Vtyjkyuk

Calculate the integral $ int_5^infty e^-5xdx $ and use this to show that $ int_5^infty e^-x^2 dx $ is negligible.

Answer:

$ int_5^infty e^-5xdx =[frace^-5x-5]_5^infty=frace^-255 $

But how to use this to show that $ int_5^infty e^-x^2 dx $ is negligible as compared to $ int_5^infty e^-5x dx $ ?

Note that for $x>5$ we have $$ e^-x^2 < e^-5x$$
Thus $$ int _5^infty e^-x^2 dx le int_5^infty e^-5x dx = frac e^-255 approx 0$$

Let
$I(a)
=int_a^infty e^-axdx
$
and
$J(a)
=int_a^infty e^-x^2dx
$.

I will show that
$dfrac12(1-dfrac12a^2+1)
lt dfracJ(a)I(a)
lt dfrac12
$.

$I(a)
=int_a^infty e^-axdx
=dfrace^-ax-a|_a^infty
=dfrace^-a^2a
$.

Since
$(e^-x^2)'
=-2xe^-x^2
$,
$-dfrac12x(e^-x^2)'
=e^-x^2
$,
so
$(dfrac12xe^-x^2)'
=dfrac12x(e^-x^2)'-dfrac12x^2e^-x^2
=-e^-x^2-dfrac12x^2e^-x^2
$
so
$int e^-x^2dx
=-dfrac12xe^-x^2-int dfrac12x^2e^-x^2dx
$
so

$beginarray\
J(a)
&=int_a^infty e^-x^2dx\
&=-dfrac12xe^-x^2|_a^infty-int_a^infty dfrac12x^2e^-x^2dx\
&=dfrac12I(a)-int_a^infty dfrac12x^2e^-x^2dx\
endarray
$

Therefore
$J(a) < frac12 I(a)
$
or
$dfracJ(a)I(a)
lt dfrac12$.

Since
$int_a^infty dfrac12x^2e^-x^2dx
lt int_a^infty dfrac12a^2e^-x^2dx
=dfrac12a^2int_a^infty e^-x^2dx
=dfrac12a^2J(a)
$,

$beginarray\
J(a)
&=dfrac12I(a)-int_a^infty dfrac12x^2e^-x^2dx\
&gtdfrac12I(a)-dfrac12a^2J(a)\
endarray
$

so
$frac12 I(a)
lt J(a)(1+frac12a^2)
= J(a)dfrac2a^2+12a^2
$
or
$dfracJ(a)I(a)
gt dfrac12dfrac2a^22a^2+1
= dfrac12(1-dfrac12a^2+1)
$.

By letting $I_1=int_5^+inftye^-x^2,dx$ and $I_2=int_5^+inftye^-5x,dx$ we have

$$ e^25 I_1 = int_0^+infty e^-x^2-10x,dx,qquad e^25I_2=int_0^+inftye^-5x,dx=frac15 $$
and by just exploiting $e^-x^2leq 1$ we get $e^25I_1leqfrac110$, such that $I_1leq frac12I_2$.

On the other hand the ratio $fracI_1I_2$ is not really negligible. We have

$$ int_0^+infty(1-e^-x^2)e^-10 x,dx leq int_0^+inftyx^2 e^-10 x,dx=frac21000$$
hence
$$ boxedfrac12-frac1100leqfracI_1I_2leq frac12.$$

Observe that $0leq e^-x^2<e^-5x$ for most positive $x$ values. You can use FTC to find $int_5^infty e^-5xdx$, but not the other integral.