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Prove uniform convergence $$\int \limits_2^{+\infty} \frac{1}{x\ln(x)^{\alpha}}dx$$ Where $ \alpha \in [\alpha_0, +\infty), \alpha_0 > 1 $

I have made

$$ \int \limits_2^{+\infty} \frac{1}{x\ln(x)^{\alpha}}dx = \int \limits_2^{+\infty}\frac{d\ln x}{\ln(x)^{\alpha}} $$ let $t = \ln x $ $$ \int \limits_{\ln2}^{+\infty}\frac{d t}{t^{\alpha}} =\int \limits_{\ln2}^{e} \frac{d t}{t^{\alpha}} + \int \limits_{e}^{+\infty} \frac{d t}{t^{\alpha}} $$

It is clear that second part has uniform convergence by Weierstrass, but I stuck with the first part. How to explore it?

Uniform convergence of I(y) = $\int \limits_a^{+\infty} f(x,y)dx$ on the segment $[c,d]$

$\forall \epsilon > 0 \exists A(\epsilon) \geq a : \forall R > A$ and $\forall y \in [c; d] \Rightarrow |\int \limits_R^{+\infty} f(x, y)dx| < \epsilon$

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    We must have $\alpha_0>1$. – 2017-02-26
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    Yes, you are right, fixed – 2017-02-26
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    What do you mean by *proving the uniform convergence* of a parametric integral? *Uniform convergence* is a concept arising in sequences of functions, and here we have a single function (of $\alpha$). – 2017-02-26
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    @JackD'Aurizio I have added definition from my course – 2017-02-26
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    @Desh: where is *uniformity* in such definition? It looks like it is simply stating that the involved parametric integral is converging for any value of the parameter in some range. Additionally, such definition has to do with parameters in closed intervals, that is not the case of the presented integral. – 2017-02-26
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    @JackD'Aurizio May be there are problems in conventional termins. Yes, segment is not only closed intervals – 2017-02-26

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Using the following definition of uniformly convergent parametric integral $$\forall\varepsilon > 0\; \exists\,A(\varepsilon): \forall R>A, \forall y\in I, \left|\int_{R}^{+\infty}\,f(x,y)\,dx\right|<\varepsilon $$ it is enough to check that

$$ \int_{\exp(k\alpha_0)}^{+\infty}\frac{dx}{x\log(x)^{\alpha}} =\int_{k\alpha_0}^{+\infty}\frac{dz}{z^{\alpha}}=\frac{(k\alpha_0)^{1-\alpha}}{\alpha-1}\leq\frac{\alpha_0^{1-\alpha_0}}{\alpha_0-1}\,k^{1-\alpha}\leq C_0 k^{1-\alpha_0}$$ so a suitable choice of $k$ in terms of $C_0$ ensures uniformity $\forall \alpha\in I = [\alpha_0,+\infty)$.

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    Thank you, but why it is enough to check? – 2017-02-26
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    @Desh: because such inequality gives a way for defining $A(\varepsilon)$ in the definition. – 2017-02-26