Re: upper bound

From: Pawel Gladki (gladki_at_NOSPAMmath.usask.ca)
Date: 11/04/04 

Date: Thu, 04 Nov 2004 11:26:39 -0600

Hello,

Student, T.U.E. wrote:

> Integrate[Exp[(-y^2)/2], {y, x, infty}] <= x^-1 * Exp[(-x^2)/2]

First, observe that the integral:

\int_x^{oo} e^{-y^2/2} dy

is convergent; indeed, if x < 1 then write:

\int_x^{oo} = \int_x^1 e^{-y^2/2} dy +
+ \int_1^{oo} e^{-y^2/2} dy.

The first integral on the right-hand side is just an
ordinary definite integral and in the second integral use
the fact that for y >= 1 we have y^2 >= y, which implies
-y^2/2 =< -y/2 and thus e^{-y^2/2} =< e^{-y/2}. Therefore
the integral \int_1^{oo} e^{-y^2/2} dy is easy to bound:

\int_1^{oo} e^{-y^2/2} dy =< \int_1^{oo} e^{-y/2} dy =
= lim_{t -> oo} \int_1^{oo} e^{-y/2} dy =
= lim_{t -> oo} -2e^{-y/2} |_1^{oo} = 2e^{-1/2}.

If x >= 1 we proceed in the similar way. Thus our integral
is convergent. Now observe that:

\int e^{-y^2/2} dy = \int (-1/y)*(-ye^{-y^2/2}) dy =
= \int (-1/y)*(e^{-y^2/2})' dy =
= -1/y e^{-y^2/2} - \int 1/y e^{-y^2/2} dy,

so that:

\int_x^{oo} e^{-y^2/2} dy = - \int_x^{oo} 1/y e^{-y^2/2} dy
+ lim_{y -> oo} 1/y e^{-y^2/2} + 1/x e^{-x^2/2} =
= - \int_x^{oo} 1/y e^{-y^2/2} dy + 1/x e^{-x^2/2}.

Since our integral is convergent, also the integral:

\int_x^{oo} 1/y e^{-y^2/2} dy

is convergent. Moreover, the function 1/y e^{-y^2/2} has
only positive values, so \int_x^{oo} 1/y e^{-y^2/2} dy is
positive and thus - \int_x^{oo} 1/y e^{-y^2/2} dy is
negative. This implies that:

\int_x^{oo} e^{-y^2/2} dy =< 1/x e^{-x^2/2}.

With regards,

Pawel Gladki