In my book it is shown how to compute the following Lebesgue integral:
$$\int e^{-x^2} =\sqrt{\pi}$$
I want to show that this result is equivalent to
$$\int e^{-x^2/2} =\sqrt{2\pi}$$
Up to now we haven't proven any change of variable formula, but we have proven the following:
Proposition. Let $f$ be a Borel measurable function which is Riemann integrable over every compact interval. Then $f$ is Lebesgue integrable over $\mathbb{R}$ if and only if $\lim_{n\to\infty}\int_{-n}^{n}|f(x)|dx $ exist in $\mathbb{R}$. In this case $\int f =\lim_{n\to\infty}\int_{-n}^{n}f(x)dx $.
Using this with $f(x)=e^{-x^2}$ and applying the change of variable rule from calculus I get
$$\sqrt{\pi}=\int e^{-x^2} =\lim_{n\to\infty}\int_{-n}^{n}e^{-x^2}dx = \frac{1}{\sqrt{2}} \lim_{n\to\infty}\int_{-\sqrt{2}n}^{\sqrt{2}n}e^{-x^2/2}dx = \frac{1}{\sqrt{2}} \lim_{n\to\infty}\int_{-n}^{n}e^{-x^2/2}dx$$
which shows, by the previous proposition, that $\int e^{-x^2/2} $ exists and equals $\sqrt{2\pi}$. The converse implication is similar.
This reasoning made use of the fact that $e^{-x^2}$ is a non-negative function because in general the convergence of $\int_{-n}^{n}f(x)dx$ does not imply Lebesgue integrability. How shoud I proceed when $f$ can be both positive and negative? Can this approach still work?
Here is an example: consider the Lebesgue integral $\int_{0}^{\infty} xe^{-x^2}=1/2$. From symmetry it should be clear that the integral $\int_{-\infty}^{0} xe^{-x^2}$ exists and its value ought to be $-1/2$. Indeed the proposition gives
$$ \int_{0}^{\infty} xe^{-x^2}=\lim_{n\to \infty} \int_{0}^{n} xe^{-x^2} dx=-\lim_{n\to \infty} \int_{-n}^{0} xe^{-x^2}dx$$
which shows that $\lim_{n\to \infty} \int_{-n}^{0} xe^{-x^2}dx$ exists and equals $-1/2$. But the existence of this limit of Riemann integrals does not imply the existence of the Lebesgue integral $\int_{-\infty}^{0} xe^{-x^2}$. Existence must be confirmed separately by checking that $$\lim_{n\to \infty} \int_{-n}^{0} |x|e^{-x^2}dx=\lim_{n\to \infty} \int_{0}^{n} xe^{-x^2}dx=\int_{0}^{\infty} xe^{-x^2}=1/2$$
This example is trivial but for more complicated cases the problem is the same.
Yes, you can just split up into positive and negative parts. As long as you make sure that they are both integrable.
It would probably be helpfull for you to give an example that you at present do not know how to handle.