TITLE: A problem about the convergence of an improper integral
QUESTION [4 upvotes]: Let $f:\mathbb R\longrightarrow\mathbb R$ be a function with 
$$f(x)=\frac{1}{3}\int_x^{x+3} e^{-t^2}dt$$
and consider  $g(x)=x^nf(x)$ where $n\in\mathbb Z$. 

I have to discuss the convergence of the integral 
  $$\int_{-\infty}^{+\infty}g(x)dx$$ at the varying of $n\in\mathbb Z$.

Any idea? We have an integral of an "integral function", and I'm a bit confused!
Thanks in advance.

REPLY [0 votes]: I interpret the question as follows: is the function $g$ integrable (in the Lebesgue sense) on $\mathbb R$? And my answer is essentially the same as TZakrevskiy's.
The function $f$ is continuous on $\mathbb R$, so $g$ is continuous on $\mathbb R\setminus\{ 0\}$ (and continuous on $\mathbb R$ is $n\geq 0$). We have to check when $g$ is integrable at $\pm\infty$ and integrable around $0$.
By the definition of $f$, we have $\vert f(x)\vert\leq e^{-x^2}$
for $x>0$ and $\vert f(x)\vert\leq e^{-(x+3)^2}$ for $x\leq -3$. It follows that 
$x^{n+2}f(x)\to 0$ as $\vert x\vert\to \infty$, so $g(x)=x^nf(x)=0(\frac1{x^2})$ as $\vert x\vert\to\infty$ and there is no integrability problem for $g$ at $\pm\infty$, for any $n\in\mathbb Z$. 
So the question is whether $g$ is integrable around $0$ or not. 
Now, $f$ is continuous with $f(0)=\int_0^3 e^{-t^2}dt\neq 0$. So $f(x)\sim f(0)$ as $x\to 0$ and $g(x)\sim f(0) x^n$. Hence, $g$ is integrable around $0$ if $n\geq 0$ and not integrable around $0$ if $n<0$.
In summary: $g$ is integrable on $\mathbb R$ if and only if $n\geq 0$.
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