A basic doubt on definite intergal involving infinity Is the following statement valid ?
$$ [x]^\infty_0 = \int^{\infty}_{0}dx$$
 A: Short answer: the statement is true.

Long answer.
Regarding the formula $\displaystyle  [x]^\infty_0 = \int \limits^{\infty}_{0}\mathrm dx$,


*

*the LHS is short for $\lim \limits_{x\to +\infty}(x-0)$, if the limit exists (finitely), and it is meaningless otherwise. In case it doesn't exist, the formula $\displaystyle  [x]^\infty_0 = \int \limits^{\infty}_{0}\mathrm dx$ isn't a statement at all, it's a string of symbols.

*the RHS is short for $\displaystyle \lim \limits_{x\to +\infty} \left(\int \limits_0^x \mathrm dt\right)$, if this limit exists. If it does, $\displaystyle \lim \limits_{x\to +\infty} \left(\int \limits_0^x \mathrm dt\right)=\lim \limits_{x\to +\infty} \left([t]_0^x\right)=\lim \limits_{x\to +\infty} \left(x-0\right)$.


Thus the formula $\displaystyle  [x]^\infty_0 = \int \limits^{\infty}_{0}\mathrm dx$ is a true statemenet whenever any of the limits exists and it isn't a statement at all if one of the limits doesn't exist.

What's below is an answer to the first version of the question and, for that reason, hidden in a spoiler.

The formula $\displaystyle  [x]^\infty_0 = \int \limits^{1}_{0}\mathrm dx$ isn't even a statement. The RHS is simply $1$, but the LHS is short for $\lim \limits_{x\to +\infty}\left(x-0\right)$ which, in this context, doesn't exist as it isn't a real number, so you can't really claim that the 'equality' is a statement.If you are to consider $+\infty$ as a proper entity, then it is a statement, but it is false because $+\infty \neq 1$.

