The integral also extends to unbounded intervals, defined as the limit of an ordinary integral: a+fdx=limt+atfdx.\int_a^{+\infty}f\,dx = \lim_{t\to+\infty}\int_a^t f\,dx. The improper integral converges if the limit is finite, otherwise it diverges.

A quick criterion much used concerns powers: 1xpdxconverge per p>1,diverge per p1.\int_1^\infty x^{-p}\,dx \quad\text{converge per } p>1,\quad\text{diverge per } p\le 1.

Example — Two cases compared

1dxx2=[1x]1=0(1)=1(converge).\int_1^\infty\frac{dx}{x^2} = \left[-\frac{1}{x}\right]_1^\infty = 0-(-1) = 1 \quad\text{(converge)}. 1dxx=[lnx]1=+(diverge).\int_1^\infty\frac{dx}{x} = [\ln x]_1^\infty = +\infty \quad\text{(diverge)}.

The same scheme (integral as a limit) applies also when the function has a vertical asymptote at an endpoint of the domain.

Topics: Integrale
Concepts: Integrale definito · Integrale improprio · Limite
Skills: Analisi casi limite · Calcolare limiti · Integrare