Encontrar la relación de $\frac{\int_{0}^{1} \left(1-x^{50}\right)^{100} dx}{\int_{0}^{1} \left(1-x^{50}\right)^{101} dx}$

Question

$$I_1=\int_{0}^{1} \left(1-x^{50}\right)^{100} dx$$ and $$I_2=\int_{0}^{1} \left(1-x^{50}\right)^{101} dx$$ Then find $\frac{I_1}{I_2}$

He intentado restando $I_1$ $I_2$

$$I_1-I_2=\int_{0}^{1}\left(1-x^{50}\right)^{100}\left(1-(1-x^{50}\right))dx$$ lo

$$I_1-I_2=\int_{0}^{1} \left(1-x^{50}\right)^{100} x^{50} dx$$ Ahora usando Integración por Partes obtenemos

$$I_1-I_2= \left(1-x^{50}\right)^{100} \times \frac{x^{51}}{51}\bigg|_{0}^{1} -\int_{0}^{1} 100 \left(1-x^{50}\right)^{99} \times -50 x^{49} \times \frac{x^{51}}{51} dx$$ Así $$I_1-I_2=\frac{5050}{51} \times \int_{0}^{1}\left(1-x^{50}\right)^{99} x^{100} dx$$

Ahora$x^{100}=\left(1-x^{50}\right)^2-(1-x^{50}-x^{50})$, por lo que

$$\frac{51}{5050}(I_1-I_2)=\int_{0}^{1} \left(1-x^{50}\right)^{99} \times \left(\left(1-x^{50}\right)^2-(1-x^{50})+x^{50}\right)dx=I_2-I_1+\int_{0}^{1} \left(1-x^{50}\right)^{99} x^{50} dx$$

Necesidad de una ayuda para seguir adelante.

Answer 1

Sugerencia: ¿integración por partes en $I_2$ primero, y luego considerar la $I_1-I_2$.

Answer 2

Factoring el integrando da $$ \int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x =\int_0^1\left(1-x^{50}\right)^{100}\,\mathrm{d}x -\int_0^1\left(1-x^{50}\right)^{100}x^{50}\,\mathrm{d}x $$ Integración por partes da $$ \int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x =101\cdot50\int_0^1\left(1-x^{50}\right)^{100}x^{50}\,\mathrm{d}x $$ La combinación de rendimientos $$ \frac{\int_0^1\left(1-x^{50}\right)^{100}\,\mathrm{d}x}{\int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x}=\frac{5051}{5050} $$

Answer 3

Otro enfoque es la nota que $$\int_0^1\left(1-x^a\right)^bdx=\int_0^1\left(1-y\right)^b\tfrac{1}{a}y^{1/a-1}dy=\frac{1}{a}\text{B}\left(\frac{1}{a},\,b+1\right)=\frac{b!\Gamma\left(\tfrac{1}{a}+1\right)}{\Gamma\left(b+\tfrac{1}{a}+1\right)},$$ so $$\frac{\int_0^1\left(1-x^{50}\right)^{100}dx}{\int_0^1\left(1-x^{50}\right)^{101}dx}=\frac{\Gamma\left(102.02\right)}{101\Gamma\left(101.02\right)}=\frac{5051}{5050}.$$

Answer 4

Aviso, cuando $\Re[n]>0$:

• $$\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^n\space\text{d}x=\frac{\Gamma(n+1)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(1+\frac{2}{n}+n\right)}$$
• $$\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^{n+1}\space\text{d}x=\frac{\Gamma(n+2)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(2+\frac{2}{n}+n\right)}$$

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Así:

$$\frac{\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^n\space\text{d}x}{\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^{n+1}\space\text{d}x}=\frac{\frac{\Gamma(n+1)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(1+\frac{2}{n}+n\right)}}{\frac{\Gamma(n+2)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(2+\frac{2}{n}+n\right)}}=\frac{2+n+n^2}{n+n^2}$$

Ahora, cuando $n=100$:

$$\frac{\int_{0}^{1}\left(1-x^{\frac{100}{2}}\right)^{100}\space\text{d}x}{\int_{0}^{1}\left(1-x^{\frac{100}{2}}\right)^{100+1}\space\text{d}x}=\frac{2+100+100^2}{100+100^2}=\frac{10102}{10100}=\frac{5051}{5050}$$