Re: sequence of tails of integrals

In article
<15142758.1212373163620.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
"jigyasu.swabhav@xxxxxxxxx" <jigyasu.swabhav@xxxxxxxxx> wrote:

I was studying the tail of a particular integral. Some numerical computations
make me believe that:
1
/
| (2 n - 2)
| n x (-a)
lim | ------------- dx = 1 - e
n->infinity | / 2 \n
| | x |
/ |1/2 + ----|
1 - a/n \ 2 /

for 'a' lying in interval with 1 as its left end point [1,2).

For example from maple I have

198
100 x
y := ---------------
/ 2 \100
| x |
|1/2 + ----|
\ 2 /

evalf(int(y,x=1-1/100..1),100);
0.63581975833643515681825593227797190668365331591453532950896967115716
evalf(int(y,x=1-2/100..1),100);
0.8700922556243790690280338534952211372840581332245395450071243926243


I feel that the obvious limit of the quantity below
(2 n - 2)
(1 - a/n)
--------------------
/ 2\n
| (1 - a/n) |
|1/2 + ----------|
\ 2 /

as n tends to infinity is exp(-a) needs to exploited, but whichever way I do
it, I end up with much cruder estimates.

Any ideas?

First replace x by (1-x) to get

int_[0, a/n] n*[(1 - x)^(n-1)]^2/(1 - x + x^2/2)^n dx,

then let x = ay/n to arrive at

int_[0, 1] a*[(1 - (ay)/n)^(n-1)]^2/(1 - ay/n + (ay/n)^2/2)^n dy.

As n -> oo, the integrand -> ae^(-2ay)/e^(-ay) = ae^(-ay) pointwise,
certainly, and I'll wager uniformly. Given the latter, the result
follows.
.