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Is there a closed-form expression for $$I = \int_{0}^{\infty}{(1+ax^{-\alpha})^{-b}}\mathrm{d}x,$$ Here, $a > 0$, $\alpha > 2$, and $b > 0$? If yes, how to get it?

I tried evaluating the expression using MATLAB, e.g., for $a = 2.3$, $\alpha = 4$, and $b = 2.2$, MATLAB gives the answer $\infty$.

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    Since the integral is divergent for α>0 we use a common procedure to try to give it a sense: analytic continuation. – 2017-01-02
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    See my solution. – 2017-01-03

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Since the integrand converges to $1$ at infinity, this seems to diverge.

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    Does diverge mean we cannot get a closed-form? – 2017-01-01
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    Divergence means that the closed form is $\infty.$ If you are happy with that, good. – 2017-01-01
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    @Mark In fact, the function $f = (1+ax^{-\alpha})^{-b}$ is always smaller than $1$, and it converges to $1$ as $x \to \infty$. So your comment is unclear to me. Am I missing something? – 2017-01-02
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    @Rohit My mistake. – 2017-01-02
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    @Mark That's fine. Can you still please elaborate it intuitively? It is still unclear to me the statement "Since the integrand converges to 1 at infinity, this seems to diverge." – 2017-01-02
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    @Rohit the big deal is that the integrand converges to a number $\neq 0$. Because of this, we know for big enough $x$ that $|f(x) - 1|<\epsilon$. It follows that $0<1-\epsilon$f(x)$ is always greater than some positive constant. But, $\int_N^\infty 1-\epsilon = \infty$, so $\int_N^\infty f(x) \geq \infty$. – 2017-01-02
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    It's possible that the indefinite integral $\int (1+ax^{-\alpha})^{-b} \, dx$ has a closed form, but the definite integral from $0$ to $\infty$ always diverges in the same way that the antiderivative of $x$ is $\frac{x^2}{2}+C$, but its definite integral from $0$ to $\infty$ is $\infty$ – 2017-01-02
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Since the integral is divergent for $\alpha>0$ we use a common procedure to try to give it a sense: analytic continuation. That is we calculate the integral for paramter values such that the integral exists and then consider the result as valid also beyond these parameter regions.

Mathematica finds this closed expression for the integral

$$f(a, b, \alpha)=\frac{a^{1/\alpha } \Gamma \left(\frac{\alpha -1}{\alpha }\right) \Gamma \left(b+\frac{1}{\alpha }\right)}{\Gamma (b)}$$

For the parameters $a = 23/10, b = 22/10, \alpha = 4$ we get $f = 1.75893$