Taylor expansion

1. Application

In this note, I am going to give you one application of the theory of power series (or Taylor expansion). Compute the following integral

I = Z 1

0

lnxln(1−x)dx.

The function ln(1−x) has Taylor expansion on|x|<1:

ln(1−x) =

∞

X

n=1

xⁿ n . The proof replies on the fact that for|x|<1,

d

dxln(1−x) =− 1

1−x =−

∞

X

n=1

xⁿ⁻¹.

Plugging ln(1−x) into the integral, we obtain Z 1

0

lnxln(1−x)dx=−

∞

X

n=1

1 n

Z 1 0

xⁿlnxdx.

Using integration by parts,

− Z 1

0

xⁿlnxdx=− xⁿ⁺¹ n+ 1lnx

1 x=0

+ 1

n+ 1 Z 1

0

xⁿdx= 1 (n+ 1)². Hence the integral becomes

I =

∞

X

n=1

1 n(n+ 1)². Notice that for each n≥1, 1

n(n+ 1)² = 1

n(n+ 1)− 1

(n+ 1)².This implies that I =

∞

X

n=1

1 n(n+ 1)

∞

X

n=1

1 n(n+ 1)−

∞

X

n=1

1 (n+ 1)²

= 1−

∞

X

n=1

1 n² −1

!

= 2−π² 6 Here we use

∞

X

n=1

1

n(n+ 1) = 1.

1