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One can easily find the integral $\int_{0}^{\infty}\exp(-x)dx$. It is equal to 1. But is there a way to understand this geometrically without integration?

If i rotate the picture i see that $\int_{0}^{\infty}\exp(-x)dx=-\int_{0}^{1}\ln(t)dt$. Maybe there is some property of exp or log which allows to avoid integration?

PS:

I would like to accept the Mamikon's method pointed out by Jim Belk. But it is impossible to accept comments... So I accept the second best.

Roah
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    Why do you want to avoid integration? The exponential is one of the simplest functions to integrate. If you want to make things a little bit simpler, change the variable $x=\ln t$. – Beni Bogosel May 30 '11 at 08:25
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    See http://math.stackexchange.com/questions/3444/intuition-for-the-definition-of-the-gamma-function for a generalized variant of this question. – Jonas Meyer May 30 '11 at 08:30
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    See http://www.its.caltech.edu/~mamikon/VisualCalc.html for a geometric explanation of this integral. – Jim Belk May 30 '11 at 08:48
  • Wow! Thanks for the links! – Roah May 30 '11 at 09:23
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    If you believe that $\int_0^{\infty} a^{-x} , dx$ is finite for any $a$, then you can regard this integral as a definition of $e$ (that is, it's the unique value of $a$ that makes the integral equal to $1$). – Qiaochu Yuan May 30 '11 at 11:27

1 Answers1

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The following argument uses only basic properties of the exponential function and the integral, but not the fundamental theorem of calculus:

Let $\int_0^\infty e^{-x}dx=:I$. By looking at a figure we see that for any $c>0$ we have $$I=\int_0^c e^{-x}dx +\int_c^\infty e^{-x}dx= \int_0^c e^{-x}dx + e^{-c} I$$ or $$(1-e^{-c}) I =\int_0^c e^{-x}dx\ .$$ Using $e^{-c}\leq e^{-x}\leq1 \ \ (0\leq x\leq c)$ we conclude that $$c \ e^{-c} \leq (1-e^{-c}) I \leq c\ .$$ Now divide by $1-e^{-c}$ and let $c\to 0+$ to get the desired result.

Christian Blatter
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