# Thread: Uniqueness of Improper Integral

1. ## Uniqueness of Improper Integral

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

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

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

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