Advanced book on Mathematics Olympiad

Real Analysis 539

=−e−xP(x)|t 0 −e−xP′(x)|t 0 −···−e−xP(n)(x)|t 0.

Because limt→∞e−tP(k)(t)=0,k= 0 , 1 ,...,n, when passing to the limit we obtain

tlim→∞

∫t

0

e−xP(x)dx=P( 0 )+P′( 0 )+P′′( 0 )+···,

hence the conclusion.

465.First, note that by L’Hôpital’s theorem,

lim

x→ 0

1 −cosnx

1 −cosx

=n^2 ,

which shows that the absolute value of the integrand is bounded asxapproaches 0, and
hence the integral converges.
Denote the integral byIn. Then

In+ 1 +In− 1

2

=

∫π

0

2 −cos(n+ 1 )x−cos(n− 1 )x

2 ( 1 −cosx)

dx=

∫π

0

1 −cosnxcosx

1 −cosx

dx

=

∫π

0

( 1 −cosnx)+cosnx( 1 −cosx)

1 −cosx

dx=In+

∫π

0

cosnxdx=In.

Therefore,

In=

1

2

(In+ 1 +In− 1 ), n≥ 1.

This shows thatI 0 ,I 1 ,I 2 ,...is an arithmetic sequence. FromI 0 =0 andI 1 =πit
follows thatIn=nπ,n≥1.

466.Integration by parts gives

In=

∫π/ 2

0

sinnxdx=

∫π/ 2

0

sinn−^1 xsinxdx

=−sinn−^1 xcos^2 x

∣∣

∣π/ 02 +(n−^1 )

∫π/ 2

0

sinn−^2 xcos^2 xdx

=(n− 1 )

∫π/ 2

0

sinn−^2 x( 1 −sin^2 x)dx=(n− 1 )In− 2 −(n− 1 )In.

We obtain the recursive formula

In=

n− 1

n

In− 2 ,n≥ 2.

This combined withI 0 =π 2 andI 1 =1 yields