In my book's table for antiderivatives of some functions, I came across the following,

$$\int{e^{ax}dx=\frac{1}{a}e^{ax}} + C, \qquad a \neq0\tag{1}$$

I can't understand the reasoning behind the condition $a\neq0$. Also,



  1. Why is the condition $a\neq0$ necessary in $(1)$?
  2. Does the condition ($a\neq0$) apply in $(2)$ as well?
  • 10
    $\begingroup$ It's definitely supposed to be $a \neq 0$, not $x \neq 0$. $\endgroup$ Jan 20, 2022 at 2:45
  • 2
    $\begingroup$ Yes, as @TheoBendit comments, definitely a typo. $\endgroup$ Jan 20, 2022 at 2:45
  • 1
    $\begingroup$ If $a =0$, the formula wouldn't hold as you're integrating $1.$ $\endgroup$ Jan 20, 2022 at 2:59
  • 3
    $\begingroup$ Just a small remark: If you let $C=-\frac{1}{a}+D$, then the formula becomes $$\int e^{ax} \, dx = \frac{e^{ax}-1}{a} +D,$$ which has the limit $x+D$ as $a \to 0$. So the special case $\int e^{0x} \, dx = \int 1 \, dx = x + D$ which occurs when $a=0$ isn't that special; it does fit the general pattern in the sense just described. $\endgroup$ Jan 20, 2022 at 6:22

3 Answers 3


The conundrum here is that $$\int_0^te^{ax}\,\mathrm{d}x=\begin{cases}\frac{e^{at}-1}{a}&a\neq0\\t&a=0\end{cases}.$$ This is because for $a\neq0,$ it is true that $$\frac{\mathrm{d}}{\mathrm{d}t}\frac1{a}e^{at}=e^{at},$$ but it is not true that $$\frac{\mathrm{d}}{\mathrm{d}t}\frac10e^{0t}=e^{0t}$$ because $\frac10$ does not exist. Instead, notice that $$\frac{\mathrm{d}}{\mathrm{d}t}e^{0t}=0e^{0t}=0,$$ which must be true, since $$e^{0t}=1,$$ and $$\frac{\mathrm{d}}{\mathrm{d}t}1=0.$$ Since $$e^{0t}=1,$$it is also true that $$\int_0^te^{0x}\,\mathrm{d}x=\int_0^t\,\mathrm{d}x=t.$$ However, you may still not be satisfied, and you may suspect that there must be some relationship between $t$ and $\frac{e^{at}-1}{a}$ for $a=0,$ your intuition tells you there must be one, even if $\frac10$ does not exist. Well, your intuition is correct! Notice that $$\lim_{a\to0}\frac{e^{at}-1}{a}=t.$$ This means that $$\lim_{a\to0}\int_0^te^{ax}\,\mathrm{d}x=t,$$ and since $$\lim_{a\to0}e^{at}=1,$$ it is also true that $$t=\int_0^t\,\mathrm{d}x=\int_0^t\lim_{a\to0}e^{ax}\,\mathrm{d}x.$$ Thus, what your intution is really telling you is the powerful fact that $$\lim_{a\to0}\int_0^te^{ax}\,\mathrm{d}x=\int_0^t\lim_{a\to0}e^{ax}\,\mathrm{d}x.$$ The reason this is a powerful fact is because $$\lim_{a\to0}\int_0^tf(a,x)\,\mathrm{d}x=\int_0^t\lim_{a\to0}f(a,x)\,\mathrm{d}x$$ is not true in general, for arbitrary $f.$


If $a = 0$ then $e^{ax} = 1$ for all $x$, in which case the correct antiderivative would be $x + C$. This doesn't match the formula $\frac1a e^{ax}$, which is why it is a special case. There must be a special case, because if $a = 0$ then $1/a$ is undefined.

However in the formula $\frac d{dx} e^{ax} = a e^{ax}$, this works even when $a = 0$ as $\frac{d}{dx} 1 = 0$.


Alright, so the question is basically: what happens at $a=0$ in the integral

$$\int e^{ax}dx$$

Well, as you noted, for $a\neq 0$ we have

$$\int e^{ax}dx=\frac{1}{a}e^{ax}+C$$

for some constant $C$. If this is confusing, just note that

$$\frac{d}{dx}\left[\frac{1}{a}e^{ax}+C\right]=\frac{1}{a}\cdot ae^{ax}=e^{ax}$$

as desired. Now, what if $a=0$? Well, then the formula given in your book doesn't make sense as you are dividing by $0$. But in this case, we have

$$\int e^{ax}dx=\int e^{0}dx=\int 1dx=x+C$$

As for your second question, for $a\neq 0$ it is obvious that


For $a=0$ we have


so the formula still holds.


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