Sensible Calculus Chapter VIII.An Important Example
Integration: An Important Example of an Improper Intergal [Application]
© M. Flashman
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Perhaps the most important integration problem for the theory and application of probability is connected to the "normal distribution." This distribution has a density function defined over the interval `(-oo, oo)` based on the function `e^{-x^2}`. What is crucial in developing this distribution is finding exactly the value of the improper integral I where I` = int_{-oo}^{oo} e^{-x^2} dx`.
Despite the fact that there is no elementary function that will serve as an indefinite integral for `e^{-x^2}`, in this section we will evaluate this integral exactly.
The exercises make the connection to the usual definition of the density function for the normal distibution, `f(x)= 1/{sqrt{2pi}}e^{-{x^2}/2}`.
We'll break up the analysis of the integral into four parts:
  1. First we'll show that I `= int_{-oo}^{oo} e^{-x^2} dx` converges.
  2. Then we'll analyze a solid of revolution obtained by rotating  `y = e^{-x^2}` for `x >= 0` about the Y-axis and finding the volume of this solid using cylindrical shells.
  3. Next we'll find the volume of this solid using cross sections.
  4. Finally we'll find `int_{-oo}^{oo} e^{-x^2} dx` by recognizing the necessary equality of the preceeding volume results.
  1. I = `int_{-oo}^{oo} e^{-x^2} dx` converges.
    For `x>=1` it is clear that `x^2>x` so `e^{-x^2} < e^{-x}`.

    But we have shown that
    `int_1^{oo} e^{-x} dx = lim_{K->oo} int_1^K e^{-x}dx = lim_{K->oo} -e^{-x}| _1^K = lim_{K->oo}-e^{-K}+ 1/e = 1/e`

    converges, so by comparison we have that `int_1^{oo} e^{-x^2} dx` also converges.
    Adding `int_0^1e^{-x^2} dx` to the convergent integral we see that `int_0^1e^{-x^2} dx + int_1^{oo} e^{-x^2} dx = int_0^{oo} e^{-x^2} dx`  converges.
    But by the even symmetry of ` e^{-x^2}`  about the Y axis,  we can see that `int_{-oo}^0 e^{-x^2} dx` also converges.
    Thus `int_{-oo}^{oo} e^{-x^2} dx = int_{-oo}^0 e^{-x^2} dx + int_0^{oo}e^{-x^2} dx` converges.
    Note: Since `int_{-oo}^{oo} e^{-x^2} dx` converges, we can now use a more symmetric integral to investigate its value:
    `int_{-oo}^{oo} e^{-x^2} dx =lim_{R->oo}int_{-R}^{R} e^{-x^2} dx`

    We can estimate I numerically with a symmetric integral which we denote as I(`a`) `= int_{-a}^{a} e^{-x^2} dx` . For example, using technology to make an estimate you should be able to find that I(10) `~~ 1.772455`. This estimate is very close to estimates with much higher values for a- so it can be used for a rough estimate of I.
    The graph in Figure 3 illustrates this integral as an area.

     We notice at this stage something remarkable....I(10)2 is approximately 3.141597, very close to `pi`, and therefore we conjecture that I = ` sqrt{ pi }`.
  2. Analysis of the solid of revolution  determined by rotating  `y = e^{-x^2}` for `x >= 0` about the Y-axis.
    We begin by rotating the graph of `y = e^{-x^2}` for `x >= 0` around the "Y axis". 

    Now we'll analyze the volume of the region enclosed below this surface and above the plane by the method of "cylindrical shells."
    In fact we can find the volume under this surface over the entire plane! This is a limit, namely, `lim_{a->oo}int_0^{a}2pi re^{-r^2} dr`.
    A numerical estimate for this infinite integral, using `a = 10`, is  `int_0^{10}2pi re^{-r^2} dr ~~` 3.141593.
    The actual integral is found easily using the substitution `u = -r^2` and the fundamental theorem of calculus.


    We see that `int_0^{a}2pi re^{-r^2} dr = pi ( 1-e^{-a^2} )` which approaches `pi` as `a->oo`.
    Aha! So the volume is `pi`!

  3. Finding the volume of this solid using cross sections.
    Now here's the key to our investigation of I. We'll find this same volume using cross sections perpendicular to the X- axis. [See Figure 4 for the graph of the same surface now organized for cross-section analysis.]
    Note each cross section is a scalar multiple of the graph of `e^{-y^2}`. This scalar at `x = a` is precisely `e^{-a^2}` . Some examples here may help convince you , so draw the graphs of `e^0e^{-y^2}`, `e^{-(1/2)^2}e^{-y^2}`, `e^{-1}e^{-y^2}` , and `e^{-2^2}e^{-y^2}`.

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    More precisely the point with coordinate `(a,y)` will be on the circle with radius of `sqrt{a^2 + y^2}`. Thus the height of the surface above the point with coordinate `(a,y)` will be `e^{-(a^2+y^2)}=e^{-a^2}e^{-y^2}`.

    Thus the area of each cross section of the region with `X=a` enclosed by our surface is precisely `int_{-oo}^{oo}e^{-a^2}e^{-y^2}dy = e^{-a^2} int_{-oo}^{oo}e^{-y^2}dy =e^{-a^2}`I where I is the integral we are trying to find.
    Thus the area of each cross section at `x` is `A(x) = e^{-x^2}`I and the volume under the entire surface can be determined by `lim_{k->oo}int_{-k}^k A(x) dx = lim_{k->oo}int_{-k}^k e^{-x^2}\I dx = \I lim_{k->oo}int_{-k}^k e^{-x^2} dx = \I^2`.
    Therefore, the volume of the region is `I ^2`. .

  4. Finding `int_{-oo}^{oo} e^{-x^2} dx` in the preceeding volume results.
    We have now computed the volume of the spatial region to be `pi` using cylindrical shells and `I ^2` using cross sections.
    Thus `pi = \I^2` and I `= int_{-oo}^{oo} e^{-x^2} dx = sqrt{pi}`.
Exercises: Evaluate the following integrals using the fact that `int_{-oo}^{oo} e^{-x^2} dx = sqrt{pi}` .
  1. `int_{-oo}^{oo} e^{-{x^2}/2} dx `. [Hint: Use a substitution, `u^2 = {x^2}/2`. ]
  2. `int_0^{oo} e^{-{x^2}/2} dx `.
  3. Show that `f(x)= 1/{sqrt{2pi}}e^{-{x^2}/2}` is a probability density function over all the real numbers.