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Integration qn
- Thread starter lpodtouch
- Start date
dan964
what
Just prove the odd case is true (just construct a series using part ii)
then you can easily prove the even case is true
then you can easily prove the even case is true
Typing out the expressions will be too tedious for me.
Using the recurrence relation, iteratively:
I_n = 2ln2 - \frac {1}{n+1} - nI_{n-1} \\ \\ = (2ln2 - \frac {1}{n+1} - [2ln2 - \frac {1}{n} - (n-1)I_{n-2}]) \\ \\ = \frac {1}{n} - \frac {1}{n+1} + \left(2ln2 - \frac {1}{n-1} - [2ln2 - \frac{1}{n-2} - (n-3)I_{n-4}]\right)\\ \\ = \frac {1}{n-2} - \frac {1}{n-1} + \frac {1}{n} - \frac {1}{n+1} + (n-3)I_{n-4}\\ \\ = \frac {1}{n-2} - \frac {1}{n-1} +\frac {1}{n} - \frac {1}{n+1} + \left( 2ln2 - \frac {1}{n-3} - [2ln2 - \frac {1}{n-4} - (n-5)I_{n-6}] \right) )
In this way, if n is odd:
The final 1I0 = 2ln2 - 1 will not be paired up within the big parentheses; its 2ln2 will cancel out the '2ln 2' from the 2I1 term so that we have no '2ln 2' left.
So the case of n-odd is shown.
If n is even:
The final 1I0 will be within the final big parentheses, so that its '2 ln 2' remains (uncancelled) so that we end up with the formula for n-even.
Using the recurrence relation, iteratively:
In this way, if n is odd:
The final 1I0 = 2ln2 - 1 will not be paired up within the big parentheses; its 2ln2 will cancel out the '2ln 2' from the 2I1 term so that we have no '2ln 2' left.
So the case of n-odd is shown.
If n is even:
The final 1I0 will be within the final big parentheses, so that its '2 ln 2' remains (uncancelled) so that we end up with the formula for n-even.
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