Evaluate: $\int_{0}^{\infty} x^{a} (1+x)^{b} dx$

Question

I want to express \begin{align} \int_{0}^{\infty} x^{a} (1+x)^{b}~ dx \end{align} with Gamma functions.

Recall that the beta function is defined as \begin{align} B(z_1, z_2) = \int_0^1 t^{z_1-1} (1-t)^{z_2-1} dt = \frac{\Gamma(z_1) \Gamma(z_2)}{\Gamma(z_1+z_2)} \end{align} and it seems the above form also has the expressions with Gammas[The above integral comes out of some derivations of Feynman loop integrals and the results are all written in terms of Gamma]

Answer 1

Note that beta function can be written as the following form,

$$B(x,y)=\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}dt$$

For you integral,

$$\int_{0}^{\infty} x^{a} (1+x)^{b}dx =\int_{0}^{\infty} \frac{ t^{a}}{ (1+t)^{-b}}dt$$

Compare the parameters

$$x-1=a,~~~x+y=-b$$

Solve it, we get

$$x=a+1,~~~y=-a-b-1$$

therefore,

$$\int_{0}^{\infty} x^{a} (1+x)^{b}dx =B(a+1,-a-b-1)=\frac{\Gamma(a+1)\Gamma(-a-b-1)}{\Gamma(-b)}$$