# Solve the integral: $\int_{0}^{\infty}\exp(-ax)dx$

I am trying to solve the integral: $$\int_{0}^{\infty}\exp(-ax)dx$$

Computing the indefinte integral gives: $$-\frac{-1}{a}\exp(-ax)$$

Now I was wondering if the following is correct: for $$\infty$$: $$-\frac{-1}{a}\exp(-a\cdot\infty)=0$$

Am I allowed to do this in this integral?

• You are just putting limits in the integral result. What are you basically asking ? Feb 5 at 11:38
• You are trying to evaluate the integral. And $\lim_{x\to\infty}-\exp(ax)/a=0$, yes# Feb 5 at 11:41
• You cannot substitute $\infty$. You have to compute the limit properly, as you probably learned in class Feb 5 at 11:45
• @FShrike i do not agree, as the result depends on the sign of $a.$ Feb 5 at 13:50
• @user376343 From context I assume $a>0$ because otherwise the question is meaningless Feb 5 at 14:31

$$\int\limits_0^\infty \exp(-ax)dx$$ $$= \frac{-1}{a}\exp(-ax)\rvert_0^\infty$$ $$=\frac{-1}{a}(\exp(-\infty)-1)$$ $$=\frac{1}{a}$$
• Wrong. It is not given that $a>0.$ Feb 5 at 13:48
• I think it's safe to suppose $a > 0$. If $a \leq 0$ then the integral diverges. @user376343 Feb 6 at 0:26