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My homework is :

For $a>0$ and $x\in \Bbb R^+ $ find $s \in \Bbb R$ such that $$\int_0^\infty x^s e^{-ax^2}dx$$ is Lebesgue integrable and than calculate it.

I have problems with the understanding of Lebesgue integral, that's what I've done so far, someone can just tell me what I have to do.

Let $f(x)=x^s e^{-ax^2}=\frac{x^s}{e^{ax^2}}$ than$$f(x)= \begin{cases} \frac{x^s}{e^{ax^2}}, & \text{if $s>0$ } \\ \frac{1}{e^{ax^2}}, & \text{if $s=0$ } \\ \frac{1}{x^se^{ax^2}}, & \text{if $s<0$ } \end{cases}$$

In all cases $f(x)<1$, so it's restricted and $f(x)$ is postive.

For $s<0$ there's a discontinuity at $x=0$ so the Null set is $\{0\}$, so Lebesgue and Riemann Integral should be the same here. So $$\int_0^\infty x^s e^{-ax^2}dx=lim_{_R\to \infty}\int_0^R x^s e^{-ax^2}dx=lim_{_R\to \infty}\begin{cases} [\frac{x^s}{e^{ax^2}}]_0^R, & \text{if $s>0$ } \\ \frac{erf(x\sqrt a )\sqrt \pi}{2\sqrt a}, & \text{if $s=0$ } \\ -\frac{1}{2}x^{1-s}(ax^2)^{(s-1)/2}\Gamma, & \text{if $s<0$ } \end{cases}$$

Edit: we know that continuous , non-negative functions on open intervalls, which are riemann integrable are also lebesgue.

For $s>0$ is $f(x)$ continuous, but as mentioned for $s=-1$ we have $\frac{1}{xe^{ax^2}}$ and this is not continuous in $x=0$ so it can't be lebesgue, yes ?

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    If $s=-1$ then look at $f$ near zero. It is not integrable (nor upper bounded by one). – 2017-01-22
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    @copper.hat: can I conclude this for all $s<-1$ ? – 2017-01-24
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    What do you think? Is $x^s$ integrable on $[0,\delta)$ for $s<-1$. – 2017-01-24
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    @cooper.hat: no, it isn't – 2017-01-24
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    Then use comparison to show that $ x \mapsto x^s e^{-ax^2}$ is not Lebesgue integrable whenever $s \le -1$. – 2017-01-24

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