About integral involving power factors. Maybe this integral is easy...but so far I have not been able to solve it:
\begin{equation}
\int_0^\infty y^{-b} (y+a)^{1-b} \, dy
\end{equation}
where $b$ is a positive constant. 
I've tried to use subtitution. Re-writing
\begin{equation}
y^{-b} (y+a)^{1-b} = \frac{y+a}{(y^2+ay)^b}
\end{equation}
By doing $y^2+ay = u^{1/b}$...I almost got it...but it failed at the end. Note that:
\begin{equation}
2y \, dy + a \, dy = \frac{1}{b}u^{\frac{1-b}{b}} \, du
\end{equation}
Any help is appreciated...thanks
 A: This integral is always divergent.
Proof. Assume $a>0$. The integrand being continuous over any compact $[b_1,b_2]$, potential issues are just as $y \to 0^+$ and as $y \to \infty$.
As $y \to 0^+$, one has
$$
y^{-b} (y+a)^{1-b} \sim \frac{a^{1-b}}{y^b}
$$ 

giving a convergent integral iff $\,b<1$.

As $y \to \infty$, one has
$$
y^{-b} (y+a)^{1-b}=y^{-b}y^{1-b} \left(1+\frac{a}y\right)^{1-b} \sim \frac1{y^{2b-1}}
$$ 

giving a convergent integral iff $\,2b-1>1$ that is iff $\,b>1$.

Remark. The case $a=0$ is clear.
A: Assuming $a>0$, by substituting $y=ax$ we get
$$ \int_{0}^{+\infty}y^{-b}(y+a)^{1-b}\,dy = a^{2-2b} \color{blue}{\int_{0}^{+\infty}x^{-b}(x+1)^{1-b}\,dx} $$
but the blue integral is not converging. The integrand function behaves like $x^{-b}$ in a right neighbourhood of the origin, hence in order to grant integrability there we must have $b<1$. In such a case, however, the integrand function behaves like $x^{1-2b}$ in a left neighbourhood of $+\infty$, that is not integrable there.
A: \begin{align}
u & = \frac 1 {1+y} \\[8pt]
dy & = \frac{-du}{u^2}
\end{align}
\begin{align}
\int_0^\infty y^{-b} (y+a)^{1-b} \, dy & = \int_1^0 \left( \frac {1-u} u \right)^{-b} \left( \frac{1-u} u + a \right)^{1-b} \left( \frac{-du}{u^2} \right) \\[10pt]
& = \int_0^1 \frac {(1-u)^{-b}( 1 - (1-a) u )^{1-b} \, du} {u^4}.
\end{align}
I'd think about comparing this with $\displaystyle \int_0^\varepsilon \frac{du}{u^4}$.
