What does $$\int_{-\infty}^{\infty} \Big( \int_{x-1/2}^{x+1/2}f(t)dt \Big) dx$$
mean? Is it the same as $$\int_{-\infty}^{\infty} f(x) dx$$ or do the boundaries "overlap"? Are there any good theorems to use here to access the limits of the integral? Is it possible to somehow change the order of integration?
I am trying to figure out this interval for a function $f$ with $$\int_{-\infty}^{\infty} f(x) dx=1$$
and I suspect that $$\int_{-\infty}^{\infty} \Big( \int_{x-1/2}^{x+1/2}f(t)dt \Big) dx=1$$
You can define the function $g(x):=\int_{x-\frac{1}{2}}^ {x+\frac{1}{2}}f(t)dt$ that assigns to every $x$ the value of the integral $\int_{x-\frac{1}{2}}^ {x+\frac{1}{2}}f(t)dt$.
So the question is to calculate $\int_{-\infty}^\infty g(x)dx$. You can observe that if you choose $y=t-x$ then
$g(x):=\int_{x-\frac{1}{2}}^ {x+\frac{1}{2}}f(t)dt= \int_{-\frac{1}{2}}^ {\frac{1}{2}}f(y+x)dy $
And so
$\int_{-\infty}^\infty g(x)dx= \int_{-\infty}^\infty \int_{-\frac{1}{2}}^ {\frac{1}{2}}f(y+x)dydx $
so (in the hypothesis in which you can change the order of integration) you have that
$ \int_{-\frac{1}{2}}^ {\frac{1}{2}} \int_{-\infty}^\infty f(y+x)dxdy= \int_{-\frac{1}{2}}^ {\frac{1}{2}} (\int_{-\infty}^\infty f(z)dz)dy$
$= \int_{-\frac{1}{2}}^ {\frac{1}{2}}1dy=\frac{1}{2}+ \frac{1}{2}=1$