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I am trying to solve an integral of the type

$\int\limits_{0}^{+\infty} x^a(x^2+b)^{-c} dx$,

Where $a$, $b$ and $c$ are positive parameters. I am not sure whether this integral is doable.

But I assume that if it is, the result should be a function of the Gamma function $\Gamma(\cdot)$.

I would be really grateful if anyone provides some help or ideas.

Thank you all!

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    $\begingroup$ $\int_0^{\infty}x^a\left[x^2+b\right]^{-c}=\frac{1}{b^c}\int_0^{\infty}x^a\left[\frac{b}{x^2+b}\right]^{c}$.Substitute $u=\frac{b}{x^2+b}$. $\endgroup$ – vnd Jun 14 '17 at 14:09
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Making $x = \sqrt{b}u$ gives $$\int_{0}^{\infty} \frac{x^a}{(x^2+b)^c} dx = b^{\frac{1+a}{2}-c} \int_{0}^{\infty} \frac{u^a}{(u^2+1)^c} du$$ and a subsequent subsitution $u=\sqrt{v}$ gives $$\int_{0}^{\infty} \frac{u^a}{(u^2+1)^c} du = \frac{1}{2}\int_{0}^{\infty} \frac{v^{\frac{a-1}{2}}}{(v+1)^c} dv$$

Using the following formula for beta function solves the integral $$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = \int_{0}^{\infty} \frac{t^{x-1}}{(t+1)^{x+y}} dt $$

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  • $\begingroup$ Thank you so much for the answer, it was really helpful! :) It seems you are quite advanced with integrals, could you please give me a hand again with another one. I thought that the approach above would work again but I might be wrong or at least I don't see the analogy. Here is the link math.stackexchange.com/questions/2323013/…. Thanks! :) $\endgroup$ – Candidate Jun 14 '17 at 23:07

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