Advanced book on Mathematics Olympiad

3.2 Continuity, Derivatives, and Integrals 161

∫ 1

0

lnxln( 1 −x)dx=

∑∞

n= 1

1

n(n+ 1 )^2

.

Using a telescopic sum and the well-known formula for the sum of the inverses of squares
of positive integers, we compute this as follows:

∑∞

n= 1

1

n(n+ 1 )^2

=

∑∞

n= 1

(

1

n(n+ 1 )

−

1

(n+ 1 )^2

)

=

∑∞

n= 1

(

1

n

−

1

n+ 1

)

−

∑∞

n= 2

1

n^2

= 1 −

(

π^2

6

− 1

)

= 2 −

π^2

6

,

which is the answer to the problem. 

Next, a problem that we found in S. Radulescu, M. R ̆ adulescu, ̆ Theorems and Prob-
lems in Mathematical Analysis(Editura Didactica ̧ ̆si Pedagogica, Bucharest, 1982). ̆

Example.Prove that for|x|<1,

(arcsinx)^2 =

∑∞

k= 1

1

k^2

( 2 k

k

) 22 k−^1 x^2 k.

Solution.The functiong:(− 1 , 1 )→R,g(x)=(arcsinx)^2 satisfies the initial value
problem

( 1 −x^2 )y′′−xy′− 2 = 0 ,y( 0 )=y′( 0 )= 0.

Looking for a solution of the formy(x)=

∑∞

k= 0 akx

k, we obtain the recurrence relation

(k+ 1 )(k+ 2 )ak+ 2 −k^2 ak= 0 ,k≥ 1.

It is not hard to see thata 1 =0; hencea 2 k+ 1 =0 for allk. Also,a 0 =0,a 2 =1, and
inductively we obtain

a 2 k=

1

k^2

( 2 k

k

) 22 k−^1 ,k≥ 1.

The series

∑∞

k= 1

1

k^2

( 2 k

k

) 22 k−^1 x^2 k

is dominated by the geometric series

∑∞

k= 1 x

2 k, so it converges for|x|<1. It therefore

defines a solution to the differential equation. The uniqueness of the solution for the
initial value problem implies that this function must equalg.