# When does the improper integral of the sum of two functions converge?

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...)??

• Your statement "so we will have ..." does not follow. Why not just say $f = g+h...$ – zhw. Jan 25 '16 at 21:57
• what i wrote is the same as you said @zhw. – YAZO Jan 25 '16 at 21:59
• No, you wrote the third line as if these integrals converge. – zhw. Jan 25 '16 at 22:01
• What's about $0=f(x), f_1(x)=x, g_1(x)=-x$? – user302982 Jan 25 '16 at 22:02
• Yes thats right @zhw. i'll fix it – YAZO Jan 25 '16 at 22:07

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.

$\infty-\infty$ is undefined, so if the integral of one of $f_1,g_1$ tends to $\infty$ and the other to $-\infty$, then the decomposition does not tell you whether $\int f_1+g_1$ goes to a limit or not. However, the decomposition will work in all other cases where one integral goes to $\pm \infty$ and the other doesn't, or where they both converge.

You ask if divergence of $\int f_1$ implies you can skip calculating $\int g_1$. No, because of the $\infty-\infty$ possibility. However, if you can show that $\int g_1$ is bounded you can infer divergence of $\int f_1+g_1$.

There is also, I suppose, the possibility that by "divergence" you include oscillation, say $f(x)=\sin(x)$. In that case, I'm not sure what can be inferred.

$$\int_a^{+\infty}{f(x)}{dx}=\int_a^{+\infty}{f_1(x)}{dx}+\int_a^{+\infty}{g_1(x)}{dx}$$ If any two of the three integrals converge, then so does the third. It can happen that any one of them converges, but the other two do not.

• Ok that was short and useful answer – YAZO Jan 25 '16 at 22:16