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

Question

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.

Answer 1

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

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

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

Answer 2

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

Answer 3

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}