# Why does $e^{-x}$ integrate to $1$ in interval $(0, \infty)$?

My question is relatively elementary, but I haven't been able to find any clear explanations despite searching online so I decided to ask my own question.

I've always learned that

$$\int_{0}^{\infty}e^{-x}dx = \int_{-\infty}^{0}e^xdx = 1$$

but I would like to specifically understand why this is the case, rather than just keeping the information stored in my head.

Would anybody be kind enough to explain why the function $$f(x) = e^x$$ integrates to $$1$$ in the given interval?

Thank you.

• Have you tried just performing the indefinite integral and taking limits? It's not very hard. – Qiaochu Yuan Nov 12 '18 at 10:25
• No, I actually haven't tried that. Perhaps my fundamental understanding of calculus is not that good yet as I've only recently started studying it again. Thanks for the tip! I'll do some research with what you suggested. – Seankala Nov 12 '18 at 10:27
• $e^0-e^{-\infty}=1-0=1$ – bof Nov 12 '18 at 10:27
• The question to you should be: "what is $e$?" what definition of $e$ are you using? Though of course, knowing how integration works in required as well, if you are asking about an integral – Yuriy S Nov 12 '18 at 10:53
• The integrand $e^{-x}$ has antiderivatives $-e^{-x}+C$. – J.G. Nov 12 '18 at 10:58

Let $$f(x) = e^x$$ and $$F(x) = e^x$$. Then $$F'=f$$ and so $$\int_{t}^{0}e^x \,dx = \int_{t}^{0}f(x) \,dx = \int_{t}^{0}F'(x) \,dx = F(x)|_{t}^{0} = F(0) - F(t) = 1 - e^t$$ Therefore, $$\int_{-\infty}^{0}e^x \,dx = \lim_{t\to-\infty} \int_{t}^{0}e^x \,dx = \lim_{t\to-\infty} 1 - e^t = 1 - \lim_{t\to-\infty} e^t = 1 - 0 = 1$$

Let's use the definition of Riemann sum for the integral:

$$\int_0^M e^{-x}dx=\lim_{N \to \infty} \frac{M}{N} \sum_{n=0}^N e^{-\frac{nM}{N}}$$

If we are familiar with the geometric sum formula (here $$e$$ is just a number and all we need to know is $$e>1$$), we can write:

$$\sum_{n=0}^N e^{-\frac{nM}{N}}=\frac{1-e^{-M \frac{N+1}{N}}}{1-e^{-\frac{M}{N}}}$$

Now we need to find the limit:

$$\lim_{N \to \infty} \frac{M}{N} \frac{1-e^{-M \frac{N+1}{N}}}{1-e^{-\frac{M}{N}}}=1-e^{-M }$$

We used the fact that for small $$p$$ we have $$e^{-p} \approx 1-p$$, which follows from any definition of the exponential function, for example the series definition.

Now we take the limit:

$$\lim_{M \to \infty} 1-e^{-M }=1$$

This may look fishy, as we brought both $$N$$ and $$M$$ to infinity, meanwhile we required before that $$M \ll N$$. However, it makes sense because Riemann sum requires us to subdivide the whole integration interval into many small pieces, so of course their number should be larger than the size of the interval, since we need their length to be $$\frac{M}{N} \ll 1$$.

• Well your explanation in the second part needs to be altered. The use of $N$ is to denote number of subintervals of $[0,M]$ and hence $N$ is an independent variable which tends to $\infty$ and $M$ remains fixed in this process. Thus you prove $\int_{0}^{M}e^{-x}\,dx=1-e^{-M}$. When you apply $M\to\infty$ there is no variable $N$ in picture. – Paramanand Singh Nov 13 '18 at 4:29
• @ParamanandSingh, would you disagree with my logic that in any case (say, approximating the integral by the Riemann sum) we should have $M \ll N$? I wanted a more intuitive picture, as it is not clear to me how we can rigorously assume $N \to \infty$ and then $M \to \infty$ while the ratio $M/N$ is not defined... – Yuriy S Nov 13 '18 at 4:46
• Well by definition we have $$\int_{a} ^{\infty} f(x) \, dx=\lim_{M\to\infty} \int_{a}^{M}f(x)\,dx=\lim_{M\to \infty} \lim_{N\to\infty} \frac{M-a} {N} \sum_{k=1}^{N}f\left(a+k\cdot \frac{M-a} {N} \right)$$ In the above $M, N$ are independent variables and $N$ is a necessarily a positive integer. These variables do not tend to $\infty$ simultaneously rather they do it one by one. – Paramanand Singh Nov 13 '18 at 5:57
• @ParamanandSingh, true, but if we try to take the $M$ limit first with $N$ fixed, the limit doesn't exist... Becaue $N$ can't be $< M$ – Yuriy S Nov 13 '18 at 6:06
• I remember, that a two variable limit only exists if it doesn't depend on the order of taking it... – Yuriy S Nov 13 '18 at 6:09

We can just integrate it indefinitely and put the limits(by using the fundamental theorem of calculus), put $$e^{-\infty} = 0$$.

The explanation for the last step is that it is not actually putting $$\infty$$ in power, it is actually $$\lim_{x \to -\infty}e^{x}$$, which evaluates to $$0$$.