# Question #055ec

##### 1 Answer
Sep 21, 2017

The integral may converge or diverge.

#### Explanation:

Example:
Let f(x) = x
Let a = 0
Let b = -1.
For this function and these values of a and b...
$f \left(a + x\right) = f \left(x\right) = x$ and $f \left(b + x\right) = f \left(x - 1\right) = x - 1$.
Therefore
$f \left(a + x\right) - f \left(b + x\right) = 1$ for all x, and we have
${\int}_{-} {\infty}^{\infty} \left(f \left(a + x\right) - f \left(b + x\right)\right) \mathrm{dx} = {\int}_{-} {\infty}^{\infty} 1 \mathrm{dx}$, which diverges.

However if:
Let f(x) = 1
Let a = 0
Let b = -1.
For this function and these values of a and b...
$f \left(a + x\right) = f \left(x\right) = 1$ and $f \left(b + x\right) = f \left(x - 1\right) = 1$.
Therefore
$f \left(a + x\right) - f \left(b + x\right) = 0$ for all x, and we have
${\int}_{-} {\infty}^{\infty} \left(f \left(a + x\right) - f \left(b + x\right)\right) \mathrm{dx} = {\int}_{-} {\infty}^{\infty} 0 \mathrm{dx} = 0$.

Were there limitations on f, or am I understanding your question correctly?