550 Real Analysis488.Expand the cosine in a Taylor series,cosax= 1 −(ax)^2
2!+
(ax)^4
4!−
(ax)^6
6!+···.
Let us forget for a moment the coefficient(−^1 )
na 2 n
( 2 n)! and understand how to compute
∫∞−∞e−x
2
x^2 ndx.If we denote this integral byIn, then integration by parts yields the recursive formula
In=^2 n 2 −^1 In− 1. Starting withI 0 =
∫∞
−∞e−x
2
dx=√
π,we obtainIn=( 2 n)!√
π
4 nn!.
It follows that the integral in question is equal to∑∞n= 0(− 1 )n
a^2 n
( 2 n)!·
( 2 n)!√
π
4 nn!=
√
π∑∞
n= 0(
−a
2
4)nn!,
and this is clearly equal to√
πe−a^2 /^4.
One thing remains to be explained: why are we allowed to perform the expansion
and then the summation of the integrals? This is because the series that consists of the
integrals of the absolute values of the terms converges itself. Indeed,∑∞n= 1a^2 n
( 2 n)!∫∞
−∞e−x
2
x^2 n=√
π∑∞
1(a 2
4)nn!=
√
πea(^2) / 4
<∞.
With this the problem is solved.
(G.B. Folland,Real Analysis, Modern Techniques and Their Applications, Wiley,
1999)
489.Consider the Taylor series expansion around 0,
1
x− 4
=−
1
4
−
1
16
x−1
64
x^2 −1
256
x^3 −···.A good guess is to truncate this at the third term and let