So this might be a stupid question, but I was wondering what the following integral evaluates to:
$$\int_{\infty}^{\infty} \, f(x) \, dx$$
And no, the lower bound is not a typo.
I would imagine that the integral is $0$, for the following reason: \begin{align*} \int_{\infty}^{\infty} \, f(x) \, dx = \lim_{k \to \infty} \, \int_k^k \, f(x) \, dx = \lim_{k\to\infty} \, 0 = 0 \end{align*}
Edit (This is my real question):
If that's true, then could we evaluate the integral $$\lim_{k\to\infty} \, \int_k^{\infty} \, f(x) \, dx$$ as \begin{align*} \lim_{k\to\infty} \, \int_k^{\infty} \, f(x) \, dx &= \lim_{k\to\infty} \, \int_k^k \, f(x) \, dx \\[8pt] &= \lim_{k\to\infty} \, 0 \\[8pt] &= 0 \end{align*}
I'm pretty sure the first equation is wrong, and should be something like $$\lim_{k\to\infty} \, \int_k^{\infty} \, f(x) \, dx = \lim_{k\to\infty} \, \lim_{j \to \infty} \, \int_k^j \, f(x) \, dx$$
If so, then I was thinking that, because $k$ and $j$ are growing increasingly large, there are points $\alpha$ where the values of $k$ and $j$ will begin being the same, so
$$\lim_{k\to\infty} \, \lim_{j \to \infty} \, \int_k^j \, f(x) \, dx = \lim_{\alpha \to \infty} \, \int_{\alpha}^{\alpha} \, f(x) \, dx$$
which equals $0$ from the above arguments.
Edit: I suppose a more formal way of expressing my thoughts concerning $\alpha$ would be to let $\alpha = \operatorname{max}\{j,k\}$.
Just wanted some thoughts or criticisms, in particular.
Thanks!