Exercise 321Calculus › Single-variable calculus › Integration › Improper integralsL2Added 31 Oct 2026

An improper integral

#improper integrals#integration by parts

Statement

Show that $\displaystyle\int_0^{+\infty}x\,e^{-x}\,dx$ converges and compute its value.

Hint

Find an antiderivative by parts, then let the upper bound go to infinity.

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An antiderivative of $xe^{-x}$ is $F(x)=-(x+1)e^{-x}$ , because
$F'(x)=-e^{-x}+(x+1)e^{-x}=xe^{-x}.$

For $A>0$ ,
$\int_0^Axe^{-x}\,dx=F(A)-F(0)=1-(A+1)e^{-A}.$

As $A\to+\infty$ , $(A+1)e^{-A}\to0$ (exponential beats polynomial). So the integral converges and equals $1$ .