{2}}y^{1004}(1-y)^{1004}dy$$

Pero %#% $ #%

Así $$I=2\int_{0}^{\frac{1}{2}}y^{1004}(1-y)^{1004}dy

Answer 2

Puede ser más instructivo ver el caso general. Definir $$I(m,n) = \int_{x=0}^1 x^n (1-x)^m \, dx, \quad J(m) = \int_{x=0}^1 x^m (1-x^{2m+2})^m \, dx.$$ Then as you wrote with your choice of substitution $$u = x^{m+1}, \quad du = (m+1) x^m \, dx,$$ we find with the binomial theorem $$J(m) = \frac{1}{m+1} \int_{u=0}^1 (1-u^2)^m \, du = \frac{1}{m+1} \int_{u=0}^1 (1+u)^m (1-u)^m \, du = \frac{1}{m+1} \sum_{k=0}^m \binom{m}{k} I(m,k).$$ Now we observe through integration by parts with the choice $u = (1-x) ^ m$, $du = -m (1-x) ^ {m-1} \, dx$, $dv = x ^ n \, dx$, $v = \frac{1}{n+1} x^{n+1}$, that $$I(m,n) = \left[-\frac{1}{n+1} x^{n+1} (1-x)^m \right]_{x=0}^1 + \frac{m}{n+1} \int_{x=0}^1 x^{n+1} (1-x)^{m-1} \, dx = \frac{m}{n+1} I(m-1,n+1).$$ With the addition that $I(0,n) = \frac{1}{n+1}$ trivially, we can inductively show that $I (m, n)$ for nonnegative integers $m, n$ is given by $$I(m,n) = \frac{m! \, n!}{(m+n+1)!}.$$ Therefore, $$J(m) = \frac{1}{m+1} \sum_{k=0}^m \frac{m!}{k!(m-k)!} \frac{m! \, k!}{(m+k+1)!} = \frac{1}{m+1} \sum_{k=0}^m \frac{(m!)^2}{(m-k)! (m+k+1)!}$$ and $%$ $\frac{J(m)}{I(m,m)} = \frac{1}{m+1} \sum_{k=0}^m \binom{2m+1}{m-k} = \frac{1}{m+1} \sum_{k=0}^m \binom{2m+1}{k} = \frac{4^m}{m+1},$la identidad última de que tengo a la izquierda como un simple ejercicio.