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.

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

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

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