Suppose we have a function $f(x)$ defined on $[a,+\infty[$ and consider the improper integral of first kind $\int_a^{+\infty}{f(x)}{dx}$ where $f$ can be written as $f=f_1+g_1$ so we will have $\int_a^{+\infty}{f(x)}{dx}=\int_a^{+\infty}{f_1(x)}{dx}+\int_a^{+\infty}{g_1(x)}{dx}$.
I have some questions:
If we prove that $\int_a^{+\infty}{f_1(x)}{dx}$(or $\int_a^{+\infty}{g_1(x)}){dx}$ is divergent then can we say directly that $\int_a^{+\infty}{f(x)}{dx}$ is divergent(i.e. without calculating $\int_a^{+\infty}{g_1(x)}{dx}$(or $\int_a^{+\infty}{f_1(x)}{dx}$))??
If not then what are the conditons on the 2 improper integrals for $\int_a^{+\infty}{f(x)}{dx}$ to be convergent or divergent?(i.e. should they be both convergent or both divergent or…)??
Best Answer
No, $\int_a^\infty f_1$ divergent does not imply $\int_a^\infty f$ diverges. Easy counterexample: Let $f\equiv 0.$ Then $f = f_1+g_1,$ where $f_1\equiv 1, g_1 \equiv -1.$ Therefore $\int_a^\infty f$ converges (obviously), even though the integrals of $f_1,g_1$ both diverge.
So what we can say is this: If the integrals of $f_1,g_1$ both converge, then so does the integral of $f.$ But this is not an iff as the above example shows.