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

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

$\implies I=B(50,1)=\dfrac{\Gamma(50)\Gamma(1)}{\Gamma(50+1)}=\dfrac{\Gamma(50)\Gamma(1)}{50\Gamma(50)}=\dfrac{\Gamma(1)}{50}=\dfrac{1}{50}$
$$\implies \boxed{I=\dfrac{1}{50}}$$
A: 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 $$I=\displaystyle \int_\infty^1-\frac{du}{u^{51}}=\frac{1}{50}\bigg[\frac{1}{u^{50}}\bigg]^{1}_{\infty} = \boxed{\frac{1}{50}}$$
