How do you deduce that

ò₀^p/2 x/sin(x) dx=2G?

for

G=

¥ å n=0

(-1)n/(2n+1)2

.
Brad

I guess you already know this, but anyway:
The integral of x/sin(x) is:
x+x³ /18+7x⁵ /1800+...+2(2^2n-1 -1)b_n x²ⁿ⁺¹ /(2n+1)!+...
Where b_n is the nth Bernoulli number.

Yatir

I'll let Yatir get back to you on that formula, but to solve your original question you can use:

1-t²+t⁴-¼ = 1/(1+t²)

which is an identity for t in (-1,1).

Integrating from 0 to x we obtain:

x-x³/3+x⁵/5-¼ = arctan x

Dividing through by x and integrating from 0 to 1 gives:

1/1¹-1/3²+1/5²-¼ = ò₀¹(arctan x)/x dx

This gives the result because the integral on the right hand side is actually half your integral (to see this substitute u=2 arctan x).