# Evaluating the Integral $\int_{0}^{\infty} \frac{x^{49}}{(1+x)^{51}} dx$

I tried evaluating the integral $$\displaystyle \int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx$$ but I wasn't able to get the result. Following is the way by which I did it- $$I=\displaystyle \int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx$$ $$\implies I=\int_{0}^{\infty} x^{49}(1+x)^{-51}dx$$ Further, I tried Integration by parts but it didn't worked. Can anyone tell that how this integral can be evaluated.

• HINT: try substitution $$u= \frac{x}{1+x}$$ Aug 18, 2021 at 3:21
• sorry, @Mathzcreator, but larger titles are discouraged on math.SE so that your question does not take up more vertical space than others, so I've edited the title. See this discussion on the meta site: math.meta.stackexchange.com/questions/9687/… Aug 18, 2021 at 3:59
• Note $$\int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx \overset{x\to\frac1x}= \int_{0}^{\infty} \dfrac{1}{(1+x)^{51}}dx =\frac1{50}$$ Feb 23, 2022 at 19:10

$$I=\displaystyle\int_{0}^{\infty}\dfrac{x^{49}}{(1+x)^{51}}dx$$ $$I=\displaystyle\int_{0}^{\infty}\dfrac{x^{50-1}}{(1+x)^{50+1}}dx$$ Using Beta Function,

$$B(x,y)=\displaystyle\int_{0}^{\infty}\dfrac{t^{x-1}}{(1+t)^{x+y}}dt$$

$$\implies I=\displaystyle\int_{0}^{\infty}\dfrac{x^{50-1}}{(1+x)^{50+1}}dx=B(50,1)$$ Using Beta Function and Gamma Function Relationship,

$$B(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

$$\implies I=B(50,1)=\dfrac{\Gamma(50)\Gamma(1)}{\Gamma(50+1)}=\dfrac{\Gamma(50)\Gamma(1)}{50\Gamma(50)}=\dfrac{\Gamma(1)}{50}=\dfrac{1}{50}$$ $$\implies \boxed{I=\dfrac{1}{50}}$$

The given integral is

$$I=\displaystyle \int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx$$

$$I=\displaystyle \int_{0}^{\infty} \dfrac{x^{51}}{x^2(1+x)^{51}}dx$$

$$I=\displaystyle \int_{0}^{\infty} \dfrac{1}{x^2(1+\frac{1}{x})^{51}}dx$$

Let $$u=1+\frac{1}{x}$$ , therefore $$\displaystyle du=-\frac{dx}{x^2}$$

Therefore $$I=\displaystyle \int_\infty^1-\frac{du}{u^{51}}=\frac{1}{50}\bigg[\frac{1}{u^{50}}\bigg]^{1}_{\infty} = \boxed{\frac{1}{50}}$$

Here is an alternative method.

First of all, for each integer $$n>1$$, we have $$\int_1^\infty x^{-n}dx=\frac1{n-1}$$.

Thus, using the binomial theorem, \begin{align} \int_0^\infty\frac{x^{49}}{(1+x)^{51}}dx&=\int_1^\infty\frac{(x-1)^{49}}{x^{51}}dx\\ &=\int_1^\infty\frac1{x^{51}}\sum_{n=0}^{49}(-1)^{n-1}{49\choose n}x^ndx\\ &=\sum_{n=0}^{49}(-1)^{n-1}{49\choose n}\int_1^\infty x^{n-51}dx\\ &=\sum_{n=0}^{49}(-1)^{n-1}{49\choose n}\frac1{50-n}\\ &=\frac1{50}\sum_{n=0}^{49}(-1)^{n-1}{50\choose n}\\ &=-\frac1{50}\left((1+(-1))^{50}-{50\choose50}(-1)^{50}\right)\\ &=\frac1{50}. \end{align}