00:01
To show that the integral from 0 to 1 of x raised to p minus 1 over 1 plus x raised to q d x equals the summation from n equals 0 to infinity of negative 1 raised to n times 1 over p plus n times q for p and q that are positive numbers, we note that 1 over 1 minus x equals summation from n equals 0 to infinity, of x raised to n.
00:32
So our 1 over 1 plus x raised to q, that's equal to 1 over 1 minus negative x raised to q.
00:44
And that's going to be equal to the summation from n equals 0 to infinity of negative x raised to q raise to n, or that's the same as summation from n equals 0 to infinity of negative 1 race to n, times x raised to n times q.
01:06
And so our x raised to p minus 1 over 1 plus x raised to q, that's the same as saying, we're multiplying x raised to p minus 1, and 1 over 1 plus x raised to q.
01:22
Since you know the summation format of 1 over 1 plus x raised to q, then this is just the same as x raised to p minus.
01:32
1 times the summation from n equals 0 to infinity of negative 1 raised to n times x raised to n q.
01:42
And because x raised to p minus 1 becomes constant in the summation because the index of the summation is n, then you can just put it inside the summation and combine it with our variable.
01:58
So that's the same as saying summation from n equals 0 to infinity of negative 1 raised to n times x raised to p minus 1 plus n q...