Indefinite integration and definite integration calculation Given the functions:

$A(x)=B(2x-1)-B(2x)$

B(x) is a function as x->infinity, limit of B(x)->0.

for $b&gt;1$ find out:

$$\int_{b}^\infty A(x) dx - \frac12\int_{2b-1}^{2b} B(x) dx$$

The answer is $0$. But why?

I attempted to do:

$\int_{b}^\infty A(x) dx=\int_{b}^\infty B(2x-1)-B(2x) dx$

$=\int_{b}^\infty \sum_{k=2x-1}^{2x} B(k) dx$

And I think that it would be possible to swap the summation and integration, however got no further...
Not a full solution, but almost. Define
$$
I(b)=\int_{b}^\infty [B(2x-1)-B(2x)]dx - \frac12\int_{2b-1}^{2b} B(x) dx\ .
$$
For a generic $B(x)$ [not necessarily the one the OP wrote], we have
$$
\frac{d}{db}I(b)=0
$$
identically, using the fundamental theorem of calculus. Indeed
$$
\frac{d}{db}\int_b^\infty dx B(2x-1)=-B(2b-1)
$$
$$
\frac{d}{db}\int_b^\infty dx B(2x)=-B(2b)
$$
$$
\frac{d}{db}\int_{2b-1}^{2b}B(x)dx=2 B(2 b)-2 B(2 b-1)\ .
$$
Therefore, $I(b)$ is a constant (as a function of $b$). So, you only need to evaluate your $I(b)$ at a value of $b$ of your choice [pick a value for which the computation is easy] to establish your claim....
This is my answer (I modified the question as the B(x)= part is not quite useful):

let m=2x-1, n=2x

Then integral of A(x)=$\int_{2b-1}^\infty B(m) dm - \int_{2b}^\infty B(n) dn$

Then equals the integral at the right hand side.