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?

  • 2
    $\begingroup$ You are just putting limits in the integral result. What are you basically asking ? $\endgroup$ Feb 5 at 11:38
  • 1
    $\begingroup$ You are trying to evaluate the integral. And $\lim_{x\to\infty}-\exp(ax)/a=0$, yes# $\endgroup$
    – FShrike
    Feb 5 at 11:41
  • 2
    $\begingroup$ You cannot substitute $\infty$. You have to compute the limit properly, as you probably learned in class $\endgroup$
    – Taladris
    Feb 5 at 11:45
  • $\begingroup$ @FShrike i do not agree, as the result depends on the sign of $a.$ $\endgroup$
    – user376343
    Feb 5 at 13:50
  • $\begingroup$ @user376343 From context I assume $a>0$ because otherwise the question is meaningless $\endgroup$
    – FShrike
    Feb 5 at 14:31

1 Answer 1


$$\int\limits_0^\infty \exp(-ax)dx$$ $$= \frac{-1}{a}\exp(-ax)\rvert_0^\infty$$ $$=\frac{-1}{a}(\exp(-\infty)-1)$$ $$=\frac{1}{a}$$

  • $\begingroup$ Wrong. It is not given that $a>0.$ $\endgroup$
    – user376343
    Feb 5 at 13:48
  • $\begingroup$ I think it's safe to suppose $a > 0$. If $a \leq 0$ then the integral diverges. @user376343 $\endgroup$ Feb 6 at 0:26

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