I've been trying to get the result of the integral
$$ \int_0^\infty \dfrac{x^{n}}{1+\left(x+a\right)^b}\, dx $$
as an aproximation to the Fermi integral:
$$ \int_0^\infty \dfrac{x^{n}}{e^{\beta(x-\mu)} + 1}\, dx $$
Which is a type of integral that appears in quantum statistics of gases in statistical mechanics.
If it's helpful to reach a solution, I'm interested in the result of the integral when $b \gg 1$ $a<1$, and $n$ is a fractional power and is on the order of 1.
Following a solution to the integral, $$ \int_0^\infty \dfrac{x^{n}}{1 + x^b}\, dx $$ I've tried contour integration using a pizza slice with the tip starting at the first pole of the function but haven't been successful. One of the integrals of the contour has the form: $$ \int_0^\infty \dfrac{(y-a)^{n}}{1+y^b}\, dx $$
and the fact that $n$ is a fractional power did not allow me to factor $a$ out of the integral.
To the nice general solution by @Claude Leibovici we can add asymptotics at $b\to\infty$. Let's consider the case $a\in[0;1); n>-1$. $$I=\int_0^\infty \dfrac{x^{n}}{1+\left(x+a\right)^b} dx=\int_a^\infty\frac{(x-a)^n}{1+x^b}dx=\int_{\ln a}^\infty\frac{(e^t-a)}{1+e^{bt}}dt$$ $$=\frac1b\int_{b\ln a}^\infty\frac{(e^\frac tb-a)^ne^\frac tb}{1+e^t}dt=\frac 1b\int_{b\ln a}^0\frac{(e^\frac tb-a)^ne^\frac tb}{1+e^t}dt+\frac 1b\int_0^\infty\frac{(e^\frac tb-a)^ne^\frac tb}{1+e^t}dt=I_1+I_2$$
Next, $$I_1=\frac 1b\int_{b\ln a}^0\frac{(e^\frac tb-a)^ne^\frac tb}{1+e^t}dt=\frac 1b\int_{b\ln a}^0(e^\frac tb-a)^ne^\frac tbdt-\frac 1b\int_{b\ln a}^0\frac{(e^\frac tb-a)^ne^\frac tb}{1+e^t}e^tdt$$ $$=\frac{(1-a)^{n+1}}{n+1}-\frac1b\int_0^{-b\ln a}\frac{(e^{-\frac tb}-a)^ne^{-\frac tb}}{1+e^{-t}}e^{-t}dt$$ $$=\frac{(1-a)^{n+1}}{n+1}-\frac1b\int_0^{b\ln \frac1a}\frac{(e^{-\frac tb}-a)^ne^{-\frac tb}}{1+e^t}dt$$ The first term is trivial:
at $b\to\infty\,\, (x+a)^b\to 0 \,\,\text{at}\,x<1-a \,\,\text{and}\,\,(x+a)^b\to\infty \,\,\text{at}\,x>1-a \,\,$, so we get the "Fermi's step", taking the interval of integration $[0;1-a]$.
For the second term we can extend the interval of integration $\to \infty$. We don't have to worry about the meaning $(e^{-\frac tb}-a)^n \,\,\text{at}\,\,e^{-\frac tb}<a$ - due to the factor $\frac1{e^t+1}$ all these additional terms will have the order $\sim a^b$ and are, therefore, exponentially small.
Hence, dropping exponentially small corrections, $$I=I_1+I_2\sim\frac{(1-a)^{n+1}}{n+1}+\frac1b\int_0^\infty\frac{(e^\frac tb-a)^ne^\frac tb-(e^{-\frac tb}-a)^ne^{-\frac tb}}{1+e^t}dt$$ We notice that at $t\sim\sqrt b\quad\frac tb\sim\frac1{\sqrt b}\ll1$; at the same time $\frac1{e^t+1}\sim e^{-\sqrt b}\ll1$. It means that we are allowed to decompose the nominator into the series: $e^\frac tb=1+\frac tb+...$ We see that all even powers of $t$ are dying out. $$I= \frac{(1-a)^{n+1}}{n+1}+\frac1b\int_0^\infty\frac{(1-a+\frac tb)^n(1+\frac tb)-(1-a-\frac tb)^n(1-\frac tb)+O(\frac1{b^3})}{1+e^t}dt$$ $$=\frac{(1-a)^{n+1}}{n+1}+\frac{(1-a)^n}b\int_0^\infty\frac{(1+\frac t{b(1-a)})^n(1+\frac tb)-(1-\frac t{b(1-a)})^n(1-\frac tb)+O(\frac1{b^3})}{1+e^t}dt$$ $$=\frac{(1-a)^{n+1}}{n+1}+\frac{2(1-a)^n(\frac n{1-a}+1)}{b^2}\int_0^\infty\frac t{e^t+1}dt+O\left(\frac1{b^4}\right)$$ $$\boxed{\,\,I=\frac{(1-a)^{n+1}}{n+1}+\frac{(1-a)^{n-1}(1-a+n)}{b^2}\frac{\pi^2}6+O\left(\frac1{b^4}\right)\,\,}\tag{1}$$ This approach allows to evaluate as many terms of the decomposition as we want. It is also interesting to check the answer, for example, for the case $a=0$.
On the one hand, we have from (1) $$I(a=0)=\frac1{n+1}+\frac{\pi^2(n+1)}{6b^2}+O\left(\frac1{b^4}\right)$$ On the other hand, $$\int_0^\infty \dfrac{x^{n}}{1+x^b} dx=\frac1b\int_0^\infty\frac{t^\frac nbt^\frac{1-b}b}{1+t}dt\overset{x=\frac1{1+t}}{=}\frac1b\int_0^1(1-x)^{\frac{n+1}b-1}x^{-\frac{n+1}b}dx$$ $$=\frac1bB\left(\frac{n+1}b;1-\frac{n+1}b\right)=\frac1b\Gamma\left(\frac{n+1}b\right)\Gamma\left(1-\frac{n+1}b\right)$$ $$=\frac\pi b\frac1{\sin\frac\pi b(n+1)}=\frac1{n+1}\left(1+\frac{\pi^2}6\frac{(n+1)^2}{b^2}+O\left(\frac1{b^4}\right)\right)$$ $$=\frac1{n+1}+\frac{\pi^2(n+1)}{6b^2}+O\left(\frac1{b^4}\right)$$