Infinite Integrals

By Andrew Hodges on Sunday, June 24, 2001 - 10:46 am:

If you integrate 1/x from 0 to infinity, you will get infinity as your answer, even though 1/x is asymptotic to the x-axis. The normal distribution function however, which is also asymptotic to the x-axis, has a definite value, 0.5, for the integral from 0 to infinity. Is there a way of telling which asymptotic functions will have a definite sum to infinity from their equations? Is it do with the rate at which they converge with the x-axis?

By Dan Goodman on Sunday, June 24, 2001 - 03:12 pm:

First of all, the integral of

1 / x

from 0 to 1 is infinity, so I'll assume we're talking about an integral from 1 to infinity (or 0.0001 to infinity, or whatever) rather than 0 to infinity. I don't know of any hard and fast rule which you can apply to test if an integral is finite, but there are some useful quick methods. For example, if

g (x)

is a positive function such that

\int_{1}^{\infty} g (x) dx

is infinite and hence

\int_{R}^{\infty} g (x) dx

is infinite for all

R > 1

, and if you have a function

f (x)

(I'll assume

f (x) > 0

) such that

f (x) > g (x)

for all

x > R

for some number

R > 1

, then

\int_{1}^{\infty} f (x) dx = \int_{1}^{R} f (x) dx + \int_{R}^{\infty} f (x) dx > \int_{R}^{\infty} g (x) dx

. So if we can say that

f (x)

is bigger than

g (x)

for big enough

x

then

\int_{0}^{\infty} f (x) dx

is infinite. Suppose

h (x)

is a function such that

\int_{1}^{\infty} h (x) dx

is finite. If

f (x) < h (x)

for

x > R

and

f (x) < M

for

x < R

and some number

M

then

\int_{1}^{\infty} f (x) dx = \int_{1}^{R} f (x) dx + \int_{R}^{\infty} f (x) dx < \int_{1}^{R} M dx + \int_{R}^{\infty} h (x) dx

which is finite.

So, let's take the function $f (x) = x^{4} e^{- x}$ as an example. For large enough $R$ , $x^{4} < e^{x / 2}$ (do you know how to prove this?) and so for $x > R$ $f (x) < e^{x / 2} e^{- x} = e^{- x / 2}$ , and we know $\int_{1}^{\infty} e^{- x / 2} dx$ is finite. If $x < R$ then $x^{4} < R^{4}$ and $e^{- x} < 1$ so $f (x) < R^{4}$ . So $\int_{1}^{\infty} x^{4} e^{- x} dx$ is finite (using the above).

Good functions $h (x)$ whose integrals from 1 to infinity converge are $x^{t}$ for $t < - 1$ , $e^{- x}$ and $e^{- x^{2}}$ . The best function $g (x)$ whose integral from 1 to infinity is infinite is $1 / x$ . This collection of functions is usually enough to prove that any integral of a positive function that you come across converges or diverges. It's more difficult to show things for functions which are sometimes positive and sometimes negative, but if you can show that $\int_{1}^{\infty} | f (x) | dx$ is finite then also $\int_{1}^{\infty} f (x) dx$ is finite.

By Andrew Hodges on Sunday, June 24, 2001 - 09:37 pm:

Could you explain how to prove e^x/2 > x⁴ for some number R?

By Dan Goodman on Monday, June 25, 2001 - 12:52 am:

Hmm, it depends on what you know. Do you know that e^x =1+x+x² /2+...+xⁿ /n!+... ? If not, write back and I'll see if I can prove it another way. If so, then for x> 0 we have e^x > xⁿ⁺¹ /(n+1)! (since this is just one term of the series). If x> (n+1)! then e^x > (x/(n+1)!)xⁿ > xⁿ since x/(n+1)!> 1. So for any integer n, if x> (n+1)! then e^x > xⁿ . Obviously if you are thinking about x/2 instead of x you have to make a slight change to that argument.