Properties of improper integral (showing that: $\int \limits_{0}^{\infty}f(x)dx=\int \limits_{0}^{1}f(x)dx+\int \limits_{1}^{\infty}f(x)dx.$)

Question

Let $f(x)$ is integrable on every segment $[r,\infty)$ where $r>0$. Let $\int \limits_{0}^{1}f(x)dx$ and $\int \limits_{1}^{\infty}f(x)dx$ converges. Why in this case we can conclude that $$\int \limits_{0}^{\infty}f(x)dx=\int \limits_{0}^{1}f(x)dx+\int \limits_{1}^{\infty}f(x)dx.$$

Can anyone explain it rigorously? Maybe I don't know a definition.

Answer 1

I think it should help to know that, given an integrable function $f$, it can be said that for $a<b$ and $c \in (a,b)$ $$\int_{a}^{b} f(x)dx = \int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx$$ So then of course we can then just take $a=0$, $c=1<b$. $$\int_{0}^{b} f(x)dx=\int_{0}^{1} f(x)dx+\int_{1}^{b} f(x)dx$$ And let $b \to \infty$.

Proof of this property is trivial. Via the fundamental theory of calculus we know that $$\int_{a}^{b} f(x)dx=F(b)-F(a), F(x)=\int f(x)dx$$ Therefore we have $$\int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx=F(c)-F(a)+F(b)-F(c)=F(b)-F(a)$$ $$=\int_{a}^{b} f(x)dx$$

Answer 2

For any fixed $D$ such that $[D,\infty)\subset$ dom$(f)$ define $$(1)....\quad \int_D^{\infty}f(x) dx=\lim_{Y\to \infty}\int_D^Y f(x) dx $$ provided that this limit exists. . If $\int_A^{\infty}f(x) dx$ exists for fixed $A$ then for fixed $B>A$ let $$\int_A^Bf(x) dx= C.$$ Then $C$ is fixed and $$(2)....\quad\int_A^{\infty}f(x) dx=\lim_{Y\to \infty}\int_A^Y f(x) dx=\lim_{Y\to \infty} \left ( \int_A^B f(x) dx+\int_B^Y f(x) dx\right )=$$ $$=\lim_{Y\to \infty}\left (C+\int_B^Y f(x) dx \right )=\left (C+\lim_{Y\to \infty}\int_B^Y f(x) dx\right )= \left (\int_A^B f(x) dx+\int _B^{\infty} f(x) dx\right ).$$The first equality in (2) is by (1) with $D=A$, and the last equality, above, is by (1) with $D=B.$ In particular let $A=0,\; B=1.$

Answer 3

I think is the case that you "don't know a definition" (as you said) because there is nothing to prove here (at least, in Tom Apostol's approach).

If the integrals $$\int \limits_{0}^{c}f(x)dx\quad \text{and}\quad \int \limits_{c}^{\infty}f(x)dx\tag{1}$$ are both convergent for some $c\in(0,\infty)$, then $$\int_0^\infty f(x)dx=\int \limits_{0}^{c}f(x)dx+ \int \limits_{c}^{\infty}f(x)dx$$ by definition. What you can prove if you want is "that the choice of $c$ is unimportant" (Apostol's words), that is, the value of $\int_0^\infty f(x)dx$ doesn't depends on $c$ (provided that both integrals in $(1)$ are convergent).

In your case, $$\int \limits_{0}^{1}f(x)dx\quad \text{and}\quad \int \limits_{1}^{\infty}f(x)dx$$ are both convergent and thus $$\int_0^\infty f(x)dx=\int \limits_{0}^{1}f(x)dx+ \int \limits_{1}^{\infty}f(x)dx$$ by definition of $\int_0^\infty$.

As I said, this answer follows the Apostol's approach (see section 10.23).