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

Question

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

Adil Ahamed · Accepted Answer · 2023-02-05 12:30:50Z

1

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

answered Feb 5 at 12:30

Adil Ahamed

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$\begingroup$ 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 $\endgroup$
– Accelerator
Feb 6 at 0:26

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