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I have a problem with this integral, I dont know what method to use to solve it. For which values of $a$ is the integral converges? $$\int _2^{\infty }\:\:\frac{\sqrt{x+3}}{\left(x^2-2x\right)^a}dx$$

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closed as off-topic by 5xum, Em., BLAZE, SchrodingersCat, Kamil Jarosz Jan 20 '16 at 11:01

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  • $\begingroup$ When $\frac{3}{4}\space<\Re(a)\space<1$ $\endgroup$ – Jan Jan 20 '16 at 9:23
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    $\begingroup$ how did you get it? $\endgroup$ – user12 Jan 20 '16 at 9:24
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    $\begingroup$ @JanEerland. Why ? Could you elaborate ? $\endgroup$ – Claude Leibovici Jan 20 '16 at 9:24
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(1) Split the integral into two pieces:

$$\int_2^\infty f(x) dx = \int_2^3 f(x)dx + \int_3^\infty f(x)dx$$

(2.a) See when the first piece converges:

$$f(x) = \frac{\sqrt{x+3}}{(x^2-2x)^a} = \frac{\sqrt{x+3}}{x^a}\cdot \frac{1}{(x-2)^a}$$

The first function is bounded and non-zero on the interval $[2,3]$, meaning that $\int_2^3 f(x) dx$ converges if and only if $$\int_{2}^{3}\frac{1}{(x-2)^a}$$ converges.

(2.b) See where the second piece converges:

$$f(x) = \frac{\sqrt{x+3}}{(x^2-2x)^a} = \frac{\sqrt{1+\frac3{x}}\cdot \sqrt x}{(1-\frac{2}{x})^a \cdot x^{2a}}$$

Can you continue?

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