An [[Improper Integrals|Improper Integral]] where one of the extents is [[Infinity]]. >[!example] >$\huge >\begin{align} >\int_{0}^\infty e^{-x} \d x >&= \lim_{t\to\infty} \int_{0}^{t}e^{-x} \d x \\ >&= \lim_{t\to\infty} >-e^{-x} >\big\rvert_{0}^{t} >\d x \\ >&= \lim_{t\to\infty} >\ba{ >-e^{-t} - \pa{-e^0} >}\\ >&= >\lim_{t\to\infty} \ba{ -e^{-t}} >+ e^{0} \\ >&= 0 >+ e^{0} \\ >&= \boxed{1} >\\ >\end{align} >$ >[!example] >$\huge >\begin{align} >\int_{1}^{\infty} \frac{1}{x^2+x}\d x &= >\lim_{t\to\infty} \int_{1}^{t} \frac{1}{x^2+x}\d x \\ >&= >\lim_{t\to\infty} \int_{1}^{t} \frac{1}{x(x+1)}\d x \\ >\\ > >\frac{1}{x^2+x} &= \frac{A}{x} + \frac{B}{x+1}\\ > > > >1&= A(x+1) + Bx \\ >1&= Ax + B(x+1) \\ >\\ >0 &= A - B \\ >A &= 1, B = -1 \\ > >\\ &= >\lim_{t\to\infty} \int_{1}^{t} \pa{ >\frac{1}{x} - \frac{1}{x+1}} >\d x \\ > >\\ &= >\lim_{t\to\infty} > >\ln|x| - \ln|x+1| \big\rvert_{1}^t >\\&= >\lim_{t\to\infty} >\pa{\ln|t| - \ln|t+1| } - >\pa{\ln|1| - \ln|2| } >\\&= >\lim_{t\to\infty} >\ln\pa{\frac{t}{t+1}} > +\ln 2 > \\&= >\ln 1 + \ln 2 >\\&= \boxed{\ln 2} >\end{align} $