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.
Best Answer
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. $$\Gamma(x+1) = x!$$