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My physics professor today wrote on the blackboard:
$$ int_infty^infty f(x) dx = 0 $$
for every function $f$.
And the proof he gave was:
$$ int_infty^infty f(x) dx = int_infty^a f(x) dx + int_a^infty f(x)dx = - int_a^infty f(x) dx + int_a^inftyf(x)dx = 0$$

However I'm still not convinced, for me an integral from infinity to infinity has no meaning. Therefore, what I'm asking is: does the above equations make sense? If not, are there cases where they do make sense? I'm thinking about functions that converge to 0 in $+infty$.

EDIT: Actually, the function f considered was a density, i.e.:
$$ int_-infty^+infty f(x)dx = 1 $$
and $f(x) geq 0$ for all $x$.

This is not necessarily true. Take the following example;
$$int_a^2afrac1xmathrmdx=[ln]_a^2a=ln(2)$$
If we take $atoinfty$ then the integral becomes
$$int_infty^inftyfrac1xmathrmdx=ln(2)$$
as the integral is constant for all $ainmathbbR$. What I guess your professor meant was that
$$lim_atoinftyint_a^a f(x)mathrmdx=0$$
which is trivially true as the LHS is constantly zero.

An improper integral with an endpoint of $infty$ means a limit of proper integrals where the endpoint approaches $infty$. Thus a reasonable definition of
$int_infty^infty f(x); dx$ would be
$$ int_infty^infty f(x); dx = lim_a, b to infty int_a^b f(x); dx $$
This is $0$ if and only if $int_a^infty f(x); dx$ converges for some $a$.

EDIT: If the double limit is $0$, there is $N$ such that
$left|int_a^b f(x); dxright| < 1$ for all $N < a < b$.
For any $epsilon > 0$ there is $M > N$ such that for $b, c > M$,
$$ left|int_b^c f(x); dx right| = left| int_a^c f(x); dx - int_a^b f(x); dx right|< epsilon$$

and this implies that $lim_b to infty int_a^b f(x); dx$ exists, i.e.
$int_a^infty f(x); dx$ converges.

Conversely, if $int_a^infty f(x); dx = L$ converges, then for any $epsilon > 0$ there is $N$ such that $left|int_a^b f(x); dx - Lright| < epsilon/2$ whenever $b > N$. Then if $b > N$ and $c > N$,
$$ left| int_b^c f(x); dxright| = left|int_a^c f(x); dx - int_a^b f(x); dx right| < epsilon $$

As has been pointed out by other answers, this is not always true because the symbol $infty$ can hide many things, even if we work with the extended reals. The actual meaning of the $infty$ is a limiting process as a certain variable becomes arbitrarily large. The upper and lower limits in the $$int_a^b f(x)mathrm d x$$ however may approach $infty$ at different rates, and this is the key point. Your professor's argument works only when $a$ and $b$ approach infinity equally fast, or in other words, when they are of equal order at infinity. In particular it is always valid if $a=b.$

So what about your professor's "proof"? Well, the ambiguity should now be obvious -- he uses the same symbol $infty$ for things that may behave differently. I think it's safe to assume he's thinking only of the variables are of equal order at infinity. Otherwise his proof breaks down since $infty-infty$ can be anything otherwise.

PS. However, you say an integral from $infty$ to $infty$ has no meaning to you. Well, I see you're thinking of ordering of the reals here. But note that we're not just dealing with the reals here, but the extended reals. As explained above, the best way to think of it is to think of the limits of the integral as approaching infinity at (not necessarily equal rates). Then it's easy to make sense of it. Another way may be to think of the one-point compactification of the real axis.

As Peter Foreman mentioned, there are some occasions where you will get an integral of this form, however the exact integral does matter. Firstly visualise the error function:
$$operatornameerf(x)=frac2sqrtpiint_0^xe^-t^2dt$$
notice that:
$$lim_xto 0operatornameerf(x)=frac2sqrtpiint_0^0e^-t^2dt$$
Normally we can visualise an integral as area, but in this form it does not make sense since the range over which the area found $to0$. The most obvious case when this integral is zero, no matter the function, would be when both the bounds are equal as the range of the integral is then zero. Overall, it comes down to notation and making sure the bounds are properly defined.