Why do we require $a\neq0$ in $\int{e^{ax}dx=\frac{1}{a}e^{ax}} + C$?

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,

$$\frac{d}{dx}e^{ax}=ae^{ax}\tag{2}$$

Questions:

1. Why is the condition $$a\neq0$$ necessary in $$(1)$$?
2. Does the condition ($$a\neq0$$) apply in $$(2)$$ as well?
• It's definitely supposed to be $a \neq 0$, not $x \neq 0$. Jan 20, 2022 at 2:45
• Yes, as @TheoBendit comments, definitely a typo. Jan 20, 2022 at 2:45
• If $a =0$, the formula wouldn't hold as you're integrating $1.$ Jan 20, 2022 at 2:59
• 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. Jan 20, 2022 at 6:22

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

$$\frac{d}{dx}e^{ax}=ae^{ax}$$

For $$a=0$$ we have

$$\frac{d}{dx}e^{ax}=\frac{d}{dx}e^{0x}=0=0e^{0x}=ae^{ax}$$

so the formula still holds.