I was working on a exercise where the goal was to compute the following limit,
$$\lim_{n\rightarrow\infty}\int_{\mathbb R}e^{-|x|n}e^{-\frac{x^2}{2}}dx$$ and some tutor of mine claimed that the limit is infinity. The problem is that i thought that I was able to show that the limit is $0$ and I can not find my mistake. Thus I hoped someone could have a quick look over my argumentation.
Here it is,
First the following holds $\forall x\in\mathbb R$: $$\lim_{n\rightarrow\infty}e^{-|x|n}e^{-\frac{x^2}{2}}=\chi_{\{0\}}(x)$$ Furthermore we have $e^{-|x|n}\in L^{\infty}( \mathbb R)$ and consequently, $$|e^{-|x|n}e^{-\frac{x^2}{2}}|\leq e^{-\frac{x^2}{2}}$$ where $e^{-\frac{x^2}{2}}\in L^1(\mathbb R)$. Hence by the Dominated convergence Theorem it follows, $$\lim_{n\rightarrow\infty}\int_{\mathbb R}e^{-|x|n}e^{-\frac{x^2}{2}}dx=\int_{\mathbb R}\lim_{n\rightarrow\infty}e^{-|x|n}e^{-\frac{x^2}{2}}dx=\int_{\mathbb R} \chi_{\{0\}}=0$$
Thanks in advance!
the integral of a characteristic function is the measure of the set. in this case the set is $\{ 0 \}$ which has $0$ Lebesgue measure, hence the answer. (you're right, your tutor was wrong) sorry, I initially thougt you concluded sth else from $\int \chi_0$, I don't know why. nevertheless your answer is correct. it's easy to see this sequence is non increasing and as you've notices the integral with $n=0$ is finite, so the limit can by no means be $\infty$ and as you've shown it is in fact $0$