วิธีการทำโดยไม่ใช้ Beta Function กับ Gamma Function ครับ
Let $$D(k) = \frac{1}{k+1}\int_{0}^{1}x^{k}(1-x^{2(k+1)})^k\, dx$$
$$D(k) = \frac{1}{2(k+1)}\int_{-1}^{1}x^{k}(1-x^{2(k+1)})^k\, dx$$
$$D(k) = \frac{2^{2k-1}}{k+1}\int_{-1}^{1}x^{k}(\frac{1-x^{2(k+1)}}{4})^k \, dx$$
$$D(k) = \frac{2^{2k-1}}{k+1}\int_{-1}^{1}x^{k}(\frac{1-x^{k+1}}{2})^k(\frac{1+x^{k+1}}{2})^k \, dx$$
$$u = \frac{1+x^{k+1}}{2}$$
$$D(k) = \frac{2^{2k}}{k+1}\int_{0}^{1} u^k(1-u)^k \, \frac{du}{k+1}$$
$$D(k) = \frac{2^{2k}}{(k+1)^2} N(k)$$
$$\frac{N(k)}{D(k)} = \frac{(k+1)^2}{2^{2k}}$$
we get $$\frac{2^{2k+1}\int_{0}^{1}x^{k}(1-x)^{k} \, dx}{\int_{0}^{1}x^{k}(1-x^{2k+2})^{k} \, dx} = \frac{2^{2k+1}N(k)}{(k+1)D(k)} = 2(k+1)$$
substitute $k=1004$ , we get $\frac{2^{2(1004)+1}N(1004)}{(1004+1)D(1004)} = 2010$
