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How to prove that following integral uniformly converge on [$\alpha_0, +\infty$], $\alpha_0 > 0$ $$I(\alpha ) = \int_0^{\infty}e^{-\alpha x^4}dx$$ Any tips, please.

How it can prove that integral uniformly converge ? $$\int_{0}^{+\infty} e^{-\alpha z^4}\,dz = \frac{\Gamma\left(\frac{5}{4}\right)}{\alpha^{1/4}}.$$

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$I(\alpha)$ is a fixed multiple of $\frac{1}{\alpha^{1/4}}$ by the change of variable $x=\frac{z}{\alpha^{1/4}}$:

$$ \int_{0}^{+\infty} e^{-\alpha z^4}\,dz = \frac{\Gamma\left(\frac{5}{4}\right)}{\alpha^{1/4}}.$$

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    What is the relationship between result of integral and uniform convergence ? – 2017-02-25
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    @marka_17: you should tell this to us, since I see no sequence in your problem, so I wonder what is the subject of such uniform convergence. (https://en.wikipedia.org/wiki/Uniform_convergence) – 2017-02-25