Uniqueness of Improper Integral

**ThePerfectHacker** · March 4th 2006, 03:38 PM

If $f(x)$ is a countinous real function, we define,
$\int^{\infty}_{-\infty}f(x)dx$ as being convergent if there exists a $a\in\mathbb{R}$ such as, $\int^{\infty}_af(x)dx$ and $\int^a_{-\infty}f(x)dx$ are convergent. Also its value is then,
$\int^{\infty}_{-\infty}f(x)dx=\int^{\infty}_af(x)dx+\int^a_{-\infty}f(x)dx$ .

Prove that if,
$\int^{\infty}_{-\infty}f(x)dx$ is convergent then for any $b\in\mathbb{R}$ the two improper integrals,
$\int^{\infty}_bf(x)dx \mbox{ and }\int^b_{-\infty}f(x)dx$ are also convergent.

Further, prove that,
$\int^{\infty}_af(x)dx+\int^a_{-\infty}f(x)dx=\int^{\infty}_bf(x)dx+\int^b_{-\infty}f(x)dx$ .
Thus, proving that the value of
$\int^{\infty}_{-\infty}f(x)dx$
is well-defined.
???

**ThePerfectHacker** · March 4th 2006, 06:20 PM

Thank you but I think I have it. Basically, you rewrite your integral.
The integral,
$\int^{\infty}_{b}+\int^{b}_{-\infty}$ if $a<b$ then write $b=a+r,r>0$ . Thus, we have,
$\int^{\infty}_{a+r}+\int^{a+r}_{-\infty}$
By the subdivision rule,
$\int_{-\infty}^a+\int^{a+r}_{a}+\int^a_{a+r}+\int_a^{ \infty }$
But, $\int^{a+r}_{a}+\int^a_{a+r}=0$
(Same when $a>b$ )
And the proof it complete.

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