Find the ratio of $\frac{\int_{0}^{1} \left(1-x^{50}\right)^{100} dx}{\int_{0}^{1} \left(1-x^{50}\right)^{101} dx}$

Hint: Do integration by parts on $I_2$ first, then consider $I_1-I_2$.

Factoring the integrand gives $$ \int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x =\int_0^1\left(1-x^{50}\right)^{100}\,\mathrm{d}x -\int_0^1\left(1-x^{50}\right)^{100}x^{50}\,\mathrm{d}x $$ Integration by parts gives $$ \int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x =101\cdot50\int_0^1\left(1-x^{50}\right)^{100}x^{50}\,\mathrm{d}x $$ Combining yields $$ \frac{\int_0^1\left(1-x^{50}\right)^{100}\,\mathrm{d}x}{\int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x}=\frac{5051}{5050} $$

Another approach is to note that $$\int_0^1\left(1-x^a\right)^bdx=\int_0^1\left(1-y\right)^b\tfrac{1}{a}y^{1/a-1}dy=\frac{1}{a}\text{B}\left(\frac{1}{a},\,b+1\right)=\frac{b!\Gamma\left(\tfrac{1}{a}+1\right)}{\Gamma\left(b+\tfrac{1}{a}+1\right)},$$ so $$\frac{\int_0^1\left(1-x^{50}\right)^{100}dx}{\int_0^1\left(1-x^{50}\right)^{101}dx}=\frac{\Gamma\left(102.02\right)}{101\Gamma\left(101.02\right)}=\frac{5051}{5050}.$$

Notice, when $\Re[n]>0$:

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

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So:

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

Now, when $n=100$:

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