Improper Integrals — Question 4

Question 4

For which real $a$ does the integral converge? Evaluate it when possible. $$\int_0^\infty e^{-ax}\,dx$$

Question 4 – Solution

Step 1: Replace infinity with a limit. $$I=\lim_{b\to\infty}\int_0^b e^{-ax}\,dx.$$ Step 2: Suppose $a\ne0$ and integrate. $$\begin{align*} I&=\lim_{b\to\infty}\left[-\frac1ae^{-ax}\right]_0^b\\ &=\lim_{b\to\infty}\frac{1-e^{-ab}}{a}. \end{align*}$$ If $a>0$, then $e^{-ab}\to0$, so $I=1/a$. If $a<0$, then $e^{-ab}\to\infty$, so the integral diverges.

Step 3: Check $a=0$. Then the integrand is $1$, and $$\int_0^\infty1\,dx$$ diverges. $$\boxed{a>0:\ 1/a;\quad a\le0:\text{diverges}}$$