How to evaluate $\int^{\infty}_0 \frac{x^{1010}}{(1 + x)^{2022}} dx$? 
How to evaluate the following integral? $$\int^{\infty}_0 \frac{x^{1010}}{(1 + x)^{2022}} dx$$


Here's my work:
$$\begin{align}I &= \int_0^\infty \dfrac{x^{1010}}{(1+x)^{2022}} dx  \\&=\int_0^\infty \dfrac{1}{x^{1012}(1 + \frac1x)^{2022}}dx\end{align}$$
Putting $( 1 + \frac1x) = t$
$$\begin{align}\implies I& =\int^1_\infty -\dfrac{1}{(\frac1{1-t})^{1010}(t)^{2022}}dx\\& =\int_1^\infty \dfrac{1}{(\frac1{1-t})^{1010}(t)^{2022}}dx \\&=\int_1^\infty \dfrac{1}{(\frac1{1-t})^{1010}\cdot t^{1010} \cdot (t)^{1012}}dx \\&=\int_1^\infty \dfrac{1}{(\frac t{1-t})^{1010} \cdot (t)^{1012}}dx\\& =\int_1^\infty \dfrac{1}{(\frac 1{1/t-1})^{1010} \cdot (t)^{1012}}dx \\&=\int_1^\infty \dfrac{(1/t-1)^{1010}}{ (t)^{1012}}dx \\&=\int_1^\infty \dfrac{(\frac{1-t}{t})^{1010}}{ t^2\cdot (t)^{1010}}dx \\& = \int_1^\infty\dfrac{1}{t^2} \cdot\left( \dfrac{1-t}{t^2}\right)^{1010} dx \end{align} $$

I don't know how to continue from here. I also thought that Integration by parts would work but not sure how to apply here.
 A: We have,
$$\begin{align}\int_{0}^{\infty} \frac {x^{1010} }{ (1+x)^{2022}}\,dx \\&=\int_{0}^{\infty} \frac {x^{1011 - 1} }{ (1+x)^{1011 + 1011}}\,dx\tag{1}\\&= B(1011, 1011)\tag{2}\\& = \dfrac{\Gamma(1011)\Gamma(1011)}{\Gamma{(2022)}}\tag{3} \\& = \dfrac{(1010!)^2}{2021!}\tag{4}\end{align}$$

Explanation:
$(1.)$, $(2.)$
Expressed the given integral in terms of  beta function
$$B(x,y) = \int_0^\infty \dfrac{t^{x-1}}{(1 + t)^{x+y}} dt$$
$(3.)$ Used gamma-beta relationship
$$B(x,y) = \dfrac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$$
$(4.)$ Relation between gamma function and factorials.
$$\Gamma(x+1) = x!$$
A: I am going to evaluate the integral in general
$$
I(m, n):=\int_{0}^{\infty} \frac{x^{m}}{(1+x)^{n}} d x
$$
by a reduction formula of $I(m,n)$.
$$
\begin{aligned}
I(m, n)&=-\frac{1}{n-1} \int_{0}^{\infty} x^{m} d\left(\frac{1}{(1+x)^{n-1}}\right)\\
&=-\frac{1}{n-1}\left[\frac{x^{m}}{(1+x)^{n-1}}\right]_{0}^{\infty}+\frac{m}{n-1} \int_{0}^{\infty} \frac{x^{m-1}}{(1+x)^{n-1}} d x\\
&=\frac{m}{n-1} I(m-1, n-1)
\end{aligned}
$$
Applying the formula repeatedly by $m$ times yields
$$
I(m, n)=\frac{m}{n-1} \cdot \frac{m-1}{n-2} \cdot \cdots \frac{1}{n-m} \int_{0}^{\infty} \frac{1}{(1+x)^{n-m}} d x=\frac{m !}{(n-1) (n-2) \cdots (n-m) (n-(m+1))}
$$
In particular,
$$
I(1010,2022)=\frac{1010 !}{2021 \cdot 2020 \cdots 1011}=\frac{(1010!)^2}{2021!}
$$
A: $$1+x=t$$ $$\int_0^\infty\frac{(x)^{1010}}{{(x+1)}^{2022}}dx=\int_1^\infty\frac{(t-1)^{1010}}{t^{2022}}dt=\frac{1010}{2021}\int_1^\infty\frac{(t-1)^{1009}}{t^{2021}}dt=\cdots=\frac{{1010}!}{2021\cdots1011}$$*Integration by parts
A: With trigonometric form $$2\int_0^{\pi/2}\sin^{2p-1}\theta\cos^{2q-1}\theta d\theta=B(p,q)$$ of Beta function:
Let $x=\tan^2\theta$ then
$$\int_0^\infty\frac{x^{1010}}{(x+1)^{2020}}dx=2\int_0^{\pi/2}\sin^{2021}\theta\cos^{2021}\theta d\theta=B(1011,1011)=\frac{(1010!)^2}{2021!}.$$
A: Another beta function manipulation:
$$\begin{align*}
I &= \int_0^{\infty} \frac{x^{1010}}{(1 + x)^{2022}} \, dx \\[1ex]
&= 2 \int_0^1 \frac{x^{1010}}{(x+1)^{2022}} \, dx \tag{1} \\[1ex]
&= \frac1{2^{2021}} \int_0^1 (1-x)^{1010} x^{-\frac12} \, dx \tag{2} \\[1ex]
&= \frac1{2^{2021}} \operatorname{B}\left(1011,\frac12\right) \tag{3} \\[1ex]
&= \operatorname{B}(1011,1011) \\[1ex]
&= \frac{(1010!)^2}{2021!}
\end{align*}$$


*

*$(1)$ : split the integral at $x=1$, and substitute $x\mapsto\frac1x$ for $x\in[1,\infty)$

*$(2)$ : substitute $x\mapsto\frac{1-x^2}{1+x^2}$

*$(3)$ : definition of beta function

